physics 12 unit 4 solns

Copyright © 2003 Nelson Unit 4 Are You Ready? 519

Unit 4 The Wave Nature of Light

ARE YOU READY? (Pages 440–441)

Knowledge and Understanding 1.

The rays are reflected so that if normals are drawn in, the angle of incidence equals the angle of reflection for each ray. 2. Rays of light travelling from air into glass at an oblique angle will slow down and bend toward the normal. 3.

4. (a) θi = 60° θR = 21° n = ?

i

R

sin sin

sin 60 sin 21

2.42

n

n

n

θθ

=

° =°

=

The index of refraction of the diamond is 2.42. (b) v1 = 3.00 108 m/s v2 = ?

1

2

12

8

82

3.00 10 m/s2.42

1.24 × 10 m/s

vnvvvn

v

=

=

×=

=

The speed of light in diamond is 1.24 108 m/s.

520 Unit 4 The Wave Nature of Light Copyright © 2003 Nelson

(c) Since the angle of incidence in diamond (60°) is greater than the critical angle, we expect that total reflection would occur.

( )( )

( )( )

2 2

1 1

2 11

2

1

sinsin

sinsin

sin 60 2.421

sin 2.10

nn

nn

θθ

θθ

θ

=

=

°=

=

Since sin θ1 > 1, θ is undefined, there is no refracted ray. If no light is reflected, then total reflection occurs. 5.

6.

7. In longitudinal waves, the particles in the medium vibrate in the same direction as the motion of the waves. In transverse

waves, the particles in the medium vibrate at right angles to the direction of motion of the waves. 8. (a) A crest; B wavelength; C amplitude; D trough (b), (c)

(d) number of cycles = 10 cycles t = 2.0 s T = ? First calculate the frequency:

number of cycles

10 cycles2.0 s

5.0 Hz

ft

f

=

=

=

Copyright © 2003 Nelson Unit 4 Are You Ready? 521

We can now calculate the period:

1

15.0 Hz 0.20 s

Tf

T

=

=

=

The period of the wave is 0.20 s. 9. f = 3.0 Hz v = 5.0 m/s λ = ?

5.0 m/s3.0 Hz

1.7 m

v fvf

λ

λ

λ

=

=

=

=

The distance between adjacent troughs is 1.7 m, which is one wavelength. 10.

Points

• Transmitted pulses are never inverted. • When travelling from a fast medium to a slow medium, the reflected pulse is inverted. • When travelling in a slower medium, the wavelength is smaller. • The reflected pulse has the same wavelength as the incident pulse. • The amplitude of the transmitted and reflected pulses is smaller than the amplitude of the incident pulse.


11. The three conditions necessary for two pulses to interfere destructively are 1. The two waves have the same wavelength (frequency). 2. The waves are in the same phase (i.e., crest on trough, trough on crest). 3. The waves have approximately the same amplitude.

12. (a) In a standing wave, the distance between adjacent nodes is 12

λ . From the diagram, 3 nodes = 6.0 m = 32

λ .

Therefore,

3 6.0 m2

4.0 m

λ

λ

=

=

The wavelength of the wave is 4.0 m. (b) number of cycles = 90 vibrations t = 1.0 min = 60 s v = ? Determine the frequency of the waves:

number of cycles

90 vibrations60 s

1.5 Hz

ft

f

=

=

=

We can now calculate the speed of the waves:

( )( )1.5 Hz 4.0 m6.0 m/s

v f

v

λ===

The speed of the waves is 6.0 m/s.

Inquiry and Communication 13. The observations suggest that as the wavelength decreases and travels through an opening that remains constant in size,

the amount of diffraction decreases.

14. Destructive interference will occur on a nodal line, when the path difference is: 1 3 5, , , etc.2 2 2

λ λ λ

Since λ = 1.2 cm. the path differences are: 0.6 cm, 1.8 cm, 3.0 cm, etc.

Math Skills 15. 1

2nx n Ldλ = −

12

nx d

n Lλ =

−

16. n = 3 λ = 632 nm = 6.32 10–6 m d = 153 µm = 153 10–6 m = 1.53 10–4 m θn = ?

6

3 4

1

3

1sin 2

1 6.32 10 msin 32 1.53 10 m

1.630 105.93

n ndλθ

θ

θ

−

−

−

= − × = − ×

= ×= °

To the correct number of significant digits, the value of θ is 5.93°.

Copyright © 2003 Nelson Chapter 9 Waves and Light 523

Technical Skills and Safety 17. The principle precaution when working with bright sources of light is not to look directly into the beam of light. The light

can affect vision either temporarily or permanently. Although some filters can shield the eyes to a certain degree from welding torches or lasers, there are no filters that are absolutely dependable when looking directly at the Sun.

CHAPTER 9 WAVES AND LIGHT

Reflect on Your Learning (Page 442)

1. The photo is the interference of white light when passing through double slits (Young's experiment). Although the student may guess that it is some form of dispersion, it is interference.

2. The wavelengths (frequencies) in the source of light determine the colours seen. 3. Nodal lines would appear as dark lines or areas where destructive interference has occurred. (Dark is the absence of light.) 4. The dark vertical lines represent complete destructive interference.

Try This Activity: Diffraction of Light (Page 443)

2. Holding the double slits vertically in front of one eye, the student should see an equally spaced vertical pattern of alternating red and black lines or areas.

3. With a green filter covering the lamp, the same pattern should be seen as in step 2, except the bright lines are green. Also, the green lines and dark lines are closer together.

4. (a) A smaller slit width produced a wider pattern (b) Green light produced a wider pattern than the red light. (c) We know that diffraction has occurred because the light spreads out after passing through the slit. We know that

interference of light has occurred because there are bright lines and black lines, which represent areas of constructive and destructive interference.

9.1 WAVES IN TWO DIMENSIONS

PRACTICE (Pages 445, 450)

Understanding Concepts 1. Using the subscript 1 for shallow water and the subscript 2 for deep water, λ2 = 2.0 cm v1 = 10.0 cm/s v2 = 18.0 cm/s λ1 = ?

1 1

2 2

1 22

2

2

(10.0 cm/s)(2.0 cm)18.0 cm/s

1.1 cm

vv

vv

λλ

λλ

λ

=

=

=

=

The wavelength of the wave in shallow water is 1.1 cm. 2. Using the subscript 1 for shallow water and the subscript 2 for deep water, v1 = 0.75v2 λ1 = 2.7 cm λ2 = ?


2 22

1

2

2

2

(2.7 cm)0.75

3.6 cm

vv

vv

λλ

λ

=

=

=

The wavelength of the wave in deep water is 3.6 cm. 3. Using subscript 1 for deep water and the subscript 2 for shallow water, v1 = 18.0 cm/s λ1 = 2.0 cm v2 = 10.0 cm/s λ2 = 1.1 cm f = ? For deep water:

1 1

1

1

18.0 cm/s2.0 cm

9.0 Hz

v fvf

f

λ

λ

=

=

=

=

For shallow water:

2

2

10.0 cm/s1.1 cm

9.0 Hz

vf

f

λ=

=

=

The frequency of the wave in both deep water and shallow water is 9.0 Hz. 4. θ1 = 60° θ2 = 45° n = ?

1

2

sinsinsin 60sin 451.225, or 1.2

n

n

θθ

=

°=°

=

(a), (b) To calculate the ratios in the two media of the wavelengths, and the velocities, we can use the following ratio:

1 1 1

2 2 2

sinsin

vv

θ λθ λ

= =

Since 1

2

sinsin

θθ

= 1.2, the ratio for both the wavelengths and the velocities is 1.2.

(c) Since the frequency remains the same in the two media, the ratio is 1.0. 5. Using the subscript 1 for shallow water and the subscript 2 for deeper water, v1 = 28 cm/s θ1 = 40° θ2 = 46° v2 = ?


1 1

2 2

2

2

sinsin

sin 40 28 cm/ssin 46

31 cm/s

vv

vv

θθ

=

° =°

=

The speed in the deeper water is 31 cm/s. 6. Using the subscript 1 for deep water and the subscript 2 for shallow water, f = 10.0 Hz v1 = 38.0 cm/s v2 = 28.0 cm/s θ1 = 30° (a) n = ?

2 1

1 2

38.0 cm/s28.0 cm/s1.36

n vnn v

n

= =

=

=

The index of refraction is 1.36. (b) λ1 = ? λ2 = ? For the deep water:

1 1 1

11

1

1

38.0 cm/s10.0 Hz

3.8 cm

v fvf

λ

λ

λ

=

=

=

=

For the shallow water:

( )( )

1 1

2 2

2 12

1

2

28.0 cm/s 3.8 cm38.0 cm/s

2.8 cm

vv

vv

λλ

λλ

λ

=

=

=

=

The wavelengths in the deeper water and the shallow water are 3.8 cm and 2.8 cm, respectively. (c) θ2 = ?

1

2

12

2

sinsin

sinsin

sin 301.36

21.6

n

n

θθ

θθ

θ

=

=

°=

= °

The angle of refraction in shallow water is 21.6°. 7. (a) Using the subscript 1 for region A and the subscript 2 for region B, θ1 = 30° θ2 = 20° n = ?


1

2

sinsin

sin 30sin 20

1.46

n

n

n

θθ

=

° =°

=

The refractive index is 1.46. (b) λ1 = 2.0 cm (by measurement) λ2 = 1.35 cm (by measurement); 1.36 cm (by calculation) f = 6.0 Hz v1 = ? v2 = ?

For region A:

( )( )1 1 1

1

6.0 Hz 2.0 cm12 cm/s

v f

v

λ===

For region B:

( )( )2 2 2

2

6.0 Hz 1.36 cm8.2 cm/s

v f

v

λ===

The speed in region A is 12 cm/s. The speed in region B is 8.2 cm/s. 8. Using the subscript 1 for crown glass and the subscript 2 for air, sin θ2 = 60° n1 = 1.52 n2 = 1.00 θ1 = ?

1 1 2 2

1

1

sin sin1.52 sin 1.00 sin 60

34.7

n nθ θθθ

== °= °

The angle of incidence in crown glass is 34.7°. 9. Using the subscript 1 for air and the subscript 2 for diamond, n1 = 1.33 n2 = 2.42 θ1 = 60° θ2 = ?

1 1 2 2

1 12

2

2

2

sin sinsin

sin

1.33 sin 60 2.42 sin28.0

n nn

n

θ θθθ

θθ

=

=

° == °

The angle of refraction in diamond will be 28.0°.

Section 9.1 Questions (Page 452)

Understanding Concepts 1. Using the subscript 1 for deep water and the subscript 2 for shallow water, v1 = 24 cm/s f1 = 4.0 v2 = 15 cm/s θ1 = 40° θ2 = ?


(a) 1 1

2 2

sinsin

vv

θθ

=

( )( )

1 22

1

2

sinsin

sin 40 15 cm/s24 cm/s

24

vvθθ

θ

=

°=

= °

The refracted wave front makes a 24° angle with the boundary. (b) λ2 = ? Calculate the wavelength in the deep water:

1 1

11

1

24 cm/s4.0 Hz

6.0 cm

v fvf

λ

λ

λ

=

=

=

=

We can now calculate the wavelength in the shallow water:

( )( )

1 1

2 2

1 22

1

2

6.0 cm 15 cm/s24 cm/s

3.75 cm, or 3.8 cm

vv

vv

λλλλ

λ

=

=

=

=

The wavelength in the shallow water is 3.8 cm. 2. Using the subscript 1 for deep water and the subscript 2 for shallow water, number of cycles = 10 waves t = 5.0 s 2λ1 = 24.0 cm

λ1 = 24.0 cm2

= 12.0 cm

2λ2 = 18.0 cm

λ2 = 18.0 cm2

= 9.0 cm

(a) v1 = ? v2 = ? In order to determine the speed of each wave, we must first calculate the frequency.

number of cycles

10 waves5.0 s

2.0 Hz

ft

f

=

=

=

For the speed of the wave in deep water:

( )( )1 1

1

2.0 Hz 12.0 cm24 cm/s

v f

v

λ===


For the speed of the wave in shallow water:

( )( )2 2

2

2.0 Hz 9.0 cm18 cm/s

v f

v

λ===

The speeds of the wave in deep water and in shallow water are 24 cm/s and 18 cm/s, respectively. (b) n = ?

1

2

24 cm/s18 cm/s1.33, or 1.3

vnv

n

=

=

=

The index of refraction is 1.3. 3. Using the subscript 1 for deep water and the subscript 2 for shallow water, f = 5.0 Hz v1 = 30 cm/s v2 = 27 cm/s θ1 = 50° θ2 = ?

(a) 1 1

2 2

sinsin

vv

θθ

=

2

2

sin 50 30 cm/ssin 27 cm/s

46.9 , or 44θθ

° =

= ° °

The angle of refraction in the shallow water is 44°. (b) n = ?

1

2

30 cm/s27 cm/s1.1

vnv

n

=

=

=

The index of refraction is 1.1. (c) λ = ?

27 cm/s5.0 Hz

5.4 cm

v fvf

λ

λ

λ

=

=

=

=

The wavelength in shallow water is 5.4 cm. 4. Using the subscript 1 for deep water and the subscript 2 for shallow water: λ1 = 2.0 cm f = 11 Hz θ1 = 30° θ2 = 60° v1 = ? v2 = ? For the wave in deep water:

( )( )1 1

1

2.0 cm 11 Hz22 cm/s

v f

v

λ===


For the wave in shallow water, we must determine the angle of incidence and the angle of refraction. The angle between the boundary and the wave front is 60°. Therefore, the angle of incidence (θ1) is 60°. Similarly, the angle of refraction (θ2) is 30°.

1 1

2 2

2

2

sinsin

sin 60 22 cm/ssin 30

12.7 cm/s, or 13 cm/s

vv

vv

θθ

=

° =°

=

The speeds of the wave in deep water and in shallow water are 22 cm/s and 13 cm/s, respectively. 5. Using the subscript 1 for cold air and the subscript 2 for warm air, v1 = 320 m/s v2 = 354 m/s θ1 = 30° θ2 = ?

1 1

2 2

2

2

sinsin

sin 30 320 m/ssin 354 m/s

33.58 , or 34

vv

θθ

θθ

=

° =

= ° °

The angle of refraction in warm air is 34°. 6. v1 = 7.75 km/s v2 = 7.72 km/s θ1 = 20° θ2 = ?

1 1

2 2

2

2

sinsin

sin 20 7.75 km/ssin 7.72 km/s

19.9

vv

θθ

θθ

=

° =

= °

The angle of refraction is 19.9°. 7. The two conditions for wave rays in water and light rays to exhibit total internal reflection are: 1. The energy (wave or light) must be travelling from a denser medium to a less dense medium. 2. The angle of incidence must be greater than the critical angle. 8. n1 = 1.30 n2 = 1.00 θ2 = 45° θ1 = ?

1 1

2 2

1

1

sin sin sin 1.30 sin 45 1.00

66.8 , or 67

nn

θθθ

θ

=

=°

= ° °

The angle of incidence is 67°. 9. n1 = 1.33 n2 = 1.63 θ1 = 30° θ2 = ?


1 1

2 2

2

2

sin sin

sin 30 1.63 sin 1.33

24

nn

θθ

θθ

=

° =

= °

The angle of refraction is 24°. 9.2 DIFFRACTION OF WATER WAVES


Understanding Concepts 1. The conditions required to maximize the diffraction of waves through a slit are: 1. long waves relative to the slit width, and/or 2. a small slit relative to the wavelength. 2. λ = 2.0 m w = 4.0 m

The condition for significant diffraction is wλ ≥ 1.

The ratio wλ = 2.0 m

4.0 m= 0.5, therefore, diffraction will not be noticeable.

3. λ = 6.3 10–4 m w = ?

The condition for significant diffraction is wλ ≥ 1.

4

4

4

6.3 10 m 1

6.3 10 m

6.3 10 m

ww

w

−

−

−

× ≥

× ≥

≤ ×

The maximum slit width that will produce noticeable diffraction is 6.3 10–4 m. 4. If there are larger slit widths (w > 6.3 10–4 m), there will only be slight diffraction at the edges of the slit, and the

majority of the wave will pass straight through without being diffracted. 9.3 INTERFERENCE OF WAVES IN TWO DIMENSIONS

PRACTICE (Page 459)

Understanding Concepts 1. λ = 1.98 m

1 2

1 2

1P S P S2112

1P S P S2

n n

n n

n λ

λ

λ

− = − = −

− =

The smallest possible path difference is 12λ.


Since = 1.98,λ

1 0.99 m2

λ =

Therefore, the smallest corresponding path length difference is 0.99 m. 2. PS1 = 35.0 cm PS2 = 42.0 cm n = 3 f = 10.5 Hz λ = ? v = ? For the wavelength:

1 21P S P S2

35.0 cm - 42.0 cm132

2.80 cm

n n n λ

λ

λ

− = −

= −

=

For the speed of the waves:

( )( )10.5 Hz 2.8 cm29.4 cm/s

v f

v

λ===

The wavelength of the sources is 2.80 cm. The speed of the waves is 29.4 cm/s. 3. PS1 = 29.5 cm PS2 = 25.0 cm n = 2 v = 7.5 cm/s For the wavelength:

n 1 n 2

n 1 n 2

1P S P S2

P S P S12

29.5 cm 25.0 cm122

3.0 cm

n

n

λ

λ

λ

− = − −

= −

−= −

=

For the frequency:

7.5 cm= 3.0 cm

2.5 Hz

v fvf

f

λ

λ

=

=

=

The wavelength of the sources is 3.0 cm. The frequency of the sources is 2.5 Hz.



Understanding Concepts 1. For a two-point interference pattern to remain stable the two sources must: - be in-phase, - have the same frequency (wavelength), - have a fixed separation.

2. For two waves from identical sources to interfere destructively their path lengths must differ by 12λ.

3. Given sin θn = 12

ndλ −

The maximum value of sin θn is 1. Therefore,

1 12

112

nd

d n

λ

λ

− ≤

≤ −

Since n = 1,

1112

2

d

d

λ

λ

≤ −

≤

A ratio of less than 2 for dλ would produce no nodal line.

4. (a) If the distance between the sources is large, the distance between the nodal lines is also large, to the point where no nodal lines are observed.

(b) If the phase of the sources is constantly changing, the interference pattern is also constantly changing and probably cannot be observed.

5. xn = 10.0 cm L = 50.0 cm d = 5.0 cm n = 1 f = 6.0 Hz

To calculate the wavelength:

12

10.0 cm 5.0 cm150.0 cm 12

2.0 cm

nx dL n

λ

λ

= −

−

=


To calculate the speed of the waves:

( )( )6.0 Hz 2.0 cm12 cm/s

v f

v

λ===

The wavelength is 2.0 cm and the speed of the waves is 12 cm/s. 6. Since the sources are in phase, an area of constructive interference is located along the right bisector. This passes through

the centre of the square whether the sources are adjacent or on opposite corners.

7. 3λ = 3.00 m λ = 1.00 m d = 2.00 m L = 5.00 m n = 1

1

1

12

12

(5.00 m)( 1.00 m) 112.00 m 2

1.25 m

nx dL n

Lx nd

x

λ

λ

= −

= − = −

=

A person would stand 1.25 m from the perpendicular bisector of the line between the openings. 8. n = 3 x3 = 35 cm L = 77 cm d = 6.0 cm θ3 = 25° 5 crests = 4.2 cm λ = ? Method 1:

5 crests = 44.2 cm

41.05 cm, or 1.0 cm

λ

λ

λ

=

=


Method 2:

1sin 2

sin=12

(6.0 cm)(sin 25 )132

1.0 cm

n nd

d

n

λθ

θλ

λ

= −

− °=

− =

Method 3:

12

35 cm 6.0 cm177 cm 32

1.0 cm

nx dL n

λ

λ

= −

= −

=

The wavelength of the waves is 1.0 cm when calculated by all three methods. 9. λ = 2.00 m v = 338 m/s (a) f = ?

338 m/s2.00 m

169 Hz

vf

f

λ=

=

=

The frequency of the sound is 169 Hz. (b) Nodal lines occur when the path difference is:

1 1 3 5 or , , , etc.2 2 2 2

n λ λ λ λ −

Since λ = 2.00 m, the path differences could be 1.00 m, 3.00 m, and 5.00 m. Three possible distances from S2 are 7.00 m + 1.00 m = 8.00 m 7.00 m + 3.00 m = 10.00 m 7.00 m + 5.00 m = 12.00 m

(c) 2 11P S P S2n n n λ − = −

112.0 m 5.00 m (2.00 m)2

7.00 m 12.00 m 24

n

n

n

− = −

= +

=

|

N is located on the 4th nodal line.

Applying Inquiry Skills 10. (a) Water waves are approximately transverse; however the water molecules move slightly back and forth as well as up

and down. In fact, an individual particle moves in a small oval path, as we see a cork move in water (see the figure below). This characteristic helps cause ocean waves to “break” if they become too large, or when they approach a beach where the depth becomes smaller and the waves slow down. Because small water ripples are very nearly


transverse, we can use the water in lakes and the water in ripple tanks in our study of waves. The ripples move slowly enough that we are able to study them directly.

(b) As the depth of the water decreases, the speed of the water and of the wavelength also decrease. The volume of the

displaced water in the wave is equal to the area under the wave. If the wavelength decreases, the amplitude increases to maintain the same volume of water (illustrated below). Also, water particles do not move straight up and down as in an ideal wave but move in small circles. As the water becomes shallow over the beach, there is bottom friction. The combination of these two effects causes the waves to “break” near the shore.

(c) Water waves are a good approximation of transverse waves because they can be observed more readily than sound,

light or electromagnetic waves. As long as the depth remains constant, water waves have a uniform speed and can be used to discover the properties of transverse waves. For a visual demonstration, see

http://www.infoline.ru/g23/5495/Physics/English/waves.htm.

Making Connections 11. d = 4.00 102 m f = 1.00 106 Hz v = 3.00 108 m/s (a) First we must calculate the wavelength of the radio signal:

8

63.00 10 m/s1.00 10 Hz3.00 m

vf

λ

λ

=

×=×

=

For interference maxima:

2

2

sin

3.00 10 m4.00 10 m

sin 0.750

n

n

nd

n

λθ

θ

=

×= ×

=

The only possible values for n are 0 and 1. Therefore, θ0 = 0, or θ1 = 49°. For maximum intensity, the directions of the radio signal would be north, 49°E of N and 49°W of N.

http://www.infoline.ru/g23/5495/Physics/English/waves.htm


9.4 LIGHT: WAVE OR PARTICLE?


Understanding Concepts 1.

2. If the speed of light changed when light was reflected, the wavelength would also change since the frequency is fixed by

the source. As a result, the colour of the reflected light might be different from the incident light, which would be difficult to observe. A better test would be to cause the reflected light to interfere with the incident light. If the wavelengths were different, interference would be impossible provided the light originated from the same source. Since we know that interference is possible, we conclude that the speed of light does not change when it is reflected.

3. Huygen’s principle is a method used to construct a succeeding wave based on the previous wave front. The procedure can be used in the same way for all waves, including sound and water waves.

4. The experimental evidence that indicates light could be a wave is: - both light and waves obey the laws of reflection - both light and waves obey Snell’s law - the speed of a wave slows down in a more dense medium, as does light - both waves and light can exhibit partial reflection-partial refraction and total internal reflection - waves exhibit dispersion, as does light 5. n = 1.50 va = 3.00 108 m/s vg = ? Using the particle theory:

( )( )

g

a

g a

8

8g

1.50 3.00 10 m/s

4.50 10 m/s

vn

vv nv

v

=

=

= ×

= ×

According to the particle theory of light, the speed of light in glass is 4.50 108 m/s. Note that this speed is greater than the speed of light in air, which is an impossible situation and will be discussed in Chapter 11.


Applying Inquiry Skills 6. Any reasonable mass moving at the speed of light has enormous momentum. When this mass is stopped by a black

absorbing surface, or reflected by a hand-held mirror, a large force would have to be applied to change this momentum. Since no force has to be applied at all (at least not one we can measure with normal instruments), we can conclude that the particles have an extremely small mass.

9.5 WAVE INTERFERENCE: YOUNG’S DOUBLE-SLIT EXPERIMENT

PRACTICE (Page 473)

Understanding Concepts 1. 6∆x = 6.0 cm

∆x = 6.0 cm6

= 1.0 cm = 1.0 10–2 m

L = 3.0 m d = 220 µm = 220 x 10–6 m = 2.2 10–4 m λ = ?

( )4

2

7

2.2 10 m1.0 10 m3.0 m

7.3 10 m

dxL

λ

λ

−−

−

= ∆ ×= ×

= ×

The wavelength of the light is 7.3 10–7 m. 2. d = 0.042 mm = 4.2 10–5 m n = 5 θ5 = 3.8°

( )5

7

1sin2

sin12

(4.2 10 m) sin 3.8152

6.2 10 m

n

n

nd

d

n

λθ

θλ

λ

−

−

= −

= −

× °=

− = ×

The wavelength of the light is 6.2 10–7 m, or 6.2 102 nm. 3. λ = 6.3 10–7 m d = 43 µm = 4.3 10–5 m L = 2.5 m ∆x = ?

7

5

2

6.328 10 m2.5 m4.3 10 m

3.7 10 m

dxL

x Ld

x

λ

λ

−

−

= ∆ ∆ =

×= ×

∆ = ×

The separation of adjacent nodal lines is 3.7 10–2 m, or 3.7 cm.


4. λr = 6.0 102 nm = 6.0 10–7 m L = 1.5 m 10∆xr = 13.2 cm

∆xr = 13.2 cm10

= 1.32 cm = 1.32 10–2 m

(a) d = ?

r

r7

2

5

(1.5 m)(6.0 10 m)1.32 10 m

6.8 10 m

Ldx

d

λ

−

−

−

=∆

××

= ×

The separation of the slits is 6.8 10–5 m, or 68 µm. (b) λb = 4.5 102 nm ∆xb = ?

λ ∝ ∆x, therefore,

b b

r r

bb r

r2

2

b

4.5 10 nm (1.32 cm)6.0 10 nm0.99 cm, or 1.0 cm

xx

x x

x

λλλλ

∆=

∆

∆ = ∆

×=×

∆ =

The spacing between adjacent nodal lines for blue light is 1.0 cm. 5. λ = 656 nm = 6.56 10–7 m d = 0.050 mm = 5.0 10–5 m L = 2.6 m ∆x = ?

7

4

2

(2.6 m)(6.56 10 m)5.0 10 m

3.4 10 m

dxL

Lxd

x

λ

λ

−

−

−

= ∆

∆ =

×=×

∆ = ×

The fringes are 3.4 10–2 m, or 3.4 cm apart. 6. λ = 6.8 102 nm = 6.8 10–7 m n = 4 x = 48 mm = 4.8 10–2 m L = 1.5 m d = ?

7

2

5

12

12

1 (1.5 m)(6.8 10 m)42 4.8 10 m

7.4 10 m

n

n

x dL n

Ld nx

d

λ

λ

−

−

−

= −

= − × = − ×

= ×

The separation of the two slits is 7.4 10–5 m, or 7.4 10–2 mm.


7. λ = 6.0 10–7 m L = 3.0 m 9∆x = 5.0 cm

∆x = 5.0 cm9

= 0.56 cm = 5.6 10–3 m

d = ?

7

3

4

(6.0 10 m)(3.0 m)5.6 10 m

3.2 10 m

dxL

Ldx

d

λ

λ

−

−

−

= ∆

=∆

×=×

= ×

The separation of the two slits is 3.2 10–4 m. 8. λr = 6.0 10–7 m L = 1.5 m 10∆x = 2.0 cm

∆x = 2.0 cm10

= 0.20 cm = 2.0 10–3 m

(a) λb = 4.5 10–7 m d = ?

b

7

4

3

(1.5 m)(4.5 10 m)4.5 10 m

1.5 10 m

Lx

d

x

λ

−

−

∆ =

×=×

∆ = ×

The spacing between adjacent nodal lines is 1.5 10–3 m. (b) d = ?

r

r

7

3

4

(6.0 10 m)(1.5m)2.0 10 m

4.5 10 m

dxL

Ldx

d

λ

λ

−

−

−

= ∆

=∆

×=×

= ×

The separation of the slits is 4.5 10–4 m.


Understanding Concepts 1. Grimaldi’s work had shown that a beam of light passing through two successive narrow slits resulted in a beam of light

slightly larger than the width of the slits. He hypothesized that the beam bent slightly outward from the edges of the second slit, and named this diffraction. Diffraction was the important property Young used to demonstrate the interference of light. The sunlight fell on a card with two closely spaced pinholes, which allowed light to pass through onto a second card. These acted as point sources in phase. Young could not have made his discovery without the knowledge of diffraction provided by Grimaldi.

2. The observation of the double-slit interference pattern was more convincing evidence for the wave theory of light than the observation of diffraction. It was more convincing because destructive interference—a null result—is predicted and observed.

3. Since λred > λblue , there would be more nodal lines if other factors were kept constant.


4. Apart from the difficulty of making measurements in water, the spacing of the interference fringes would decrease by a factor of n (the index of refraction) of the water. For example, the first dark fringe would occur when the path difference was one-half the wavelength in water.

5. θn = 5.4°

dλ

= ?

1sin2

12

sin 122

sin 5.4

15.9 or 16

n

n

nd

nd

d

λθ

λ θ

λ

= − − =

− =°

=

The ratio of the slit separation to the wavelength of the light is 16. 6. d = 0.15 mm = 1.5 10–4 m L = 2.0 m ∆x = 0.56 cm = 5.6 10–3 m (a) λ = ?

3 4

7

(5.6 10 m)(1.5 10 m)2.0 m

4.2 10 m

dxL

λ

λ

− −

−

= ∆ × ×=

= ×

The wavelength of the source is 4.2 10–7 m, or 4.2 102 nm. (b) λ = 6.0 102 nm = 6.0 10–7 m ∆x = ? λ ∝ ∆x, therefore,

1 1

2 2

11 2

27

37

31

(6.0 10 m) (5.6 10 m)(4.2 10 m)

8.0 10 m

xx

x x

x

λλλλ

−−

−

−

∆=

∆

∆ = ∆

×= ××

∆ = ×

The spacing of the dark bands would be 8.0 10–3 m, or 0.80 cm apart. 7. d = 0.80 mm = 8.0 10–4 m L = 49 cm = 4.9 10–1 m ∆x = 0.33 mm = 3.3 10–4 m (a) λ = ?

4

41

7

8.0 10 m3.3 10 m4.9 10 m

5.4 10 m

dxL

λ

λ

−−

−

−

= ∆ ×= ×

× = ×

The wavelength of the light is 5.4 10–7 m.


(b) d = 0.60 mm = 6.0 10–4 m ∆x = ?

7 1

4

4

(5.4 10 m)(4.9 10 m)6.0 10 m

4.4 10 m

dxL

Lxd

x

λ

λ

− −

−

−

= ∆

∆ =

× ×=×

∆ = ×

The separation of the nodal lines would be 4.4 10–4 m. 8. d = 0.040 mm = 4.0 10–5 m L = 5.00 m ∆x = 5.5 cm = 5.5 10–2 m λ = ?

( )5

2

7

4.0 10 m5.5 10 m5.0 m

4.4 10 m

dxL

λ

λ

−−

−

= ∆ ×= ×

= ×

The wavelength of the light is 4.4 10–7 m. 9. λ = 6.3 102 nm = 6.3 10–7 m d = 3.3 10–5 m

θ1 = ? θ2 = ? θ3 = ?

1

7

5

1

sin (maxima)

(1)(6.3 10 m)=3.3 10 m

1.1

ndλθ

θ

−

−

=

××

= °

Similarly, we find that θ2 = 2.2° and θ3 = 3.3°. The angles, with respect to the slits, that locate the first-, second-, and third-order bright fringes on the screen, are 1.1°, 2.2°, and 3.3°, respectively.

Applying Inquiry Skills 10. Since the sources are now 180˚ out-of-phase, the pattern would shift so that a nodal (dark) line along the right bisector

would occur, whereas previously there was an area of constructive interference (bright).


9.6 COLOUR AND WAVELENGTH

PRACTICE (Page 478)

Understanding Concepts

1. λ = 6.0 10–7 m f = ? v = c = 3.00 108 m/s

8

7

14

3.00 10 m/s6.0 10 m

5.0 10 Hz

c fcf

f

λ

λ

−

=

=

×=×

= ×

The frequency of the orange light is 5.0 1014 Hz. 2. f = 3.80 1014 Hz

λ = ?

8

14

7

3.00 10 m/s3.80 10 Hz7.89 10 m

cf

λ

λ −

=

×=×

= ×

The wavelength is 7.89 10–7 m, or 789 nm. 3. Using the subscript 1 for air and the subscript 2 for alcohol,

λ1 = 4.4 10–7 m n1 = 1.00 n2 = 1.40 λ2 = ?

( )( )

2 1

1 2

1 12

2

7

72

1.00 4.4 10 m

1.403.1 10 m

nn

nn

λλ

λλ

λ

−

−

=

=

×=

= ×

The wavelength of the violet light in alcohol is 3.1 10–7 m. 4. Using the subscript 1 for air and the subscript 2 for turpentine,

n1 = 1.00 n2 = 1.47 λ1 = 6.5 10–7 m θ1 = 40.0°

(a) λ2 = ?

( )( )

2 1

1 2

1 12

2

7

72

1.00 6.5 10 m

1.474.4 10 m

nn

nn

λλλλ

λ

−

−

=

=

×=

= ×

The wavelength of the red light in the turpentine is 4.4 10–7 m.


(b) θ2 = ?

( )( )

1 1 2 2

1 12

2

2

sin sin sin

sin

1.00 sin 401.47

26

n nn

n

θ θ

θ

θ

=

=

°=

= °

The angle of refraction is 26°. 5. d = 0.12 mm = 1.2 10–4 m

L = 0.80 m x = 9.0 mm = 9.0 10–3 m n = 3 λ = ?

3 4

7

(9.0 10 m)(1.2 10 m)(3)(0.80m)

4.5 10 m

x nLd

xdnL

λ

λ

λ

− −

−

=

=

× ×=

= ×

The wavelength of the light used was 4.5 10–7 m. The colour used was blue. 6. d = 0.20 mm = 2.0 10–4 m

L = 2.0 m λ1 = 4.0 102 nm = 4.0 10–7 m λ2 = 6.0 102 nm = 6.0 10–7 m ∆x = ?

To find ∆x, we must first find xv and xr. For maxima (bright) for the violet light:

v

7

4

3v

4.0 10 m(2.0 m)(1)2.0 10 m

4.0 10 m

nx dL n

x Lnd

x

λ

λ

−

−

−

= =

×= × = ×

For the second-order band of the red light:

( )( )

r

7

4

3r

6.0 10 m2.0 m 12.0 10 m

6.0 10 m

x Lnd

x

λ

−

−

−

= ×= ×

= ×

We can now calculate ∆x:

r v

3 3

3

6.0 10 m 4.0 10 m

2.0 10 m

x x x

x

− −

−

∆ = −

= × − ×

∆ = ×

The distance separating the violet from the red light is 2.0 10–3 m.



Understanding Concepts 1. The index of refraction for violet light in glass is slightly larger than that for red light. As a result, the violet light will be

refracted slightly more than the red light, and the focal lengths will be different in each type of lens. 2. White light passing through a flat piece of window glass is not broken down into colours as it is by a prism. This is

because window glass has parallel sides and is thin relative to a prism. As a result, there is not sufficient refraction for dispersion to be visible.

3. Using the subscript 1 for air and the subscript 2 for alcohol, λ1 = 7.50 10–7 m n1 = 1.00 n2 = 1.40

λ2 = ?

2 1

1 2

1 12

27

72

(1.00)(7.5 10 m)1.40

5.4 10 m

nn

nn

λλλλ

λ

−

−

=

=

×=

= ×

The wavelength of the red light in alcohol is 5.4 10–7 m. 4. λr = 4.00 102 nm = 4.00 10–7 m

λv = 7.50 102 nm = 7.50 10–7 m fr = ? fv = ?

To determine the range of frequencies, we first calculate the frequency of red light:

rr

8

7

14r

3.00 10 m/s4.00 10 m7.50 10 Hz

cf

f

λ

−

=

×=×

= ×

Calculate the frequency of the violet light:

v

8

7

14v

3.00 10 m/s7.50 10 m4.00 10 Hz

v

cf

f

λ

−

=

×=×

= ×

The range of frequencies of visible light is from 4.00 1014 Hz to 7.50 1014 Hz. 5. λ = 5.8 102 nm = 5.8 10–7 m

n = 3 d = 0.10 mm = 1.0 10–4 m θn = θ3 = ?

7

4

3

sin

5.8 10 m(3)1.0 10 m

1.0

n ndλθ

θ

−

−

= ×=

× = °

The angle for the third-order maximum is 1.0°.


6. λ = 6.10 102 nm = 6.10 10–7 m n = 1 θn = 3.0° d = ?

7

5

sin

sin

(1)(6.10 10 m)sin 3.0

1.2 10 m

n

n

nd

nd

d

λθ

λθ

−

−

=

=

×=°

= ×

The separation between the two slits is 1.2 10–5 m. 7. In questions 5 and 6, it is difficult to observe and analyze the interference pattern produced by a double slit. The angle

predicting the direction of the fringes is so small, and therefore is difficult to measure. This problem will be remedied in the next chapter when we study the diffraction grating.

Making Connections 8. To produce a rainbow, all of the colours in white light must be present for reflection, refraction, and dispersion to occur

inside suspended raindrops in the atmosphere. Violet and red rays intersect inside the raindrop, and when the rays leave the raindrop, we see violet at the top, red at the bottom, and the other colours of the spectrum in between. This is known as a primary rainbow. Just after sunrise and just before sunset, the sunlight must pass through more of the Earth's atmosphere and the green-blue end of the spectrum is removed, primarily by scattering. Since all the colours are not present, a rainbow cannot be formed.

CHAPTER 9 LAB ACTIVITIES

Investigation 9.1.1: Transmission, Reflection and Refraction of Water Waves in a Ripple Tank (Pages 480–481)

Observations Part 1: Transmission 2–3. When the surface of the water was lightly touched, the wave moved out in an expanding circle as illustrated. The arrows

indicate the direction of the motion of the wave front.


4. Rolling the dowel produced straight waves as illustrated. The arrows indicate the direction of the wave movement.

Part 2: Reflection 7. Reflection straight on

8. Reflection at an angle


Part 3: Refraction 11. Straight waves moving from deep to shallow water with an angle of incidence of 0˚.

14. Straight waves moving from deep to shallow water with an angle of incidence > 0˚.

15. Trial 1:

1

2

sinsinsin 45sin 301.41

n

n

θθ

=

°=°

=

16. Trial 1:

1

2

sinsinsin 40sin 281.38

n

n

θθ

=

°=°

=

Analysis (a) From a point source the wave front is circular, indicating that each part of the wave front has travelled the same distance

in the same time interval. Therefore, we can conclude that the speed of the wave is the same in all directions.


(b) An increase in the frequency results in a decrease in the wavelength. Any change in the frequency does not affect the speed of the wave in the ripple tank.

(c) As the waves pass from deep to shallow water the wavelength decreases and the wave ray is bent towards the normal.

(d) deep

shallow

6.0 cm4.3 cm

λλ

=

deep

shallow1.39

λλ

=

The Snell's law value calculated in step 15 was 1.41. The values are compatible, within experimental error.

(e) To determine the ratio deep

shallow

vv

, we must first calculate the frequency of the generator.

number of cycles

10 cycles5.0 s

2.0 Hz

ft

f

=

=

=

Using this value, we can calculate the speed of the wave in deep water:

( )( )deep

deep

2.0 Hz 6.0 cm12 cm/s

v f

v

λ=

==

We then calculate the speed of the wave in shallow water:

( )( )shallow

shallow

2.0 Hz 4.3 cm8.6 cm/s

v f

v

λ===

Using these values, we can determine the ratio:

deep

shallow

deep

shallow

12 cm/s8.6 cm/s

1.39

vvv

v

=

=

The ratio of deep

shallow

vv

is 1.39. This compares favourably with the ratio of the wavelengths, as one would expect.

(f) The appraisal of the Prediction will vary for each student, depending on the prediction that was made.

Evaluation (g)

• In the investigation we used the shadow wave, not the actual wave. Through appropriate scaling, the actual wavelengths could be used in the measurements and calculations.

• More care could be taken when measuring the frequency of the generator by measuring the number of cycles over a longer time period or by using the strobe to 'stop' the waves and measuring the strobe frequency instead. In either case, a more accurate value for the frequency and the calculated speed could result.

• The speeds of a wave front in deep and shallow water could be measured directly on the screen by recording the time interval for a wave front to travel a fixed distance (e.g., 30 cm). A more accurate value for the speeds would give a

more accurate value for the ratio of the speeds, deep

shallow

vv

.

(h) Both the waves are “stopped” by the stroboscope because the frequency of the wave is the same in both deep and shallow water.


Synthesis (i) (i) Waves passing from shallow to deep water would be refracted away from the normal. (ii) Snell's law would predict their behaviour as follows

1

2

1

2

sin 1sin

11.38

sin0.72

sin

nθθ

θθ

=

=

=

(iii) Partial internal reflection should be observed.

Investigation 9.2.1: Diffraction of Water Waves (Page 482)

Procedure (typical example) Diffraction around an Obstacle 1. Fill the tank with water to a depth of 1 cm. 2. Place a full wax block in the tank approximately 10 cm from the straight wave generator. The block should be standing on

its end, and parallel to the wave generator. Position the light source directly above the block. Using the wave generator, generate a straight periodic wave with a long wavelength. Make a sketch showing a series of wave fronts on both sides of the wax block.

3. Gradually reduce the wavelength of the wave by increasing the frequency.

Diffraction around an Edge 4. Place a wax block, with one side bevelled, at the centre of the wave tank, about 10 cm away from the wave generator.

Line up the other wax blocks so they prevent wave fronts from passing by the other side of the first wax block. 5. Generate a straight periodic wave with a long wavelength, and observe how the wave fronts pass by the bevelled edge of

the block. Record your observations in a sketch showing a series of wave fronts on both sides of the block. 6. Slowly decrease the wavelength of the wave, noting any changes in the amount of diffraction. Draw a sketch to illustrate

your answer.

Diffraction through an Opening 7. Using two bevelled pieces of wax, create a barrier about 10 cm from the wave generator, with an opening of

approximately 4 cm. Generate a wave with a long wavelength, observing the amount of bending that occurs. Increase the frequency gradually. Record your observations in the form of sketches.

8. While generating a wave with a constant frequency, slowly decrease the size of the opening by moving one of the blocks, noting how the size of the opening affects the diffraction. Observe the relationship between the size of the opening and the wavelength to achieve significant diffraction.

Observations Diffraction around an Obstacle 2. Waves were observed moving behind the edges of the obstacle as illustrated. Longer wavelength waves were diffracted

more than shorter wavelength waves.


Diffraction around an Edge 5–6. Waves were observed moving behind the edge of the obstacle as illustrated. Again, waves with longer wavelengths were

diffracted more than waves with shorter wavelengths.

Diffraction through an Opening 7. Waves were observed moving behind the edges of the opening, as illustrated. Increasing the frequency decreased the

wavelength and the amount of diffraction.

8. For a fixed wavelength, if the size of the opening was reduced, the amount of diffraction increased. It appears that to

maximize diffraction, the size of the opening must be smaller than the wavelength.

Analysis (b) There is more diffraction behind the obstacle or opening if the wavelength is longer. (c) Diffraction around an edge increases when the wavelength increases. (d) To keep diffraction minimal, the aperture width should be larger than the wavelength. (e) For maximum diffraction the wavelength should be large and the aperture should be smaller than the wavelength.

(f) When λ is approximately the same as the width w, diffraction is noticeable. In other words, 1wλ = .

If λ > w, noticeable diffraction will be observed. (g) The accuracy of the prediction will depend on the prediction made.

Evaluation (h) The evaluation of the design will depend on the design used in the Procedure.


Investigation 9.3.1: Interference of Water Waves in Two Dimensions (Pages 482–484)

Observations, Data, and Calculations 2. As the two circular waves pass through one another neither is affected. At the point where they interfere for an instant, the

line is darker on the screen. 4. The nodal pattern is symmetrical on either side of the right bisector of the line connecting the two sources.

5. The frequency was determined for a fixed number of cycles of the two-point generator as follows:

number of cycles

20 cycles5.0 s

4.0 Hz

ft

f

=

=

=

The frequency was 4.0 Hz. 6. As the separation of the sources increased, the number of nodal lines increased and the spacing between the lines was

smaller. 7. As the phase of the sources was changed, the nodal pattern shifted. At 180˚, when the two sources were out of phase, a

nodal line ran down the right bisector to the line joining the two sources. No matter how the phase changed, the number of nodal lines remained constant, as long as the frequency of the sources was kept constant.

10. Points P1, P2, and P3 are on the first nodal line (n = 1) to the left side of the right bisector. Method 1:

12

12

x nL d

x dL n

λ

λ

= − = −


Method 2:

1sin2

sin12

nd

d

n

λθ

θλ

= −

= −

n = 1, d = 16.8 cm (This is the shadow distance on the screen, actual = 6.0 cm)

Method 1 Method 2

L (cm) x (cm) θ λ (cm) λ (cm) 18.0 4.0 13˚ 7.5 7.6 38.0 8.0 12˚ 7.1 6.0 45.0 23.5 14˚ 7.8 8.1

11. Points P1, P2, and P3 are on the second nodal line (n = 2) to the right side of the right bisector. n = 2, d = 16.8 cm

Method 1 Method 2

L (cm) x (cm) θ λ (cm) λ (cm) 21.0 13.5 40˚ 6.8 7.2 35.0 23.5 42˚ 7.7 7.5 45.0 29.0 40˚ 7.2 7.2

12. For the first prediction:

( )

avg

avg

67.5 7.1 7.8 7.6 6.0 8.1 cm

644.1 cm

67.4 cm

λλ

λ

=

+ + + + +=

=

=

∑

The average predicted wavelength is 7.4 cm. For the second prediction:

( )

avg

avg

66.8 7.7 7.2 7.2 7.5 7.2 cm

643.6 cm

67.3 cm

λλ

λ

=

+ + + + +=

=

=

∑

The average predicted wavelength is 7.3 cm. 13. These values were obtained by measuring the wavelength directly on the screen.

4 45.5 cm30.5 cm

47.6 cm

λ

λ

=

=

=

The wavelength was measured to be 7.6 cm. Both predictions were acceptable, within experimental error.


14. When the waves passed through the two openings, each opening acted as a point source and a two-point interference pattern was created on the other side of the paraffin block.

15.

Analysis (c) When the frequency of the source increases, the wavelength of the waves decreases and more nodal line are created in the

interference pattern. (d) When the sources are separated, the number of nodal lines on each side of the right bisector increases. (e) When the phase changes, the number of nodal lines remains constant and the pattern shifts position.

Evaluation (f) The accuracy of the prediction will depend on the prediction made. (g) The values for the mathematical predictions for the wavelength are close to those measured directly.

Synthesis (h) The measurements on the screen are measurements made on the shadows of the waves on the ripple tank surface. Since all

measurements are scaled up by the same factor, we are justified in using these values to test the mathematical relationships.

(i) If the relative phase of the identical sources is constantly changing phase, the interference pattern will be constantly shifting. Therefore, the pattern will be unstable, difficult to observe, and difficult to use to make measurements.

(j) The pattern created by diffraction of the waves from a single source through two slits is more stable than the pattern produced by two sources. This occurs because the waves from the two slits are always in phase since they originate from a single source. With two separate sources, there will always be some variations in the phase, which will shift the pattern every time the phase changes. We observed this in the experiment.


Investigation 9.5.1: Young's Double-Slit Experiment (Pages 484–485)

Observations and Calculations 1. The pattern on the screen is made up of a series of equally spaced, narrow, vertical red bars with dark areas between. 2. The distance between the first and the eighth bright line was 4.2 cm

8 bright lines = 7 4.2 cm4.2 cm

70.60 cm

x

x

x

∆ =

∆ =

∆ =

3. d = 3.20 10–4 m (on the slit plate)

4 2

7

(3.20 10 m)(0.60 10 m)3.00 m

6.4 10 m

xL d

d xL

λ

λ

λ

− −

−

∆ =

∆=

× ×=

= ×

The first predicted wavelength of the helium-neon laser light is 6.4 10–7 m. 4. First pair of slits: 7∆x = 7.0 cm

∆x = 7.0 cm7

= 1.00 cm

d = 2.00 10–4 m

7 2

7

(2.00 10 m)(1.00 10 m)3.00 m

6.67 10 m

xL d

d xL

λ

λ

λ

− −

−

∆ =

∆=

× ×=

= ×

The second predicted wavelength of the helium-neon laser light is 6.67 10–7 m. Second pair of slits: 7∆x = 13.3 cm

∆x = 13.3 cm7

= 1.90 cm

d = 1.00 10–4 m

4 2

7

(1.00 10 m)( 1.90 10 m)3.00 m

6.33 10 m

xL d

d xL

λ

λ

λ

− −

−

∆ =

∆=

× ×=

= ×

The third predicted wavelength of the helium-neon laser light is 6.33 10–7 m.


5. Accepted value for the wavelength of a helium neon laser is 633 nm, or 6.33 10–7 m The average value of the calculated wavelengths is:

avg

7

7avg

3(6.40 + 6.67 + 6.33) 10 m

36.46 10 m

λλ

λ

−

−

=

×=

= ×

∑

Therefore, the average wavelength is 6.46 10–7 m.

Analysis (a) The average value of the wavelength that was calculated is correct to two significant digits. Since the accepted value is

6.33 10–7 m, we will use 6.46 10–7 m for the calculation of experimental error.

7 7

7

accepted value experimental value percent difference 100%

accepted value

6.33 10 m 6.46 10 m100%

6.33 10 mpercent difference 2.05%

− −

−= ×

× − ×= ×

×=

This is an acceptable value for experimental error.

Evaluation (b) Factors that contribute to errors in measuring the wavelength of the light would include the given value of d and the

measurements of L and ∆x. (c) To obtain a more accurate value for the wavelength of the laser light we would have concentrated on the measured values

of ∆x, since d and L are fixed. More accurate values could be achieved by measuring the distance between a larger number of bright lines. Also, a magnifier may assist in achieving more accurate measurements.

Synthesis (d) As the value for d decreased, the interference pattern of bright and dark lines became more spread out as indicated by the

values for ∆x. In other words as d decreased, ∆x increased, as indicated by the proportionality statement 1xd

∆ ∝ .

(e) This relationship 1xd

∆ ∝ was exactly the same observation for water wave interference in the ripple tank.

(f) This investigation strongly supports the wave theory of light. The predicted wavelength of light, based on the two-point interference pattern, compares well with the predicted wavelength of light for water wave interference. Also, the mathematical relationships hold for both interferences, and can be used when making a prediction for the wavelength of light.

Investigation 9.6.1: Wavelengths of Visible Light (Pages 485–486)

Prediction (a) Red light has a longer wavelength than green light.

Observations, Data, and Calculations 1. A pattern of nodal and bright spectral lines is observed on either side of the light filament. 2. The red light interference pattern is more spread out than that for green. In other words, ∆xred > ∆xgreen. 3. With the red filter covering the filament and the paper sliders 25 cm apart, 16 nodal lines were observed. 4. Prediction: The green light should have fewer nodal lines than the red light. 5. With the green filter covering the filament and the paper sliders 25 cm apart, 12 nodal lines were observed. 6. The ratio of the number of nodal lines can be found using the following relationship:

red

green

16 1.3312

λλ

= =


7. Red light: Number of nodal lines between the paper sliders was 9. Sliders positioned at 25.0 cm and 45.0 cm. Distance between sliders is 45.0 cm – 25.0 cm = 19.0 cm.

8 19.0 cm19.0 cm

82.38 cm

x

x

x

∆ =

∆ =

∆ =

The average separation ∆x of adjacent nodal lines for red light is 2.38 cm, or 2.38 10–2 m. 8. ∆x = 2.38 10–2 m L = 1.00 m

d = 2.61 10–5 m

red

2 5

7red

(2.38 10 m)(2.61 10 m)1.00 m

6.21 10 m

dxL

λ

λ

− −

−

= ∆

× ×=

= ×

The predicted wavelength of red light is 6.21 10–7 m. 9. Green light: (Step 6) Number of nodal lines between the paper sliders was 7.

Sliders positioned at 25.0 cm and 32.0 cm. Distance between sliders is 36.0 cm – 32.0 cm = 12.0 cm.

6 12.0 cm12.0 cm

62.00 cm

x

x

x

∆ =

∆ =

∆ =

The average separation ∆x of adjacent nodal lines for green light is 2.00 cm, or 2.0 10–2 m.

(Step 7) L = 1.00 m d = 2.61 10–5 m

green

2 5

7green

(2.00 10 m)(2.61 10 m)1.00 m

5.22 10 m

dxL

λ

λ

− −

−

= ∆

× ×=

= ×

The predicted wavelength of green light is 5.22 10–7 m. 10. The comparison of results from other groups will vary. 11. To find the wavelength of each source, we use a green helium-neon laser and the same procedure as followed in

Investigation 9.5.1. d = 1.0 10–4 m L = 3.00 m 7∆x = 11.3 cm

∆x = 11.3 cm7

= 1.61 cm = 1.61 10–2 m

4 2

7

(1.0 10 m)( 1.61 10 m)3.0 m

5.37 10 m

d xL

λ

λ

− −

−

∆=

× ×=

= ×

The wavelength of the green helium-neon laser is 5.37 10–7 m, or 537 nm. This value is comparable to 543 nm, the value provided with the laser specifications.


Analysis (b) The spectral colours are observed when white light is viewed through a double slit. This occurs because the various

colours of the visible spectrum have unique wavelengths and therefore interfere constructively at different positions in the interference pattern.

(c) The interference pattern for green light is more closely spaced than the interference pattern for red light because green light has a shorter wavelength.

(d) The ratio red

green

16 1.3312

λλ

= = indicates that red light has a longer wavelength than green light. Since v = fλ, red has a lower

frequency than green light. (e) This agrees with the prediction that the wavelength for red light is longer than the wavelength for green light.

(f) The ratio from steps 8 and 9 is 7

red7

green

6.21 10 m 1.195.22 10 m

λλ

−

−×= =×

. This compares favourably with the ratio 1.33 from step 6.

Evaluation (g) difference in measurementspercent difference 100%

average measurement= ×

1.33 1.19

1.26percent difference 11.1%

−=

=

The percent difference is 11.1%. (h) The values from group to group will vary because of precision in measuring the number of nodal lines. Values will also

vary because the red and green filters will produce a range of wavelengths for that sector of the visible spectrum.

Synthesis (i) LEDs and lasers emit monochromatic (one wavelength) light, not a range of wavelengths. Since there would not be as

much variation from group to group, the observations and calculations should be more accurate.

CHAPTER 9 SUMMARY (Page 487)

Making a Summary

Property of Light Particle Theory Wave Theory transmission in a straight line

very strong weaker – waves tend to spread out

reflection strong – also supported by mechanics strong – with both straight and circular waves refraction weak – obeys Snell's law, but light must

speed up to bend towards the normal strong – obeys Snell's law, wave rays bend towards the normal as the waves slow down

partial reflection/ refraction

very weak – some particles reflect, some refract at the interface

strong – can be demonstrated with water waves

dispersion very weak – some particles bend more than others

strong – demonstrated with waves of differing wavelengths

diffraction very weak – edge affects direction of particles

strong – can be demonstrated for obstacles and openings with water waves

interference does not explain very strong – interference of waves predict interference of light

CHAPTER 9 SELF QUIZ (Pages 488–489)

True/False 1. F The wave equation, v = fλ, can be applied to all waves. 2. T 3. F Waves with long wavelengths experience more diffraction than waves with shorter wavelengths.


4. F For a given slit, the amount of diffraction depends on the ratio wλ . For observable diffraction, 1

wλ ≥ .

5. F In a two-point, in phase interference pattern, increasing the wavelength of the two sources decreases the number of nodal lines.

6. F Decreasing the separation of the two-point interference pattern sources decreases the number of nodal lines. 7. T 8. F Early attempts to demonstrate the interference of light were unsuccessful because the two sources were too far apart and

out of phase, and the frequency of light is very large. 9. F Dispersion occurs because the refractive index of light is slightly higher for violet light than it is for red light. 10. F Young’s experiment validated the wave theory of light by showing the interferences of light waves.

Multiple Choice 11. (b) 2

1

vnv

=

8

83.00 10 m/s2.13 10 m/s1.41n

×=×

=

12. (d) Option (ii) is not valid because decreasing the depth of the water decreases the wavelength, which therefore decreases the amount of diffraction.

13. (a) (a) increasing the frequency of the source, decreases the wavelength and decreases the diffraction (b) increasing the amplitude of the waves has no affect on diffraction (c) decreasing the width of the slit increases the diffraction (d) decreasing the distance between the wave generator and the slit produces a wider area of diffraction (e) using a longer wavelength increases the amount of diffraction 14. (d) λ = 0.024 m = 2.4 cm n = 2

1path difference212 2.4 cm2

path difference 3.6 cm

n λ = − = −

=

15. (b) d = 4.5 cm n = 5

1sin2n n

dλθ = −

The maximum value for sin θ is 1. Therefore,

( )

( )

1 12

112

1 4.5 cm152

1.0 cm

nd

d

n

λ

λ

λ

λ

− ≤

≤ −

≤ −

≤

16. (e) For a point to be on a nodal line in a two-point interference pattern, where the sources are in phase, the condition is:

sin θ = xL

= 12

ndλ −


In the diagram, the point is on the second nodal line, so the expression reduces to

122

23

xL d

dxL

λ

λ

= −

=

17. (d) 12

x nL d

λ = −

12

dxnLλ

= +

The value of n can be increased by increasing either d or the separation of the sources, or by decreasing λ. In turn, λ can be decreased by increasing the frequency of the waves. A phase change would change the position of the nodal lines but would not increase their number.

18. (c) See the arguments made in 17. 19. (c) According to the model that Newton proposed, the particle theory cannot explain interference. 20. (c) All the statements are true except c. A nodal line only will be found down the centre line if the two point sources of

light are 180˚ out of phase 21. (a) x1 = 0.30 cm

L1 = 1.50 m = 1.5 102 cm L2 = 1.00 m = 1.00 102 cm x2 = ?

( )( )

1 1

2 2

1 22

1

2

2

2

0.30 cm 1.0 10 cm

1.50 10 cm0.20 cm

xL dx Lx L

x LxL

x

λ∆ =

=

=

×=

×=

22. (d) λ = 340 nm = 3.40 10–7 m = 3.40 10–5 cm L = 2.0 m = 2.0 102 cm ∆x = 3.4 cm

5 2

3

(3.4 10 cm)(2.0 10 cm)3.4 cm

2.0 10 cm, or 0.002 cm

xL d

Ldx

d

λ

λ

−

−

∆ =

=∆

× ×=

= ×

23. (e) The pattern has doubled or the value of ∆x is half of its original value. Given that

xL

∆ = dλ

(i) If the frequency of the source was decreased, the wavelength would increase and ∆x would increase. (ii) If the frequency of the source was increased, the wavelength would decrease and ∆x would decrease. (iii) If the width of each slit was increased, it would no affect on ∆x. (iv) If the separation of the slits was increased, ∆x would decrease. (v) If the separation of the slits was decreased, ∆x would increase.


CHAPTER 9 REVIEW (Pages 490–491)

Understanding Concepts 1. Since v = fλ, v ∝ λ, provided the frequency remains constant. Thus, if v decreases, λ also decreases. 2. Similarities between refraction and diffraction

• both bend or change the direction of a ray of light Differences between refraction and diffraction

• there is a change in speed with refraction; there is no change in speed with diffraction • there is a wavelength change with refraction; there is no wavelength change with diffraction • diffraction requires a slit or obstacle; refraction does not require a slit or obstacle

3. ng = n1 = 1.52 nw = n2 = 1.33

1 2

2 1

1

2

1.331.52

0.88

nn

λλ

λλ

=

=

=

To accommodate the same number of wavelengths, the water would have to be thicker than the glass since the wavelengths in water are longer. Thus, the thickness ratio (glass to water) is 0.88.

4. The measurements of the speeds of light in various media were not available to Newton. These measurements would have shown him that light slows down when it bends toward the normal, and does not speed up as he predicted.

5. Experimentally, reflection, refraction, partial-reflection, total internal reflection/refraction, interference, and dispersion can all be demonstrated with waves.

6. The light from the two headlights of a car does not produce an interference pattern because the headlights are too far apart, their light is not in-phase, and the sources are not point sources.

7. Young’s experiment is a pivotal event in the history of science because the experiment produced a null result (total destructive interference) providing the most important validation of the wave theory of light. Once established, the wave concept could be applied to a host of applications and concepts including electromagnetic waves.

8. λ1 = 6.33 10–7 m λ2 = 3.30 10–7 m nz = 1.92 nd = 2.42

1 1

2 2

1

27

7

6.33 x 10 m3.30 x 10 m1.92

vnv

n

n

λλ

λλ

−

−

= =

=

=

=

Therefore, the material is zircon. 9. θ1 = 10.0°

34

n = 1.00

n = ( )4 1.00

3 = 0.75


1

2

12

2

sinsin

sinsin

sin 10.00.75

13.4

n

n

θθ

θθ

θ

=

=

°=

= °

According to a particle theorist, the angle in water would be 13.4°. 10. d = 0.50 mm = 5.0 10–4 m ∆x = 7.7 mm = 7.7 10–3 m

L = 6.50 m λ = ?

( )4

3

7

5.0 10 m7.7 10 m6.50 m

5.9 10 m

dxL

λ

λ

−−

−

= ∆ ×= ×

= ×

The wavelength of the light is 5.9 10–7 m. 11. θn = 2.0°

d = 0.038 mm = 3.8 10–5 m n = 1 λ = ?

( )

1

5

7

sin (maxima)

sin

(3.8 10 m) sin 2.01

6.6 10 m

n nd

dn

λθ

θλ

λ

−

−

=

=

× °=

= ×

The wavelength of the light is 6.6 10–7 m. 12. d = 0.15 mm = 1.5 10–4 m λ = 482 nm = 4.82 10–7 m L = 2.0 m

( )7

1 4

31

(minima)12

12

1 4.82 10 m2.0 m 12 1.5 10 m

3.8 10 m

n

n

x dL n

x L nd

x

x

λ

λ

−

−

−

= −

= − × = − ×

= ×

The second-order dark bands are 3.8 10–3 m, or 3.8 mm away from the central axis on either side. 13. λ = 5.50 102 nm = 5.50 10–7 m

n = 7 L = 2.00 m d = 0.10 mm = 1.0 10–7 m x7 = ?


( )( )7

7 4

27

(maxima)

5.50 10 m2.00 m 71.0 10 m

7.7 10 m

n

n

x dL n

x Lnd

x

x

λ

λ

−

−

−

= =

×=×

= ×

The distance between the two 7th bright lines on either side of the right bisector would be 2 7.7 cm = 15.4 cm.

14. From the relationship sin θn = 12

ndλ −

we can see that the maximum value of sin θn is 1. Since n is the number of the

fringe, we can write dλ ≤ 1, or λ ≤ d. For this to be true, λ cannot be greater than d.

15. λ = 632.8 nm = 6.328 10–7 m L = 2.00 m ∆x = 1.0 cm = 1.0 10–2 m

(a) d = ?

7

2

4 4

(6.328 10 m)(2.00 m)1.0 10 m

1.27 10 m, or 1.3 10 m

LxdLdx

d

λ

λ

−

−

− −

∆ =

=∆

×=×

= × ×

The separation of the slits is 1.3 10–4 m. (b) n = 1 θ = ?

7

4

1sin2

1 6.328 10 m12 1.27 10 m

0.43

n

n

ndλθ

θ

−

−

= − × = − ×

= °

The angle of the first-order dark fringes is 0.43°. 16. θn = 2.0°

d = 3.8 10–5 m λ = ?

5

7

sin

sin

(3.8 10 m)sin2

6.6 10 m

n

n

n

nd

dn

λθ

θλ

θ

λ

−

−

=

=

×=

= ×

The wavelength of the light is 6.6 10–7 m.

17. The maximum is determined by the relationship sin θn = ndλ . The largest value for sin θn is 1. Thus, n

dλ ≤ 1.

Re-arranging the relationship we get: dλ ≤ 61.20 10 mλ −≤ × A wavelength less than 1.20 10–6 m, or 12 nm, is invisible in the deep ultraviolet section of the spectrum.


18. f = 4.75 1014 Hz λ = ?

8

14

7

3.00 10 m/s4.75 10 Hz6.32 10 m

v fvf

λ

λ

λ −

=

=

×=×

= ×

The wavelength is 6.32 10–7 m, or 632 nm. By referring to Table 1 in Section 9.6, we see that this radiation is found in

the red part of the electromagnetic spectrum. 19. λ = 4.60 102 nm = 4.60 10–7 m

n = 2 λmin = ?

7

2

sin (maxima)

4.66 10 msin 2

n nd

d

λθ

θ−

= ×=

2

1sin (minima)2

1sin 2 22

n nd

d

λθ

λθ

= − = −

Since in the same position,

7

7

4.66 10 m 12 22

6.21 10 m

d dλ

λ

−

−

× = − = ×

The wavelength of visible light that would have a minimum at P is 6.21 10–7 m, or 621 nm. 20. d = 0.158 mm = 1.58 10–4 m

λr = 665 nm = 6.65 10–7 m λyg = 565 nm = 5.65 10–7 m L = 2.2 m ∆x = ?

We must first calculate the maxima for red light:

( )( )

r

7

4

2r

6.65 10 m2 2.2 m1.58 10 m

1.85 10 m

x nLd

x

λ

−

−

−

= ×=

× = ×

We can then calculate the maxima for yellow–green light:

yg

7

4

2yg

5.65 x 10 m(2)(2.2 m)1.58 x 10 m

1.57 10 m

x nLd

x

λ

−

−

−

=

=

= ×


Finally, we calculate the distance between the two fringes:

r yg

2 2

2

3

1.85 10 m 1.57 10 m

0.28 10 m

2.8 10 m

x x x

x

− −

−

−

∆ = −

= × − ×

= ×

∆ = ×

The distance between the third-order red fringe and the third-order yellow-green fringe is 2.8 10–3 m, or 28 mm. 21. Using the subscript 1 for air and the subscript 2 for water,

λ1 = 4.00 10–7 m d = 5.00 10–5 m L = 40.0 cm = 0.40 m n2 = 1.33 n1 = 1.00 ∆x2 = ?

We must first calculate the wavelength in water:

( )( )

2 1

1 2

1 12

2

7

72

4.00 10 m 1.00

1.333.00 10 m

nn

nn

λλλλ

λ

−

−

=

=

×=

= ×

Using this value, we can calculate the distance between the fringes:

22

7

5

32

(3.00 10 m)(0.40 m)5.00 10 m)

2.4 10 m

Lxd

x

λ

−

−

−

∆ =

×=×

∆ = ×

The fringes are 2.4 10–3 m, or 2.4 mm apart.

Applying Inquiry Skills 22. L = 1.00 m

d = 6.87 10–4 m λr = ? λb = ? Measurements on the photograph for red light (Figure 1, Section 9.6): 14∆xr = 5.1 cm

∆xr = 5.1 cm14

= 3.64 10–1 cm = 3.64 10–3 m

actual ∆xr = 31 (3.64 10 m)4

−× = 6.94 10–4 m

rr

4 4

7r

(6.87 10 m)(9.10 10 m)1.00 m

6.25 10 m

d xL

λ

λ

− −

−

∆=

× ×=

= ×


Measurements on the photograph for blue light (Figure 1, Section 9.6): 9∆xb = 2.5 cm

∆xb = 2.5 cm9

= 2.78 10–1 cm = 2.78 10–3 m

actual ∆xb = 31 (2.78 10 m)4

−× = 6.94 10–4 m

bb

4 4

7b

(6.87 10 m)(6.94 10 m)1.00 m

4.77 10 m

d xL

λ

λ

− −

−

∆=

× ×=

= ×

The wavelengths of the red and blue light are 6.2 10–7 m and 4.77 10–7 m, respectively. 23. Measured quantities: d = 2.2 cm

PS1 = 4.9 cm PS2 = 4.6 cm Given: n = 3

2 1

2 1

1PS PS2

|PS PS |12

4.9 mm 4.6 mm132

1.2 mm

n

n

λ

λ

λ

− = − −

=−

−=−

=

Since the scale is 1.0 mm = 1.0 cm, the true wavelength is 1.2 cm.

Making Connections 24. In their research the students will find that:

• The storms with the large waves occur off the east coasts of Canada and the United States • They usually occur in the fall. • The largest waves and rogue waves are the result, not of one storm but two or more storms, often coming from

different directions. • Large rogue waves can occur in relative calm water, sometimes a large distance for any storm. • Constructive interference is the probably cause. • Research is centering on past storm information which is being put into a computer model to simulate and predict

storms in the future. 25. d1 = 7.00 km = 7.00 103 m

d2 = 8.12 km = 8.12 103 m f = 536 kHz = 5.36 105 Hz To calculate wavelength:

8

5

2

3.00 10 m/s5.36 10 Hz5.6 10 m

cf

λ

λ

=

×=×

= ×


To calculate the difference:

2 1

3 3

3

8.12 10 m 7.00 10 m

1.12 10 m

d d d

d

∆ = −

= × − ×

∆ = ×

The difference is 1.12 103 m, which is equivalent to 2 (5.6 103 m), or 2λ. Since constructive interference occurs when the path difference is nλ, there is constructive interference at your house.

26. d = 7.00 m f = 85 Hz v = 346 m/s θn = ?

First, we must calculate the wavelength of the bass tone from the speakers:

346 m/s842 Hz4.07 m

vf

λ

λ

=

=

=

We can then calculate the angle:

1

1

1sin21 4.07 msin 12 7.00 m

17

n ndλθ

θ

θ

= − = −

= °

The smallest angle where the audience has trouble hearing the bass tone is 17°. 27. (a) Since the speakers are 180° out-of-phase, there will be a nodal line running down the right bisector. Therefore, you are

sitting at a low intensity point in the interference pattern. (b) d = 4.00 m

L = 5.0 m f = 842 Hz

83.00 10 m/s

842 Hz0.41 m

cf

λ

λ

=

×=

=

Since the sources are 180° out-of-phase, the pattern shifts so that an intensity peak is now where a null area would have been and visa versa. Thus, the first loud intensity peak will be located using the following relationship for n = 1:

1

1

121 0.41m1 (5.0 m)2 4.00m

0.26 m

nx n Ld

x

x

λ = − = −

=

You should move 0.26 m, or 26 cm due east along the opposite wall. 28. f = 1.0 MHz = 1.0 106 Hz

d = 585 m L = 19 km = 1.9 103 m xn = ?


First, we must calculate the wavelength:

8

6

2

c

3.00 10 m/s1.0 10 Hz3.0 10 m

λλ

λ

=

×=×

= ×

Using the relationship (maxima)nx dL n

λ = :

2

3

2

3.0 10 m(1)(1.9 10 m)585 m

9.7 10 m

n

n

x nLd

x

λ = ×= ×

= ×

The receiver should be moved 9.7 102 m, or 9.7 km north. 29. λ = 488 nm = 4.88 10–7 m θ2 – θ1 = 1.0°

1

71

1

sin

sin

1(4.88 10 m)sin

n

n

nd

nd

d

λθ

λθ

θ

−

−−

=

= ×=

and 7

12

2(4.88 10 m)sind

θ−

− ×=

Given θ2 – θ1 = 1.0°

7 71 1

7 7

7

5

2(4.88 10 m) 1(4.88 10 m)sin sin 1.0

2(4.88 10 m) 1(4.88 10 m) sin1.0

4.88 10 msin1.0

2.8 10 m

d d

d d

d

d

− −− −

− −

−

−

× ×− = °

× ×− = °

×=°

= ×

The slit separation is 2.8 10–5 m. 30. λ1 = 4.80 102 nm = 4.80 10–7 m

λ2 = 632 nm = 6.32 10–7 m d = 0.52 mm = 5.2 10–4 m L = 1.6 m n = 2 ∆xn = ?

Using the equation xn = 12

n Ldλ −

, we can calculate the value for x480:

7

480 4

3480

1 4.80 10 m2 (1.6 m)2 5.2 10 m

2.2 10 m

x

x

−

−

−

× = − × = ×


Similarly, we can also calculate the value for x632:

7

632 4

3632

1 6.32 10 m2 (1.6 m)2 5.2 10 m

2.9 10 m

x

x

−

−

−

× = − × = ×

Using these values, we can calculate the distance between the second-order fringes:

632 480

3 3

2

2.9 10 m 2.2 10 m

7.0 10 m

x x x

x

− −

−

∆ = −

= × − ×

∆ = ×

The second-order fringes are 7.0 10–2 m, or 7 cm apart. 31. d = 2.0 m

f = 3.0 109 Hz L = 1.0 102 m t = 0.20 s v = ?

First we must calculate the value of the wavelength of the radio waves:

8

9

1

3.00 10 m/s3.0 10 Hz

1.0 10 m

cf

λ

λ −

=

×=×

= ×

We can now calculate the distance between the point sources:

1

2 1.0 10 m(1.0 10 m)2.0 m

5.0 m

dxL

x Ld

x

λ

λ

−

= ∆ ∆ =

×= ×

∆ =

Finally, we calculate the speed of the car:

5.0 m0.20 s25 m/s

dvt

v

=

=

=

The car is moving at a speed of 25 m/s. 32. The radio will respond to the direct signal and to the reflected signal. If they are in-phase, constructive interference will

occur, and if out-of-phase, destructive interference will occur. When the girl moves a distance of 9.0 m, the intensity goes

from a maximum to a minimum. Since the 9.0 m move produces a path difference of 18 m, 12λ = 18 m and λ = 36 m.

8

6

3.00 10 m/s=36 m

= 8.3 10 Hz

vf

f

λ=

×

×

The frequency of the radio transmitters is 8.3 MHz.

Copyright © 2003 Nelson Chapter 10 Wave Effects of Light 569

CHAPTER 10 WAVE EFFECTS OF LIGHT

Reflect on Your Learning (Page 492) 1. Reflected light is polarized in the horizontal plane on some flat surfaces. Polaroid glasses polarize light in the vertical

plane. Since little light polarized in the vertical plane is received from the surface, the reflected light (glare) is significantly reduced.

2. The pits and bumps on the surface of the CD act as lines and the clear spaces between the adjacent lines reflected light. Together they act as a reflective diffraction grating producing the spectral colours since different wavelengths are diffracted constructively different amounts.

3. A transparent thin coating is placed on the lens to reduce unwanted internal reflection in the lens system. These losses by reflection reduce the transmitted light. Light directed at the thin surface coating constructively reflects light of a specific wavelength, in this case in the blue violet region of the spectrum.

4. As in the colours produced by soap films, some of the surface feathers have thin films that selectively produce reflective constructive interference. The thickness of the film determines how interference occurs constructively in the blue and green segments of the electromagnetic spectrum. Light soap bubbles, the thickness of the films is not uniform causing changes in the colours, depending on the angle of the incident light.

Try This Activity: Thin Film on Water (Page 493) (a) Dark areas in the oil film represent destructive interference. (b) The pattern is caused because the reflected light from the surface of the oil and from the oil-water interface interfere. The

thickness of the oil film determines whether constructive or destructive interference occurs for each colour, since each colour has a different wavelength.

(c) The pattern changes from broad bands of spectral colours to single bright and dark areas because the different filters change the light from white light to monochromatic light. Since a small range of wavelengths are striking the surface of the oil and water, the reflected light can be either diminished (black) or increased (bright) depending on the thickness of the oil film.

10.1 POLARIZATION OF LIGHT

Try This Activity: Polaroid Sheets (Page 494)

Observations 1. No change should be observed when the Polaroid is rotated through 180°. 2. As the Polaroid is rotated, the intensity of light should diminish almost to zero, and then increase again as the rotation

continues. 3. As the Polaroid is rotated, the glare from the disk diminishes. 4. When rotating the Polaroid film and looking at various locations in the sky, some darkening should be observed

depending on the direction. This may bring clouds into more prominence.

Try This Activity: Polaroid Sunglasses (Page 496) • You should see less reflection from the points where the light is reflected from a flat surface. Some types of autoglass may

show patterns of dark lines. In this case, the safety glass is under tension, causing polarization.



Understanding Concepts 1. Double refraction: certain crystals doubly refract the light creating two rays from one. Each ray is polarized in a different

direction. • Polarizing filter: polarizing filters, such as Polaroid, polarize the light in one plane. • Reflection: light striking a flat surface is polarized in a plane parallel to the surface. • Scattering: small particles in the atmosphere scatter the light causing the transmitted light to be polarized.

2. Using two Polaroid filters, the intensity can be changed from maximum to zero, by rotating one of the Polaroids through 90°.

3. Since there is no absorption and polarization from a mirrored surface, the Polaroid will not reduce the intensity of the light. Also, since the Sun is low in the sky, even glare will not be reduced as much, because of the large angle of incidence.

Applying Inquiry Skills 4. (i) Hold two pairs of sunglasses together. Holding them both up to the light so they are overlapping, keep one pair in a

fixed position and rotate the other one 90°. If they are polarized, when they cross the light intensity should be zero (black).

(ii) Wear the sunglasses outside on a sunny day. Look at the glare coming off some flat surface, such as a car hood. If the glasses are Polaroid, the glare will be reduced.

(iii) Look at various points in a blue sky. If the glasses are Polaroid, you should see some darkening. 5. If you rotate the filter, the two “A’s” will disappear and reappear since each “A” is polarized in a different plane.

Making Connections 6. Students may list some of the following points:

• liquid crystals are affected by electric current • nematic liquid crystal, called twisted nematics, (TN), is naturally twisted; applying an electric current to these liquid

crystals untwists them to varying degrees, depending on the current's voltage • LCDs use these liquid crystals because they react predictably to electric current • a special polymer is added to two pieces of polarized glass that creates microscopic grooves in the surface; grooves are

on the side of the glass that does not have the polarizing film on it; a coating of nematic liquid crystals is added in to the grooves

• a second piece of glass with the polarizing film and a grooved coating is lined up at a right angle to the first piece • light that strikes the first filter is polarized and guided to the next layer; the liquid crystal layers change the light's plane

of vibration to match their own angle • light reaches the far side of the liquid crystal substance and vibrates at the same angle as the final layer of molecules; if

the final layer is matched up with the second polarized glass filter, the light will pass through • when liquid crystal molecules straighten out, they change the angle of the light passing through them so that it no

longer matches the angle of the top polarizing filter • no light passes through that area of the LCD, making that area darker than the surrounding areas; controlling the

current in various parts of the film, black and clear areas are created that can be combined to create images (i.e., the numbers on a calculator)

• small and inexpensive LCDs usually reflect light from external light sources; an LCD watch displays numbers where small electrodes charge the liquid crystals and make the layers untwist so that light is not transmitting through the polarized film.

• computer displays have built-in fluorescent tubes above, beside, and sometimes behind the LCD; a white diffusion panel behind the LCD redirects and scatters the light evenly to ensure a uniform display

• colour displays can be passive or active 7. (a) Until 1980, sugar beet farmers were paid by weight. The quality of the product (sugar level) was not a factor. The

polarimeter was called a saccharimeter. This was an important innovation for farmers because higher prices could be obtained for smaller bushels of higher quality sugar beets.

(b) Industries that widely use polarimetry to determine the active purity of raw materials such as vitamins, steroids, and antibiotics. The most common application is for sugar content in such products as chocolate, wine, jellies, flour, and lactose in milk.


10.2 DIFFRACTION OF LIGHT THROUGH A SINGLE SLIT

PRACTICE (Page 504) Understanding Concepts 1. λ = 7.50 × 102 nm = 7.50 × 10–7 m n = 2 w = 2.0 µm = 2.0 × 10–8 m θ = ?

( )( )7

2 8

2

sin (minima)

2 7.50 10 msin

2.0 10 m49

nnwλθ

θ

θ

−

−

=

×=

×= °

The light produces a second minimum at an angle of 49°. 2. θ = 15° λ = 580 nm = 2.80 × 10–7 m w = ?

1

17

6

sin (minima)

sin

5.80 10 msin 15

2.2 10 m

nwnw

w

λθ

λθ

−

−

=

=

×=°

= ×

The width of the slit is 2.2 × 10–6 m, or 2.2 µm. 3. λr > λb

From the equation λ = w yL∆ , we know that ∆y ∝ λ.

Therefore, spacing for red light will be larger than spacing for blue light. 4. λ = 6.328 10–7 m w = 43 µm = 4.3 10–5 m L = 3.0 m ∆y = ?

( )( )7

5

2

3.0 m 6.328 10 m

4.3 10 m4.4 10 m

Lyw

y

λ

−

−

−

∆ =

×=

×∆ = ×

Other than the central maxima, the separation of adjacent minima is 4.4 10–2 m, or 4.4 cm. 5. w = 3.00 × 10–6 m θ = 25.0° λ = ?


( )6

7

sin (minima)

sin

3.00 10 m sin12.5

16.49 10 m

n

n

nww

n

λθ

θλ

λ

−

−

=

=

× °=

= ×

The wavelength is 6.49 × 10–7 m. 6. w = 1.5 × 10–5 m λ = 694.3 nm = 6.943 × 10–7 m

m = 2 θ = ?

( )7

5

12sin (maxima)

12 6.943 10 m21.5 10 m

sin 6.6

m

m

m

w

λθ

θ

−

−

+ =

+ × =×

= °

The angular position of the second maximum is 6.6°. 7. θa = 56° n = 1

θb = 34° a

b

ww

= ?

Since sin (minima),nnwλθ = wa =

asinnλ

θ and wb =

bsinnλ

θ.

Therefore, a

b

ww

= a

b

sin

sin

λθ

λθ

. Cancelling out the λ, we get:

a b

b a

a

b

sinsinsin 34sin 56

0.67

ww

ww

θθ

=

°=°

=

The ratio a

b

ww

of the two slits is 0.67.

Try This Activity: Resolution (Page 505) • As the size of the hole increased, it was easier to see the two filaments separately. When viewing through the smallest hole,

the two filaments appeared as one. • The larger the opening, the better the resolution.



Understanding Concepts 1. From the equation λ = w y

L∆ , we know that ∆y ∝ 1

λ. If wavelength is doubled, the spacing of the pattern is doubled.

We also know that ∆y ∝ w and ∆y ∝ λ. Therefore, if both wavelength and slit width doubled at the same time, the spacing of the pattern would increase by a factor of 4.

2. w = 2.60 × 10–6 m θ1 = 12° λ = ?

( )( )6

7

sin (minima)

sin

sin12 2.60 10 m

15.41 10 m

n

n

nw

wn

λθ

θλ

λ

−

−

=

=

° ×=

= ×

The wavelength is 5.41 10–7 m, or 541 nm. 3. n = 9

θ9 = 6.4° λ = 6.94 × 10–7 m w = ?

( )( )7

5

sin

sin

9 6.94 10 m

sin 6.45.57 10 m

n

n

nwnw

w

λθ

λθ

−

−

=

=

×=

°= ×

The width of the slit is 5.57 × 10–5 m. 4. n = 2

w = 2.25 × 10–6 m θ2 = 25° λ = ?

( )( )6

7

12sin

sin12

sin 25 2.25 10 m112

6.34 10 m

n

n

n

ww

n

λθ

θλ

λ

−

−

+ =

=+

° ×=

+

= ×

The wavelength of the light is 6.34 10–7 m, or 634 nm.


5. w = 6.0 × 10–6 m λ = 482 nm = 4.82 × 10–7 m L = 2.0 m

(a) θ = ? The first nodal (minima) line on both sides of the central line defines the width of the central maximum. Using the

relationship for the angle to the minima:

( )7

1 6

1

sin

1 4.82 10 msin

6.00 10 m4.6

nnwλθ

θ

θ

−

−

=

×=

×= °

Therefore, the angular width is (2)(4.6°) = 9.2°. (b) w = ?

( )

1

1

2

sin

sin2.00 sin 4.6

9.2 10 m

xL

x L

x

θ

θ

−

=

== °

= ×

Therefore, the width is (2)(9.2 cm) = 18.4 cm. 6. nw = 1.33

na = 1.00 λw = 4.82 × 10–7 m

( )

a a

w w

ww a

a

7

7w

1.33 4.82 10 m1.006.41 10 m

nnnn

λλ

λ λ

λ

−

−

=

=

= ×

= ×

Using the wavelength, we can now calculate the angular width:

( )7

1 6

1

sin

1 6.41 10 msin

6.00 10 m6.1

nnwλθ

θ

θ

−

−

=

×=

×= °

The angular width in water is (2)(6.1°) = 12.2°. 7. λ = 632.7 nm = 6.327 × 10–7 m

w = 1.00 × 10–5 m L = 10.0 m ∆y = ?

( )7

5

10.0 m 6.327 10 m

1.00 10 m

Lywλ

−

−

∆ =

×=

×

∆y = 6.33 × 10–1 m Other than the central maximum, the separation of adjacent maxima is 6.33 × 10–1 m, or 63.3 cm.


8. λ = 461.9 nm = 4.619 × 10–7 m 2y1 = 4.0 cm

y1 = 4.0 cm2

= 2.0 cm = 2.0 × 10–2 m

w = ? Since we know the central maximum (y1), we can calculate the value of the angular width:

11

2

1

sin

2.0 10 msin1.50 m

yL

θ

θ−

=

×=

We can plug this expression into the equation for minima:

( )( )

2 7

7

2

5 5

sin (minima)

2.0 10 m 4.619 10 m1.50 m

1.50 m 4.619 10 m

2.0 10 m3.46 10 m, or 3.5 10 m

nnw

w

w

w

λθ

− −

−

−

− −

=

× ×=

×=

×= × ×

The width of the slit is 3.5 × 10–5 m. 9. λ = 589 nm = 5.89 × 10–7 m

w = 7.50 × 10–6 m (a) θ2 = ?

( )7

2 6

2

sin

2 5.89 10 msin

7.50 10 m9.0

nnwλθ

θ

θ

−

−

=

×=

×= °

The second minimum is at an angle of 9.0°. (b) Since sin θ1 ≤ 1, the maximum value of sin θn = 1.

6

7

sin

1

7.50 10 m5.89 10 m12.73

nnwnwwn

n

λθ

λ

λ−

−

=

=

=

×=×

=

Therefore, the highest-order minimum produced is 12. 10. Two factors are important when trying to distinguish between two objects that are far away—the size of the aperture and

the degree of magnification. When looking through a telescope or binoculars, the aperture has vastly increased relative to the unaided eye. This increases the resolution. The magnification of the telescope separates the images making it easier to separate the two stars.

Applying Inquiry Skills 11. The paper clip image will appear on the screen with interference fringes around the image. 12. By squinting, you are decreasing the aperture through which your eye is seeing the image. The resolution is decreased,

which causes the pixels to overlap and makes the image easier to distinguish. By holding the photograph further away from your eyes, this also decreases the resolution, causing the pixels to merge together.


10.3 DIFFRACTION GRATINGS

Try This Activity: Grated Rulers (Page 508) • On the wall should be seen a diffraction interference pattern with equally spaced bright lines of constructive interference. • The equally spaced lines etched on the metal ruler create a ruled grating. The lines themselves do not reflect the laser light;

the shinier spaces between the lines reflect the laser light and create an interference pattern on the screen.

PRACTICE (Page 509)

Understanding Concepts 1. grating = 4000 line/cm

d = 1 cm4000

= 2.5 × 10–6 m

n = 2 θ = 23.0° λ = ?

( )6

7

sin (maxima)

sin

2.5 10 m sin 23.0

24.88 10 m

m

m

md

dm

λθ

θλ

λ

−

−

=

=

× °=

= ×

The wavelength is 4.88 × 10–7 m. 2. d = 1.00 × 10–5 m

λ = 6.00 × 102 nm = 6.00 × 10–7 m θ3 = ?

( )7

3 5

3

sin (maxima)

3 6.00 10 msin

1.00 10 m10.4

mmdλθ

θ

θ

−

−

=

×=

×= °

The angle of the third-order maximum is 10.4°. 3. m = 4

θ = 22° λ = 694.3 nm = 6.943 × 10–7 m d = ?

( )7

6

sin (maxima)

sin

4 6.943 10 m

sin 227.4 10 m

m

m

mdmd

d

λθ

λθ

−

−

=

=

×=

°= ×

The spacing of the lines is 7.4 × 10–6 m, or 7.4 × 10–4 cm.


4. 62

1 cm 1 m 1.61 10 m6200 line 10 cm

d − = = ×

λ = 633 nm = 6.33 × 10–7 m m = ?

The maximum value of sin θm is 1. Therefore:

6

7

sin (maxima)

1

1.61 10 m6.33 10 m2.54

mmd

md

dm

m

λθ

λ

λ−

−

=

=

=

×=×

=

The highest spectral order possible is 2.

Try This Activity: Crossed Gratings (Page 510) • The pattern is similar in that it a symmetrical pattern with points of constructive interference arranged in planes around a

central bright point. Usually the light will have streaks of light radiating out from the source of light. • When the fabric is pulled at the corners the pattern is no longer symmetrical, but distorted. • Using different fibres produces different patterns. Fabrics where the fibres are closer together (e.g., nylon) produce a larger

interference pattern than those where the fibres are further apart. This occurs because there is more diffraction through smaller openings than through larger, provided the wavelength of the light in the same.

• When looking at a bright star through umbrella fabric on a dark night you should see the starlight diffracted by the fabric in the umbrella. The crossed fibres in the umbrella create a crossed grating and the diffracted light produces an interference pattern similar to that seen on page 509. Usually you will see streaks of light emanating out from the star.

Try This Activity: Using a Grating Spectroscope (Page 510) • Calculations for the PSSC apparatus are found in Investigation 12.5.1. • Using more sophisticated spectrometers should produce better results, but analysis will vary. Some hand spectroscope will

give approximate readings within the spectroscopes. The values that can be determined should be: λr = 656 nm λbg= 486 nm λb = 434 nm λv= 410 nm


Understanding Concepts 1. On the surface of CDs are grooves, or very closely spaced lines, creating a reflection diffraction grating. 2. d = 1.15 × 10–3 cm = 1.15 × 10–5 m

λ = 6.50 × 102 nm = 6.50 × 10–7 m θ2 = ?


( )7

2 5

2

sin (maxima)

2 6.50 10 msin

1.15 10 m6.5

mmdλθ

θ

θ

−

−

=

×=

×= °

A second-order maximum will be produced at an angle of 6.5°.

3. d = 1 cm10 000 line

2

1 cm10 m

= 1.00 × 10–6 m

θ1 = 31.2° θ2 = 36.4° θ3 = 47.5° λ1, λ2, λ3 = ?

Using the expression sin (maxima)mmdλθ = :

( )( )6

1

71

sin

1.00 10 m sin 31.2

15.18 10 m

mdm

θλ

λ

λ

−

−

=

× °=

= ×

Similarly,

( )( )6

2

72

1.00 10 m sin 36.4

15.93 10 m

λ

λ

−

−

× °=

= ×

( )( )6

3

73

1.00 10 m sin 47.5

17.37 10 m

λ

λ

−

−

× °=

= ×

The colours of the spectral lines are green (518 nm), yellow-orange (593 nm), and red (737 nm). 4. λα = 656 nm = 6.56 × 10–7 m

λδ = 4.10 × 102 nm = 4.10 × 10–7 m

d = 1 cm6600 line

2

1 cm10 m

= 1.52 × 10–6 m

θα = ? θδ = ?

First we must calculate the values for θα and θδ:

( )( )7

6

sin

1 6.56 10 m

1.52 10 m25.6

ndα

α

λθ

θ

−

−

=

×=

×= °

( )( )7

6

sin

1 4.10 10 m

1.52 10 msin 15.6

ndδ

δ

λθ

θ

−

−

=

×=

×= °

We can now calculate the angular separation:

25.6 15.610.0

α δθ θ θ

θ

∆ = −= ° − °

∆ = °

Therefore, the angular separation is 10.0°. 5. θ = 25.0°

λ = 4.70 × 102 nm = 4.70 × 10–7 m number of lines = ?


First we must determine the distance d between adjacent gratings:

( )( )7

6

sin (maxima)

sin

1 4.70 10 m

sin 25.01.11 10 m

m

m

mdmd

d

λθ

λθ

−

−

=

=

×=

°= ×

We can now calculate the number of lines in the grating:

6 2

3

1 line 1.0 mgrating1.11 10 m 1.0 10 cm

grating 9.00 10 lines/cm

− −= ×× ×

= ×

Therefore, the number of lines in the grating is 9.00 × 103 lines/cm.

Applying Inquiry Skills 6. You should see a spectral halo effect around the source of light. If you look at the Sun on a clear day with a blue sky you

may see a similar effect. In each case, particles create spaces on a diffraction grating. In the case of the glass, the space is created between water molecules on the glass. In the case of the sky it is high-altitude-floating ice crystals. Different wavelengths of light are diffracted in differing amounts causing the separation of colours of the spectrum.

Making Connections 7. λ = 5.00 × 102 nm = 5.00 × 10–7 m

θ1 = 20° θ2 = 18° n = ?

1

2

sinsinsin 20sin181.10

n

n

θθ

=

°=°

=

The index of refraction of the planet’s atmosphere is 1.10. 10.4 INTERFERENCE IN THIN FILMS

Try This Activity: Soap Bubbles (Page 512) This is an observation activity. Try This Activity: Interference in Soap Bubbles (Page 513)

• In reflected red light the soap bubble will show a dark area at the top of the soap film with alternating bright and dark lines, approximately horizontal.

• With white light, the same pattern will appear except for the colours of the spectrum, which will appear in the bright areas. • With transmitted light, a bright area will appear at the top of film, with the order of the interference pattern exactly opposite

to that for the reflected pattern.


PRACTICE (Page 516)

Understanding Concepts 1. Light is transmitted from air to soap – less to more. Destructive interference occurs at 0,

2λ , λ, etc., in the soap.

λa = 645 nm = 6.45 × 10–7 m λs = ?

a

s

as

s

645 nm1.33

484 nm

n

n

λλλ

λ

λ

=

=

=

=

Since destructive interference occurs at 0, 2λ , and λ, the thicknesses are 0, 242 nm, and 485 nm.

2. ng = 1.50 nl = 1.35 λa = 5.80 × 102 nm λg = ?

Destructive interference will occur at t = 2λ after t = 0. In order to calculate the minimum thickness of the glass, we must

first calculate the wavelength of the light in glass:

ag

2

g

5.80 10 nm1.50

387 nm

nλ

λ

λ

=

×=

=

Using this value, we can calculate the minimum thickness:

2387 nm

2193 nm

t

t

λ=

=

=

The minimum thickness of the glass is 193 nm.

3. t = 177.4 nm na = 1.00 nc = 1.55 ng = 1.48 λa = ?


Destructive interference occurs at 0, 2λ , λ, etc. First we must determine the wavelength of the coating:

( )

c

c

c

222 177.4 nm354.8 nm

t

t

λ

λ

λ

=

===

We can now calculate the wavelength in air:

( )

a

c

a c

a

1.55 354.8 nm549.9 nm

n

n

λλ

λ λ

λ

=

===

The wavelength in air that is minimally reflected is 549.9 nm, or 5.50 102 nm. 4. no = 1.29

nw = 1.33 λa = 6.00 × 10-7 m t = ?

Constructive interference occurs at 0, 2λ , λ, etc. First we must calculate the wavelength in oil:

a

o

ao

7

7o

6.00 10 m2

4.65 10 m

n

n

λλλ

λ

λ

−

−

=

=

×=

= ×

We can now calculate the thickness:

7

7

24.65 10 m

22.33 10 m

t

t

λ

−

−

=

×=

= ×

The thickness of the oil slick is 2.33 10–7 m, or 233 nm.


PRACTICE (Page 518)

Understanding Concepts 5. L = 9.8 cm

t = 1.92 10–3 cm 7∆x = 1.23 cm

11.23 cm 1.76 10 cm7

x −∆ = = ×

λ = ?

( )( )3 1

5

22

2 1.92 10 cm 1.76 10 cm

9.8 cm6.9 10 cm

x Lt

t xL

λ

λ

λ

− −

−

∆ = ∆=

× ×=

= ×

The wavelength of the light is 6.9 × 10–5 cm. 6. λ = 6.40 102 nm

L = 7.7 cm ∆x = 0.19 cm t = ?

( )( )

( )

5

3 3

2

27.7 cm 6.40 10 cm

2 0.19 cm

1.29 10 cm, or 1.30 10 cm

x Lt

Ltx

t

λ

λ

−

− −

∆ =

=∆

×=

= × ×

The thickness of the paper is 1.3 × 10–3 cm. 7. L = 4.0 cm

λ = 639 nm = 6.39 10–5 cm 56∆x = 4.0 cm

24.0 cm 7.14 10 cm56

x −∆ = = ×

( )( )

( )5

2

3 3

2

24.0 cm 6.39 10 cm

2 7.14 10 cm

1.78 10 cm, or 1.8 10 cm

x Lt

Ltx

t

λ

λ

−

−

− −

∆ =

=∆

×=

×

= × ×

The thickness of the paper is 1.8 × 10–3 cm.



Understanding Concepts 1. A: light reflected at the air/soap interface changes phase

(less to more) B and C: transmitted light – no phase change D: reflected light – no phase change (more to less) E: reflected light – phase change (less to more) F: transmitted light – no phase change Rays 1 and 2 are back in phase, producing constructive interference (bright).

2. In a thin film, the light from the front and back surfaces is still physically close together and of equivalent intensity

making interference possible. In a thick film both of these factors break down and interference is not apparent. (There are other factors such as the coherence length of a wave train, which we cannot discuss here.)

3.

As the diagram indicates, when t = 0, destructive interference occurs. Constructive interference occurs at:

a

244.50 10 nm

4112 nm

t

t

λ=

×=

=

Destructive interference occurs at:

a

224.50 10 nm

2225 nm

t

t

λ=

×=

=

4. ng = 1.40 nw = 1.33 λ = 5.80 × 102 nm


(a) t = ? First we must calculate the wavelength of the film of gasoline:

ag

2

g

5.80 10 nm1.40

414 nm

nλ

λ

λ

=

×=

=

Bright (constructive) occurs with a path difference of 2λ or t = g

4λ

.

g

4414 nm

4103.5 nm, or 104 nm

t

t

λ=

=

=

The minimum nonzero thickness of the film is 104 nm. (b) nglass = 1.52

λgas = 414 nm (from part (a)) t = ?

Since the light is going from gasoline to glass (less to more), there is a phase change and the minimum thickness would

be 2λ .

gas

2414 nm

2207 nm

t

t

λ=

=

=

If the light were to go from gasoline to glass, the minimum nonzero thickness of the film would be 207 nm. 5. L = 10.0 cm = 1.00 10–2 m

t = 1.5 10–3 mm = 1.5 10–6 m ∆x = 0.20 cm = 2.0 10–3 m

( )( )6 3

2

7

22

2 1.5 10 m 2.0 10 m

1.00 10 m6.0 10 m

x Lt

t xL

λ

λ

λ

− −

−

−

∆ = ∆=

× ×=

×= ×

The wavelength of the light is 6.0 × 10–7 m. 6. L = 12.0 cm

λ = 6.30 10–5 cm 8∆x = 1.0 cm

11.0 cm 1.25 10 cm8

x −∆ = = ×


( )( )

( )5

1

3 3

2

212.0 cm 6.30 10 cm

2 1.25 10 cm

3.02 10 cm, or 3.0 10 cm

x Lt

Ltx

t

λ

λ

−

−

− −

∆ =

=∆

×=

×

= × ×

The thickness of the paper is 3.0 × 10–3 cm.

Making Connections 7. λa = 5.50 × 10–7 m

nSiO = 1.45 nSi = 3.50 t = ?

In order to determine the thickness, we must first calculate the wavelength of the coating of SiO:

a

SiO

aSiO

7

7SiO

5.50 10 m=1.45

3.79 10 m

n

n

λλλ

λ

λ

−

−

=

=

×

= ×

Since rays 1 and 2 are in phase at t = 0, the next minimum is where the path difference is 2λ , or t =

4λ .

7

8 8

43.79 10 m

49.475 10 m, or 9.48 10 m

t

t

λ

−

− −

=

×=

= × ×

The minimum thickness of the film is 9.48 10–8 m, or 94.8 nm. 8. Students should find some of the following points:

• standard window glass allows the sun’s energy to pass through it • at night it emits infrared heat energy back through the glass to the exterior—radiation heat loss • low-E coating is a thin film layer applied directly to glazing surfaces; normally to the exterior face of the interior

glazing • low-E coatings are transparent to short-wave solar energy, and opaque to long-wave infrared energy; it allows most of

the sun’s solar spectrum (including visible light) to pass through the window to the interior • the coating reflects most heat energy (from room temperature objects) back to its source


• many different types of low-E coatings with different performance characteristics: northern low-E coatings maximize solar heat gains and reduce heat loss at night; solar control low-E coatings are good for west-facing windows because they reduce solar heat gain as well as visibility, and are often tinted

10.5 APPLICATIONS OF THIN FILMS


Understanding Concepts 1. For t → 0, rays 1 and 2 are out of phase and the path difference approaches zero. Thus, there is destructive interference

and a dark area at the centre where the glass nearly touches glass.

2. The air layer between the two surfaces is not uniform. The wider the separation of the dark fringes, the closer together the surfaces are. This can be seen in the oval shape of the interference pattern to the left, indicating that the two surfaces are closer together in this region.

3. λ = 521 nm = 5.21 × 10–7 m number of rings = 15 t = ?

Each dark fringe represents destructive interference and the fringes will have a path difference of 2

nλ . The air wedge will

have a dark fringe at the centre, where t = 0. The first dark fringe will occur at 2λ , and the 15th dark circle will occur at

152λ .


( )7

6

152

15 5.21 10 m

23.9 10 m

t

t

λ

−

−

=

×=

= ×

The air wedge is 3.9 × 10–6 m thicker at the 15th dark ring than at the centre. 4. nMgF = 1.38

ng = 1.66 λa = 5.50 × 102 nm t = ? First we must calculate the wavelength of the magnesium fluoride:

aMgF

2

2MgF

5.50 10 nm1.38

3.98 10 nm

nλ

λ

λ

=

×=

= ×

For destructive interference, the minimum thickness of the path difference is λ, or t = 2λ .

2

2

23.98 10 nm

21.99 10 nm, or 199 nm

t

t

λ=

×=

= ×

The minimum thickness of the layer of magnesium fluoride is 199 nm. 5. λa = 5.70 × 102 nm

nc = 1.38 ng = 1.50 t = ?

At this thickness, rays 1 and 3 that emerge in the glass will cancel out one another. This is the purpose of the coating,

which is to reduce unwanted internal reflections and therefore reduce the amount of light that passes through the lens.


First we must calculate the wavelength of the coating:

a

c

ac

2

c

5.70 10 nm1.38

413 nm

n

n

λλλ

λ

λ

=

=

×=

=

For constructive interference at minimum thickness, the path difference is 2λ or t =

4λ .

c

4413 nm

4103 nm

t

t

λ=

=

=

The minimum thickness of the coating is 103 nm. 6. λ = 7.00 × 102 nm

nMgF = 1.38 t = ? First we must calculate the wavelength of the coating:

a

c

ac

2

c

7.00 10 nm1.38

507 nm

n

n

λλλ

λ

λ

=

=

×=

=

The thickness of the nonreflective coating for the first thinnest coating (see question 5) would be 4λ . Therefore, the

second thinnest coating is t = 34λ .

( )

c34

3 507 nm4

380.25 nm, or 380 nm

t

t

λ=

=

=

The second thinnest coating of magnesium fluoride would be 380 nm. 7. nacr = 1.50

λa = 7.80 × 102 nm t = ? We must first calculate the wavelength of the acrylic coating:

a

acr

aacr

2

2acr

7.80 10 nm1.50

5.20 10 nm

n

n

λλλ

λ

λ

=

=

×=

= ×


For destructive interference, the path difference is 2λ , and the thickness is

4λ .

acr

2

2

45.20 10 nm

41.30 10 nm

t

t

λ=

×=

= ×

The heights of the bumps needed to produce destructive interference is 1.3 × 102 nm. 8. Since the tracking starts at the centre of the CD, and not at the outer edge, CD players can play smaller discs. 9. Since the extended DVD records two layers and the top layer is a thin layer of gold, reflected light from the extended

DVD will have a golden shade. Since a regular CD only has the first layer recorded, reflected light will have a silver shade.

Applying Inquiry Skills 10. If the slides stick, air wedges are created causing interference fringes similar to Newton’s rings, but with a random

pattern.

Making Connections

11. A think layer on the surface of the glass with a thickness of 4λ will reduce reflections from the glass, making the picture

or photo easy to see. 12. Students may find some of the following points:

• CD-recordable disks (CD-Rs) don't have any bumps or flat areas; they have a smooth reflective metal layer, which rests on top of a layer of photosensitive dye

• for blank discs, the dye is translucent; light can shine through and reflect off the metal surface • when the dye layer is heated with concentrated light of a particular frequency and intensity, the dye turns opaque; it

darkens to the point that light can't pass through • points along the CD track are selectively darkened, and other areas are left translucent; this creates a digital pattern that

a standard CD player can read • light from the laser beam will only bounce back to the sensor when the dye is left translucent, in the same way that it

will only bounce back from the flat areas of a conventional CD • the CD burner has a “read laser,” same as a conventional CD player, as well as a “write laser” • the write laser is more powerful than the read laser, so it interacts with the disc differently; it alters the surface instead

of just bouncing light off it • read lasers are not intense enough to darken the dye material; playing a CD-R in a CD drive will not destroy any

encoded information • write lasers move outward while the disc spins; the bottom plastic layer has grooves pre-pressed into it, to guide the

laser along the correct path • to record data, the burner turns the laser writer on and off in synch with the pattern of 1s and 0s; the laser darkens the

material to encode a 0 and leaves it translucent to encode a 1 • in a CD-RW (read-write) disc, the reflecting lands and non-reflecting bumps of a conventional CD are represented by

phase shifts in a special compound; when the compound is in a crystalline state, it is translucent and therefore reflective; when the compound is melted into an amorphous state, it becomes opaque, making the area non-reflective; encoding is erased when the laser melts the crystal back into its original state

13. One reason is design – the golden look enhances the bank’s image. Secondly, the film reduces transmitted light, reducing the intensity of the sunlight passing through the glass and keeping the building cooler on a bright sunny day.

14. When applying thin films the thickness can be determined by directing a beam of monochromatic light at the surface. Depending on the thickness, the reflected light will go from dark to bright. By observing the transitions, the thickness of the film can be determined and the depositing of the coating terminated.

15. The acetate sheet can create air wedges between it and the glass surface of the scanner, resulting in patterns of destructive interference.


10.6 HOLOGRAPHY

Try This Activity: Viewing Holograms (Page 527)

There are no answers for this activity. The students are observing holograms on paper money and credit cards.


Understanding Concepts 1. To produce a good holograph, a distinct image, crated by interference, requires monochromatic light, which is provided

by the laser. 2. A normal film negative records the image of the light from the object. A hologram records the light coming from the

illuminated object. 3. The hologram is a record of the object recorded as an interference pattern. Thus, the smudges are points of destructive

interference.

Applying Inquiry Skills 4. Using a hologram (holographic image) of the ghost, you would set it up so the laser and hologram were hidden but the

image would appear, suspended in the air, in the darkened display.

Making Connections 5. Student reports should include some or all of the following points:

• holographic wills are solely prepared by and signed by the person involved with the will • holographic wills are often not witnessed at all, or not witnessed properly • wills are usually prepared without the assistance of a lawyer • states in the U.S. and provinces in Canada have different laws regarding the validity of holographic wills • some places allow holographic wills if the person is overseas, serving in the military, or are in other circumstances

preventing the aid of a lawyer to write the will • holographic documents are hand-written documents or objects • historians are concerned with the authenticity of the document • handwriting must match the person thought to have written the document • ink, paper, and appropriate water marks are analyzed to determine if they match the time period the document was

thought to be written in 6. Students may find some of the following points:

• CDs, DVDs and magnetic storage all store bits of information on the surface of a recording medium. In order to increase storage capabilities, scientists are now working on a new optical storage method, called holographic memory that will go beneath the surface and use the volume of the recording medium for storage, instead of only the surface area.

• Instead of photographic film, the holographic image is stored in the specific area of a crystal. • An advantage of a holographic memory system is that an entire page of data can be retrieved quickly and at one time.

Each page of data is stored in a different area of the crystal, based on the angle at which the reference beam strikes it. During reconstruction, the beam will be diffracted by the crystal to allow the re-creation of the original page that was stored. This reconstructed page is then projected onto the charge-coupled device (CCD) camera, which interprets and forwards the digital information to a computer

• Early holographic data storage devices will have capacities of 125 GB and transfer rates of about 40 MB per second. It is expected that these devices could have storage capacities of 1 TB and data rates of more than 1 GB per second -- fast enough to transfer an entire DVD movie in 30 seconds.

• There are still some technical problems that need to be worked out. For example, if too many pages are stored in one crystal, the strength of each hologram is diminished. If there are too many holograms stored on a crystal, and the reference laser used to retrieve a hologram is not shined at the precise angle, a hologram will pick up a lot of background from the other holograms stored around it. It is also a challenge to align all of these components in a low-cost system.


10.7 MICHELSON’S INTERFEROMETER

PRACTICE (Page 528)

Understanding Concepts 1. The mirrors must move

2λ to create a path difference of λ. Thus, the mirror must have moved 500 nm

2 = 2.50 × 102 nm.

2. For each shift, the mirror has moved 2λ .

path difference 10002638 nm1000

2path difference 0.319 mm

λ = =

=

For 1000 shifts the mirror moves 0.319 mm.

3. 598 fringes = 5982λ

= 299λ

path difference = 0.203 mm = 2.03 × 10–4 m 4

7

299 2.03 10 m

=6.79 10 m

λ

λ

−

−

= ×

×



Understanding Concepts 1. The counting of fringes is exact. The measurements are accurate to the nearest half wavelength of the source of light

utilized. 2. λ = 638 nm

number of fringes = 262 t = ?

( )

5

2622

262 638 nm2

8.36 10 m

t

t

t

λ

−

=

=

= ×

The thickness of the foil is 8.36 × 10–5 m. 3. d = 2.32 × 10–5 m

number of fringes = 89 λ = ?

( )5

7

2 number of fringes2

number of fringes

2 2.32 10 m

895.21 10 m

d

d

λ

λ

λ

−

−

=

=

×=

= ×

The wavelength of the incident beam is 5.21 × 10–7 m.


4. number of fringes = 580 λ = 589 nm d = ?

( )

( )( )

2 number of fringesnumber of fringes

2580 589 nm

2171 nm

d

d

d

λ

λ

=

=

=

=

The mirror M in Michelson’s interferometer must be moved 171 nm. 5. number of fringes = 595

d = 2.14 × 10–4 m λ = ?

( )4

7

2 number of fringes2

number of fringes

2 2.14 10 m

5957.19 10 m

d

d

λ

λ

λ

−

−

=

=

×=

= ×


Applying Inquiry Skills 6. An evacuated cell could be set up in one arm of the interferometer. Air or water could be slowly added to the cell and the

interference fringes counted. The change in the number of wavelengths in the cell length could be used to calculate the refractive index.

Making Connections 7. Students may find some of the following points:

• In 1960, the standard metre length in wavelengths of light was measured since the standard bar of metal, used previously, could undergo minute changes in length over a period of time. The metre was defined as the length of 650 763.73 wavelengths of the orange-red spectral line emitted by krypton-86. The advantages of this system were that it was portable, and not dependent on a length of metal in Paris, and an interferometer, using krypton-86 light, could be used to accurately measure the length of objects in metres.

• In 1983, the metre was again redefined as the distance light travels in 1/299 792 458 s. This was based on the speed of light, where c = 299 792 458 m/s. This method was chosen not to change the length of the metre as previously defined, but to express it with more precision. It possible to measure frequencies with more accuracy than wavelengths because of the precise definition of the second. This definition is 10 000 times more precise than the previous krypton-86 definition of the standard unit of length.

8. By placing the moveable mirror at one point on Earth’s surface and the rest of the apparatus on another, minute movements in Earth, say along a fault line, could be measured by counting fringes. Currently, a laser and mirrors can accomplish the same measurement to the accuracy of the wavelength of the laser and the speed of light.

10.8 ELECTROMAGNETIC WAVES AND LIGHT

PRACTICE (Page 533)

Understanding Concepts 1. f = 107.1 MHz = 1.071 × 108 Hz

v = c = 3.00 × 108 m/s λ = ?


8

83.00 10 m/s1.071 10 Hz2.80 m

cf

λ

λ

=

×=×

=

The wavelength of the signal is 2.80 m. 2. f = 3.00 × 1017 Hz

v = c = 3.00 × 108 m/s λ = ?

8

17

9

3.00 10 m/s3.00 10 Hz1.00 10 m

cf

λ

λ −

=

×=×

= ×

The wavelength of the X rays produced is 1.00 × 10–9 m. 3. λ = 638 nm = 6.38 × 10–7 m v = c = 3.00 × 108 m/s

Since c = fλ and T = 1f

,

7

8

15

6.38 10 m3.00 10 m/s2.13 10 s

Tc

T

λ

−

−

=

×=×

= ×

The period of the light is 2.13 × 10–15 s. 4. λ = 6.0 × 101 Hz

v = c = 3.00 × 108 m/s d = 5.0 × 103 km = 5.0 × 106 m

8

1

6

3.00 10 m/s6.0 10 Hz

5.0 10 m

cf

λ

λ

=

×=×

= ×

To determine the number of wavelengths to span the North American continent, we divide the wavelength by the distance across the continent.

6

6

5.0 10 mnumber of wavelengths5.0 10 m1.0 wavelengths

×=×

=

It would take one wavelength of the radiation to span the North American continent.


Understanding Concepts 1. (a) λ = 1.80 cm = 1.80 × 10–2 m

v = c = 3.00 × 108 m/s f = ?


8

2

10

3.00 10 m/s1.80 10 m

1.67 10 Hz

cf

f

λ

−

=

×=×

= ×

The frequency is 1.67 × 1010 Hz. (b) f = 3.20 × 1010 Hz

v = c = 3.00 × 108 m/s λ = ?

8

10

3

3.00 10 m/s3.20 10 Hz9.38 10 m

cf

λ

λ −

=

×=×

= ×

The wavelength is 9.38 × 10-3 m. (c) f = 60.0 Hz

v = c = 3.00 × 108 m/s λ = ?

8

6

3.00 10 m/s60.0 Hz

5.00 10 m

cf

λ

λ

=

×=

= ×

The distance between adjacent maxima is 5.00 × 106 m. (d) λ = 6.50 × 10-7 m

v = c = 3.00 × 108 m/s f = ?

8

7

14

3.00 10 m/s6.50 10 m4.62 10 Hz

cf

f

λ

−

=

×=×

= ×

The frequency is 4.62 × 1014 Hz. 2. d1 = 6.0 × 103 km

d2 = 3.6 × 104 km ∆t = ?

Distance travelled by microwave signal:

( ) ( )1 2

2 23 4

3

7

6.00 10 km 3.6 10 km

3.6 10 km3.6 10 m

d d d

d

∆ = +

= × + ×

= ×∆ = ×


We can now calculate the time delay.

7

8

3.6 10 m3.00 10 m/s0.12 s

dtc

t

∆∆ =

×=×

∆ =

The fan in Montreal will hear results of any play 0.12 s sooner than the fan in Inuvik. 3. w = 6.0 cm = 6.0 × 10–2 m

f = 7.5 GHz = 7.5 × 109 Hz v = c = 3.00 × 108 m/s θ = ?

First we must calculate the wavelength:

8

9

2

3.00 10 m/s7.5 10 Hz

4.0 10 m

cf

λ

λ −

=

×=×

= ×

To calculate the angle:

( )( )2

2

sin

1 4.0 10 m

6.0 10 m41.8 , or 42

nndλθ

θ

−

−

=

×=

×= ° °

The angle of the first minimum of the diffraction pattern is 42°. 4. The frequency of the arcs was not fixed, so a large range of frequencies were transmitted. This soon became unworkable

as more radio amateurs started transmitting. Their signals interfered with one another making it almost impossible to distinguish one radio message from another.

Applying Inquiry Skills 5. Since the spark plus in tact will transmit a radio signal, these can be detected by a radio. If the discharges are inside the

engine, shielding created by the iron engine will shield most of the radiation. If there is a break in the insulation of a high voltage wire in the ignition system, an arc will jump to any metallic part of car. By moving the portable radio around, the “leak” can be detected. The radio should be tuned to the lower frequencies (low end of the AM band).

Making Connections 6. All electromagnetic radiation travels at the speed of light (3.00 × 108 m/s). The signal from the correspondent must travel

to a satellite and then back to a ground station and through the ground transmission system to the station. The distances are significant, creating a time delay for the correspondent to hear the question, before he or she replies.

7. (a) f = 75 MHz = 75 × 106 Hz v = c = 3.00 × 108 m/s d = 134 m

First we must calculate the wavelength of the signal:

8

6

3.00 10 m/s75 10 Hz

4.0 m

cf

λ

λ

=

×=×

=


Since the path difference is 134 m, that is the extra distance travelled by signals.

interference = 134 m4.0 m/λ

interference = 33.5λ Normally, destructive interference would occur between the incident and reflected waves since the path difference is

33.5λ, but there is a phase change upon reflection, so there is constructive interference. (b) If the airplane is 42 m closer, the path difference is now 134 m – 42 m = 92 m.

interference = 92 m4.00 m/λ

interference = 23 λ With the phase change, the reflected waves will interfere destructively with the incident waves. 10.9 SOME APPLICATIONS OF ELECTROMAGNETIC WAVES

Try This Activity: Scattering Laser Light (Page 537)

When the laser beam travels through the air it is invisible unless scattered by small particles in the air. The chalk dust provides more of these particles and thus the laser beam is more easily seen because the light is scattered by the particles.

Explore an Issue: Cell Phones (Page 538)

As illustrated below, there is a host of material on the Internet reporting the 'latest' studies ad rehashing the old. Nevertheless, this area is under constant change and students can backup the arguments on either sides of the issue, or in-between. The sources and reliability of the reports are serious considerations in making an evaluation. But, this fact makes the topic excellent for debate. Selections from the Literature • cell phone handsets and base antenna towers are thought to pose health risks in human users • fears based on electromagnetic radiation from cell phones and its potential link to cancer • no proven link between cell phone transmissions and human health risks • information about risk is misleading; the exposure of human tissue to certain types of radiation causes abnormal tissue

growth and a cell phone causes some level of radiation • with no link, you can't say that cell phone radiation causes abnormal tissue growth • policy introduced requiring cell phone makers to disclose information on radiation levels produced by their phones • 1990s fears have turned to cell phones • neither human, animal nor cellular studies has "proven any reason to worry" about the effects of mobile phones on cancer

rates (Dr. John Moulder, professor at the Medical College of Wisconsin) • Australian report found that lymphoma-prone mice exposed for 18 months to strong RF fields similar to those generated by

digital cell phones showed a higher than normal incidence of lymphomas • human already exposed to radiation—the Sun, Earth—natural low-level background of EMR • electromagnetic environment altered with technological advances • human are exposed to electromagnetic radiation from a variety of sources • exposure to these unnatural sources of EMR has definitely been shown to increase the risk of problems such as leukemia,

brain cancer and birth defects


Understanding Concepts 1. Larger waves diffract more than shorter waves. AM waves are longer than TV and FM waves.


2. Maxwell stated that electrical and magnetic fields are always perpendicular to one another. Therefore, the direction of oscillation of the magnetic field is in the north-south plane.

3. The strong electromagnetic field created around the powerlines will deflect the electromagnetic filed of the radio signal. If the car radio is in a steel-reinforced structure, the electromagnetic wave from the radio station is shielded or absorbed reducing the signal strength, particularly for lower wavelengths, such as AM radio. A similar problem occurs with cell phones.

Applying Inquiry Skills 4. The modulated laser light can be received and demodulated to hear the signal. This makes an excellent demonstration, if

one has a modulated laser.

Making Connections 5. The higher the number, the larger the amount of protection from UV radiation, preventing sunburn and the resulting cell

damage. 6. There are three types of ultraviolet rays: UV-A, UV-B, and UV-C. UV-A is the weakest form. It causes skin-aging,

wrinkles and can also damage outdoor plastics and paint. UV-B, which is stronger than UV-A, is the most harmful to human and other species. UV-B causes skin cancer and cataracts – a permanent clouding of the eye, which reduces vision. Both UV-B and UV-A cause suntans and sunburns. UV-B also reduces the growth of plants, and may affect the health of wildlife and other animals. UV-C, which is even stronger than UV-B, never reaches Earth’s atmosphere because it is filtered out by the atmosphere.

7. Example of research:

How Ear Thermometers Work • The eardrum is an extremely accurate point to measure body temperature because it is recessed inside the head (just

like your tongue). The problem with the eardrum is that it is so fragile. You don't want to be touching the eardrum with a thermometer.

• The remote sensing of an object's temperature can be done using its infrared radiation. This technique is a very good way to detect the temperature of a person's eardrum.

• One common sensor is the thermopile, which can be accurate to the tenth of a degree. The thermopile sees the eardrum and measures its infrared emissions. The emission is converted into a temperature and displayed on an LCD.

Other topics include: • Infrared blood testing of glucose testing for diabetics • Using heat in physiotherapy to promote healing. • Using infrared thermographs of the human body to detect 'hot' areas that could be cancerous. • Using infrared radiation to kill cancer cells containing photosensitive chemicals.

8. Some favourable: • radio and television communication • mobile communication (e.g., cell phones, global positioning satellites (GPS)) • microwave cooking • radiation for cancer treatment • infrared heating • infrared healing • microwave, infrared, ultraviolet mapping of the Earth by remote sensing from satellites • wireless communication systems (e.g., computer linkage) • remote control of TV, VCR, CD and DVD machines • security detectors Some adverse: • possible cell damage from high voltage power lines • possible damage from microwave and radio transmissions (e.g., cell phones) • loss of privacy through remote sensing of various types • dangers to human tissue by X ray and gamma ray radiation • dependence on radio and television for human interaction.


9. Since the cell phone is a radio transmitter, the emitted radiations could interfere with other electromagnetic devices used in the hospital. For the same reason they are banned on airplanes.

10. Infrared heats up tissue, ultraviolet light can cause cell changes and possibly kill cells (sunburn). This occurs because the ultraviolet has a higher frequency and therefore more energy (E = hf) and thus more penetrating power than infrared radiation.

11. (a) In the case of microwave ovens, the commonly used radio wave frequency is roughly 2,500 MHz or 2.5 GHz. (b) A high-voltage transformer increases the typical household voltage, of about 115 V, to approximately 3000 V. This

powers a magnetron tube that converts this high electrical potential into electromagnetic waves. The microwave energy is transmitted into a metal channel called a wave guide, which feeds the energy into the cooking area.

(c) Microwaves absorb water, fats and sugars. When they are absorbed they are converted directly into atomic and molecular motion—heat. Most plastics, glass or ceramics does not absorb microwaves. Metal reflects microwaves, which is why metal pans do not work well in a microwave oven.

(d) Food with low water, fats or sugars. Examples include all dehydrated foods, some meats with a low water content. Canning of foods requires temperatures of over 100˚C, whereas the maximum temperature in a microwave oven is 100˚C.

(e) When the microwaves enter the cooking area they encounter the slowly revolving metal blades of the stirrer blade. Some models use a type of rotating antenna while others rotate the food through the waves of energy on a revolving carousel. In any case, the effect is to evenly disperse the microwave energy throughout all areas of the cooking compartment. Some waves go directly toward the food, others bounce off the metal walls and flooring; and, because of the metal screen in the glass, microwaves also reflect off the door. So, the microwave energy reaches all surfaces of the food from every direction. All microwave energy remains inside the cooking cavity. When the door is opened, or the timer reaches zero, the microwave energy stops.

CHAPTER 10 LAB ACTIVITIES

Investigation 10.2.1: Diffraction of Light in a Single Slit (Pages 540–541)

Procedure, Observations, and Calculations Part 1: Single-Slit Diffraction 3. As pressure on the fingertips was varied, the light from the single filament source changed from a single line of light to a

bright central region with adjacent bright bars on either side spreading out in a horizontal pattern.


4. Using the prepared single slit, there was a much clearer central maximum with a nodal line and then a brighter area of lower intensity on each side of the central maximum.

5. As the slit width increased the central maximum contracted and as did the whole interference pattern. When the slit width

decreased the reverse occurred with the central maximum becoming larger and the interference pattern spreading out horizontally.

6. As the distance between the slit and the source decreased the pattern remained the same, becoming slightly brighter. When the slit and the observer increased the pattern was more spread out, with a larger central maximum.

7. The pattern was less spread out for green light than it was for red light, and distance y to the centre line was smaller for green light than it was for red light.

Part 2: Using a Laser as a Source of Monochromatic Light 8. When the razor blade entered the laser beam a shadow was cast on the screen by the razor blade edge. Using s diverging

lens to spread out the shadow, it was noted that the shadow was not distinct with some light on the screen in the “shadow” region. A pattern of faint bright and dark areas was noted.

9. As the slit width w decreased, the central maximum became larger on the screen, as did the total interference pattern. 10. Using a prepared small slit, there was large central area with equally spaced nodal lines on both sides followed by a series

of bright and dark areas, with successively decreasing intensity.


11. If the equipment is available the graph should be a similar shape that is found in Figure 6 in Section 10.2. 12, 14. 4 1.0 10 mw −= ×

1 1

2.20 m1.5 cm, 1.3 cm

Ly y

== =

1

1

1.5 cm 1.3 cmaverage 2

average 1.4 cm

y

y

+=

=

The average for y1 is 1.4 cm, or 1.4 10–2 m.

15. 1wyL

λ =

4 2

7 7

(1.0 10 m)(1.4 10 m)= 2.20 m

6.36 10 m, or 6.4 10 mλ

− −

− −

× ×

= × ×

The wavelength of the light is 6.4 10–7 m. 16. (Step 14)

7

1 1

= 6.33 10 m (helium-neon laser)2.00 m2.5 cm, 2.6 cm

Ly y

λ −×== =

1

1

2.5 cm + 2.6 cmaverage 2

average 2.55 cm

y

y

=

=

The average for y1 is 2.55 cm, or 2.55 10–2 m. (We do not round off the average of y1 to ensure accurate calculation of the width of the slit.)

(Step 15)

17

2

5 5

(6.33 10 m)(2.00 m)2.55 10 m

4.96 10 m, or 5.0 10 m

Lwy

w

λ

−

−

− −

=

×=×

= × ×

(Step 14)

7

1 1

= 6.33 10 m 2.00 m1.6 cm, 1.8 cm

Ly y

λ −×== =

1

1

1.6 cm + 1.8 cmaverage 2

average 1.7 cm

y

y

=

=

The average for y1 is 1.7 cm, or 1.7 10–2 m. (Step 15)

17

2

5

(6.33 10 m)(2.00 m)=1.7 10 m

7.5 10 m

Lwy

w

λ

−

−

−

=

××

= ×

The width of the slit is 7.5 10–5 m.


Part 3: Diffraction by a Human Hair 18. Diagram of the interference pattern created by a human hair:

Analysis (b) The width of the slit must be larger than the wavelength of light but small enough to create significant diffraction. Since

very narrow slits are required the wavelength on light must be very small. The follows from the relationship sinwλθ = .

The maximum value for sin θ is 1. Thus, 1 , and .wwλ λ≤ ≥

(c) Since the central maximum was larger, as was the separation of the nodal lines for red light, than for green light, red light

must have a longer wavelength. This follows from the relationship 1wyL

λ = , where 1yλ ∝ .

(d) The diffraction pattern for a single slit shows the bright areas for specific colours located at separate location, just as was the case for double slit interference. This indicates the white light is made up of a range of wavelengths.

(e) (i) The wider the width of the slit w, the smaller the value for y1 or 11yw

∝ , if λ and L are constant.

(ii) The larger the value for L, the larger the larger the larger the value of y1 or 1 λy ∝ , if L and w are constant. (iii) The longer the wavelength λ, the larger the value of y1 or 1 λy ∝ , if L and w are constant.

(iv) Combining the proportionality statements we produce the equation 1wyL

λ = .

(f) This will depend on the prediction made and results produced experimentally.

Evaluation (g) The graph should be analogous to Poisson’s Bright Spot except the interference pattern will be vertical, not round. (h) The experimental value was 6.36 × 10–7 m which was close to the accepted value of 6.328 × 10–7 m. The experimental

error would be

7 7

7

measured value accepted value percent error = 100 %accepted value

6.36 10 m 6.328 10 m= 100%6.328 10 m

= +0.51%

− −

−

− ×

× − × ××

(i) Calculated value of w was 4.5 10–5 m. The given value was 5.2 10–5 m.

7 7

7


5.0 10 m 5.2 10 m= 100%5.2 10 m

3.8%

− −

−

− ×

× − × ××

= −

Calculated value of w was 7.5 10–5 m. The given value was 7.9 10–5 m.


7 7

7


7.5 10 m 7.9 10 m= 100%7.9 10 m

percent error 5.1%

− −

−

− ×

× − × ××

= −

(j) The most likely error occurs in measuring y1. Other errors can occur with the measurement of L. One must assume the value of w and λ must be taken as the accurate to the number of significant figures provided.

(k) One way to improve the measurement of y1 would be to the measure the total width of the central maximum by projecting the interference pattern on a ruled grating and using a magnifying glass. The value would have to be divided by 2 to obtain y1. (If the pattern in bright enough, the distance between adjacent nodal lines would provide a value for y1.)

(l) The evaluation of the prediction will depend on the prediction made. Investigation 10.4.1: Interference in Air Wedges (Page 542)

Procedure (sample) 1. Construct an air wedge, using two microscope slides attached together with elastic bands, as illustrated. Separate the ends

of the glass plates at one end and insert a single human hair between them. 2. Place the diverging lens in front of the laser beam so that a diverging laser light illuminates the air wedge. 3. Using the microscope, measure the distance covered by 50 or more bright fringes. (Consider one bright fringe to be from

the middle of one dark fringe to the middle of the next dark fringe.) Find the average separation between adjacent dark fringes (∆x).

4. Measure the total length of the air wedge L. Using the relationship 2

x Lt

λ ∆ = , determine the thickness of the human

hair. 5. With a micrometer, measure the thickness of a human hair. Compare the micrometer thickness with that measured using

the air wedge. 6. Repeat the procedure for two other hairs, from the same head.

Data and Analysis 1. λ = 6.328 10–7 m

L = 75.0 mm First hair:

4

100 = 2.40 cm2.40 cm

100 2.40 10 m

x

x

x −

∆

∆ =

∆ = ×

2 7

4

5

2

= 2(7.50 10 m)(6.328 10 m)

2(2.40 10 m)

9.89 10 m

Lxt

Ltx

t

λ

λ

− −

−

−

∆ =

∆× ×=

×

= ×

Micrometer measurement = 0.091 mm


Second hair:

4

80 = 2.1 cm2.1 cm=

80 2.62 10 m

x

x

x −

∆

∆

∆ = ×

2 7

4

5

2

= 2(7.5 10 m)(6.328 10 m)

2(2.62 10 m)

= 9.06 10 m

Lxt

Ltx

t

λ

λ

− −

−

−

∆ =

∆× ×=

×

×

Micrometer measurement = 0.084 mm Third hair:

4

122 = 2.6 cm2.6 cm

122 2.13 10 m

x

x

x −

∆

∆ =

∆ = ×

2 7

4

4

2

= 2(7.50 10 m)(6.328 10 m)

2(2.13 10 m)

1.11 10 m

Lxt

Ltx

t

λ

λ

− −

−

−

∆ =

∆× ×=

×

= ×

Measurement of the thickness with micrometer = 0. 076 mm Average value for the thickness of a human hair using the air wedge is:

0.0989 mm + 0.0906 mm + 0.111 1.00 mm3

=

Average value for the thickness of a human hair using the micrometer is:

0.091 mm + 0.084 mm + 0.076 mm 0.0837 mm3

=

Evaluation (a) The values for thickness of a human hair vary because of the type of method used and for different hairs. The air wedge is

a more accurate method since the number of significant figures that can be carried throughout the calculation is more than is the case with the micrometer. Also, the micrometer pressure on the hair could have caused the lower readings than those for the air wedge, because the hairs could have been slightly compressed.

The accuracy of the counting of the number bright lines in the air wedge interference pattern is compromised by the sophistication of the equipment used.

(b) Modifications the may improve the accuracy of the air wedge method could include: • Use a longer air wedge increasing the value of L and creating more nodal line to be counted. • Set up the air wedge with mm scale attached to the slide so the measurements could be made more accurately. • Design a set-up so the air wedge is illuminated under a microscope improving the counting of the bright lines and the

calculation of ∆x.


CHAPTER 10 SUMMARY

Make A Summary (Page 544)

Property Sound Electromagnetic Waves

Equation

transmission requires a medium

no medium required v f λ=

reflection obeys the laws of reflection

obeys the laws of reflection

i r∠ = ∠

refraction sound is refracted when the speed changes ( new medium)

waves are refracted when the speed changes ( new medium)

Snell’s law 1 2

2 1

sinsin

nn

θθ

=

total internal reflection

applies applies 1sin c nθ =

double-slit interference

applies

applies 1sin ( ) (minima)2n n

dλθ = −

single-slit interference

does not apply λ >> w

applies sin (minima)n

nwλθ =

diffraction grating does not apply λ >>> d

applies sin (maxima)m

mdλθ =

(a) Both sound and electromagnetic waves obey the same relationships and behaviour for reflection, refraction, total internal reflection, simple interference and two-point interference (see table). But, sound originates from a vibrating object, whereas electromagnetic waves originate from an oscillating electric or magnetic field. The speed of sound is 332 m/s and it varies with temperature and the medium. Electromagnetic waves have a speed of c in a vacuum, but it too can be lower in a different medium. Sound requires a material medium for its transmission, whereas electromagnetic waves do not.

(b) Some answers should include any five of the following points: • Double and single-slit interference patterns to find the λ of light or the size of a small opening • Spectroscopy using a diffraction grating • Making objects to small tolerances – Newton rings • Measuring very small distances – interferometer • Measuring the size of small objects – air wedge • CD and DVD – thin films • Coated eyeglasses, cameras and high-efficiency windows – thin films • Holograms and holographs • Transmission and receiving of radio and television broadcasts- modulation and de-modulation

CHAPTER 10 SELF QUIZ (Page 545)

True/False 1. F Polarization provided the proof that light is a transverse wave. 2. (a) F The smaller the slit width, the larger the distance between adjacent maxima and minima. (b) T (c) T 3. F The larger the aperture of an optical instrument, the better its resolution. 4. T 5. T


6. F Maxwell postulated and Hertz proved that light and radio waves and light travel through space at the speed of light (3.00 × 108 m/s).

7. T 8. T 9. F Ionizing radiation originates in light with a wavelength shorter than that of visible light.

Multiple Choice 10. (d) Not to be confused with the two-point interference pattern, this is the interference pattern formed by a single slit. In the

single slit the path difference is determined by comparing the rays from points in the slit in pairs, beginning with the wave ray from the edge of slit and the wave ray at the center of the slit. For the second intensity minimum the path

difference for these rays and the point P is 2λ , making the path difference for the wave rays at both edges 2λ.

11. (c) Using the same argument as in 10, the path difference to the first maximum would be 3dλ .

12. (c) The effect is created by the interference between the rays reflected from the top and bottom surfaces. 13. (d) In I and II, the ray reflected from the top surface is inverted while that reflected at the bottom interface is not inverted.

Thus, constructive interference is observed when the thickness t is 2λ , 3

4λ , 5

4λ , etc., making option I correct. For

transmission, there is no inversion of either ray and the thickness t for constructive interference is 2λ , λ, 3

2λ , etc.,

making III a correct option. 14. (d) The ray reflected from the surface of the coating will be inverted, as will the ray reflected from the coating-glass

interface. For destructive interference (minimum reflection), the path difference must be 2λ or t =

2λ . But, the

wavelength is measured in the coating, making it smaller by a factor of the refractive index 1.2. Therefore the correct

answer is ( )4 1.2λ .

15. (e) All electromagnetic waves travel at the same speed (3.00 × 108 m/s). 16. (e) Only (e) has the correct order from lower to higher frequency.

CHAPTER 10 REVIEW (Pages 546–547)

Understanding Concepts 1. Fix one Polaroid sheet to the window. Make the other Polaroid sheet adjustable so it can be rotated. By varying the

rotation, the intensity could be changed to zero, if necessary. 2. w = 4.30 × 10–5 m

L = 1.32 m 2y1 = 3.8 cm

y1 = 3.8 cm2

= 1.9 cm

( )( )5 2

7

4.30 10 m 1.9 10 m

1.32 m6.2 10 m

ywL

λ

λ

− −

−

= × ×

=

= ×

The wavelength of the light is 6.2 × 10–7 m. 3. λ = 5.50 × 102 nm = 5.50 × 10–7 m

n = 2 θ = 25° w = ?


( )( )7

6

sin

sin

2 5.50 10 m

sin 251.3 10 m

n

n

nwnw

w

λθ

λθ

−

−

=

=

×=

°= ×

The spacing between the lines is 1.3 × 10–6 m. 4. θ = 36.9° w = 1.00 µm = 1.00 × 10–6 m

( )( )6

7

sin

1.00 10 m sin 36.9

16.00 10 m

nwn

θλ

λ

−

−

=

× °=

= ×

The wavelength of the light is 6.00 × 10–7 m. 5. λ = 694.3 nm = 6.943 × 10–7 m

y = 42 mm = 42 × 10–3 m n = 1 L = 2.50 m w = ?

( )( )( )7

3

5

1 2.50 m 6.943 10 m

21 10 m8.3 10 m

y nL w

nLwy

w

λ

λ

−

−

−

=

=

×=

×= ×

The width of the slit is 8.3 × 10–5 m. 6. (a) θ = 28.0°

λ = 6.00 × 10–7 m n = 1 w = ?

( )( )7

6

sin

1 6.00 10 m

sin 28.01.28 10 m

n

nw

w

λθ

−

−

=

×=

°= ×

The width of the slit is 1.28 × 10–6 m. (b) θ = 67.0°

n = 2 w = 1.28 × 10–6 m λ = ?

( )( )6

7

sin

1.28 10 m sin 67.0

25.89 10 m

nwn

θλ

λ

−

−

=

× °=

= ×



7. n = 1 θ = 18.0° θ3 = ?

1

2

1sin2

1sin 121sin 32

n nw

w

w

λθ

λθ

λθ

= + = + = +

( )

( )1

23

1.5sin183.5sin

w

w

λ

λθ

=°

=

But, w1 = w2

( ) ( )

( )( )3

3

3

1.5 3.5sin18 sin

3.5 sin18sin

1.546.1

λ λθ

θ

θ

=°

°=

= °

The angle for the third-order maximum is 46.1°. 8. w = 3.00 × 10–6 m

λ = 694.3 nm = 6.943 × 10–7 m L = 2.0 m y1 = ?

( )( )( )

1

1

7

6

1

12sin

12

1.5 6.943 10 m 2.00 m

3.00 10 m0.694 m

m

my

w L

m Ly

w

y

λθ

λ

−

−

+ = =

+ =

×=

×=

Each spot is 0.694 m, or 69.4 cm away from the central axis. 9. λ = 7.80 × 102 nm = 7.80 × 10–7 m

2y = 1.2 mm

y = 1.2 mm2

= 6.0 mm = 6.0 × 10–4 m

L = 3.0 mm = 3.0 × 10–3 m d = ?

( )( )7 3

4

6

7.80 10 m 3.0 10 m

6.0 10 m3.9 10 m

yL d

Ldy

d

λ

λ

− −

−

−

=

=

× ×=

×= ×

The spacing between the slits of the grating is 3.9 × 10–6 m.


10. d = 2

1 cm 1 m2400 lines 10 cm

× = 4.17 × 10–6 m

n = 1 L = 0.625 m y = 8.94 × 10–2 m λ = ?

( )( )

( )( )

2 6

7

sin

8.94 10 m 4.17 10 m

1 0.625 m

5.96 10 m

y mL dydmL

λθ

λ

λ

− −

−

= =

=

× ×=

= ×

The wavelength of the light is 5.96 10–7 m, or 596 nm. 11. λ = 638 nm = 6.38 × 10–7 m

n = 3 θ = 19° number of lines = ? First we must calculate the distance between the lines:

( )7

6

4

sin

sin

3 6.38 10 m

sin19.05.88 10 m

5.88 10 cm

m

m

mdmd

d

λθ

λθ

−

−

−

=

=

×=

°= ×

= ×

Now we can calculate the number of lines:

4

5

1number of lines5.88 10 cm

number of lines 1.70 10 lines/cm

−=×

= ×

The number of lines is 1.70 × 105 lines/cm. 12. Ray 1 is reflected with a 480° phase change (less to more). Ray 2 is reflected without a phase change. At t = 0 the two

rays undergo constructive interference. At t = 4λ the path difference is

2λ . The two rays undergo constructive

interference, which improves reflection.


13. Ray 1 is reflected with a phase change (less to more). Ray 2 also has a phase change (less to more). As t → 0, the two rays undergo constructive interference and a bright area is seen.

Ray 1 is reflected with a phase change (less to more). Ray 2 is reflected without a phase change (more to less). As t → 0,

destructive interference occurs and a dark area is observed.

14. The minimum thickness for constructive interference is 4λ since there is no phase change for ray 2. Assume the

wavelengths are in air: λr = 6.40 × 10–7 m λv = 4.50 × 10–7 m

The wavelength in the coating would be:

7

r

7r

6.40 10 m1.40

4.57 10 m

λ

λ

−

−

×=

= ×

7

v

7v

4.50 10 m1.40

3.21 10 m

λ

λ

−

−

×=

= ×

The thickness would be 4λ .

tr = 74.57 10 m

4

−×

tr = 1.14 × 10–7 m

tv = 73.21 10 m

4

−×

tv = 8.02 × 10–8 m The thickness could be 1.14 × 10–7 m and 8.02 × 10–8 m.


15. no = 1.25 nw = 1.33

λ = 556 nm t = ?

Constructive interference occurs when the path difference is 0, λ, 2λ etc., or when t = 0, 2λ , 3

4λ , etc.

ao

o

556 nm1.25

445 nm

nλλ

λ

=

=

=

For minimum thickness:

2445 nm

2222 nm

t

t

λ=

=

=

Therefore, the minimum thickness of the oil is 222 nm. 16. ns = 1.34

λ = 5.80 × 102 nm t = ?

a

s

as

2

s

5.80 10 nm1.34

439 nm

n

n

λλλ

λ

λ

=

=

×=

=

For maximum reflection, path difference is 2λ , 3

2λ , etc., and thickness would be 3,

4 4λ λ , etc.

4439 nm

4110 nm

t

t

λ=

=

=

( )

34

3 439 nm4

329 nm

t

t

λ=

=

=

Two possible values for the thickness of the film are 110 nm and 329 nm. 17. t = 122 nm

n = 1.40 λo = ?

ao

o

122 nm1.40

871 nm

nλλ

λ

=

=

=


Path difference for bright areas is 2λ or

4t λ= .

4871 nm

4218 nm

t

t

λ=

=

=

The wavelength of the oil slick is out of range of the spectrum. Therefore, the oil has no colour. 18. λ = 546 nm ∆x = ?

( )

2

262

26 546 nm2

7.10 10 nm

x

x

λ∆ =

=

∆ = ×

The surface receded from the fixed surface. The closer the rings are together, the smaller the path difference (and thickness of the air layer).

19. nMgF = 1.38 t = 1.25 × 10–5 cm nc = 1.55 λ = ?

Path differences for bright areas are 2λ , 3

2λ , etc. For bright thickness, t =

4λ .

Since a cc and

4t

nλ λ

λ = = or c 4tλ =

( )( )

a

a

5

5a

4

4

4 1.25 10 cm 1.38

6.9 10 cm

tn

tn

λ

λ

λ

−

−

=

=

= ×

= ×

The wavelength is 6.9 10–5 cm. 20. nm = 1.50

ng = 1.50 λ = 689 nm t = ?

Rays 1 and 2 are in phase. Ray 3 is out of phase. For ray 3 to be in phase with rays 1 and 2 it requires a path difference of

2λ or a thickness of

4λ in the liquid.


First we need to calculate the wavelength of the liquid:

a

L

aL

L

689 nm1.76

391 nm

n

n

λλλ

λ

λ

=

=

=

=

Now we can calculate the thickness:

4391 nm

497.8 nm

t

t

λ=

=

=

The thickness of the liquid layer is 97.8 nm.

21. For a dark fringe to move to a dark fringe requires a path difference of ∆λ or a distance movement of 2λ∆ .

For a dark fringe to move into a position previously occupied by a bright fringe requires a mirror position of 4λ .

22. (a) w = 81 cm = 8.1 × 10–2 m λ = 489 nm = 4.89 × 10–7 m θ = ?

( )( )7

2

5 51

sin

1 4.89 10 m

8.1 10 m3.46 10 , or 3.5 10

nndλθ

θ

−

−

− ° − °

=

×=

×= × ×

The angle is 3.5 × 10–5°. (b) f = 512 Hz

v = 342 m/s θ = ?


1

342 m/s512 Hz6.68 10 m, or 66.8 cm

vf

λ

λ −

=

=

= ×

Now we can calculate the angle:

sin

66.8 cm0.81 cm56

dλθ

θ

=

=

= °

The angle is 56°. (c) The answers in part (a) and part (b) are so different because of the wavelengths and the size of the opening. In the case

of light, the wavelength is much smaller than the width of the opening, therefore there is little diffraction. For a sound wave, the wavelength is only slightly smaller than the opening, therefore there is significant diffraction and interference.


Making Connections 23. Students may find some of the following points:

• The Blackwall spider (Drassodes cupreus), a grey-brown northern European arachnid has a pair of secondary eyes, specifically to analyse the polarization of light from the sky.

• The researchers who made this discovery, Marie Dacke of the University of Lund, Sweden and colleagues, believe that "a similar organisation of the secondary eyes in several spider families indicates that this may not be an isolated phenomenon."

• The Blackwall has a pair of oval-shaped eyes that are close together and roughly at right angles to one another on the upper surface of its front end. These do not form images. Instead, they make up a filter, which can calculate precisely the direction of polarization of light falling upon them. This is done by comparing the discrepancy between the planes-of-vibration of the light waves going into each eye. For this purpose they are highly reflective, hence their brilliant blue glow.

• Within each eye is a long V-shaped organ made up of two mirror-like surfaces, splayed to catch the light so that light polarized along the long axis of one of the eyes, for example, is reflected four times more efficiently than light polarized along the short axis. These do not meet at the bottom, but rather are separated by a narrow gap, through which poke light-receptive nerve fibres. These fibres are arranged so as to maximize the contrast between light of differing levels and directions of polarization.

• This compass organ, which has no real imaging optics, is quite crude compared to others- - in insect compound eyes, for instance. It works best at dusk and dawn, the researchers have found, probably because at these times the sky is polarized in only one direction and there is no direct unpolarized sunlight to confuse things.

• There is some evidence that bees and some ant species also use polarize light for navigation. 24. The wavelength of the sound waves cannot be larger than the size of the target or the wavelengths will be diffracted

instead of reflected. Thus, the wavelength should be equal to or smaller than the target of 3.00 mm. v = 3.40 102 m/s L = λ = 3.00 mm = 3.00 10–3 m f = ?

2

3

5

3.40 10 m/s3.00 10 m

1.13 10 Hz

vf

f

λ

−

=

×=×

= ×

The typical frequency of a bat’s sonar is estimated to be 1.13 × 105 Hz. 25.

Source: National Research Council Canada • The Algonquin Radio Observatory’s 46-m telescope began operations in May 1966. The telescope was built and

operated by the National Research Council (NRC) of Canada. At the current time the ARO is operated by the Geodetic Survey Division of Natural Resources Canada in partnership with the Space Geodynamics Laboratory, CRESTech.


• The ARO was used in the first successful VLBI experiment in 1967 and was involved as early as 1968 in Geodesy, when the baseline length between the ARO and a telescope in Prince Albert, Saskatchewan was measured to be 2143 km (sigma = 20 m). The long base line effectively created a radio telescope with a very large aperture (2143 km) vastly improving the resolution for the locating strong sources of electromagnetic radiation in the universe.

• The antenna participated in the NASA CDP and DOSE programs beginning in 1984. A long-term loan of a Mark III terminal from NASA in 1989 makes it possible to participate regularly in VLBI observations in support of the maintenance of the celestial and terrestrial reference frames in the CORE and NEOS series.

26. (a) Biomimicry (from bios, meaning life, and mimesis, meaning to imitate) is a new science that studies nature's best ideas and then imitates these designs and processes to solve human problems. Studying a leaf to invent a better solar cell is an example.

The core idea is that nature, imaginative by necessity, has already solved many of the problems we are grappling with. Animals, plants, and microbes are the consummate engineers. They have found what works, what is appropriate, and most important, what lasts here on Earth. Like the viceroy butterfly imitating the monarch, we humans can imitate the best and brightest organisms in our habitat. We are learning, for instance, how to harness energy like a leaf, grow food like a prairie, build ceramics like an abalone, self-medicate like a chimp, compute like a cell, and run a business like a hickory forest.

The occurrence and distribution of iridescent colours and the various theories regarding their manifestation in the natural world were for a long time discussed by scientists. Even Sir Isaac Newton in his book Opticks, published in 1704, put forward a reason for the iridescent nature of the colour from the feathers of peacock tails. The word iridescence is itself defined as the change in hue of the colour of an object as the observer changes viewing position.

An example of this iridescent colour is the appearance of light reflected from an oil film on the surface of water. If the oil film is viewed from different angles the colours appear to shift and change. The iridescence of the peacock, the grackle, some butterflies, and the hummingbird is caused by the same phenomena – thin films on the scales or feathers of the insect or bird.

(b) Iridescent colours are particularly striking for the observer. They can be among the purest and most brilliant and cannot be matched by even the brightest pigmented colours in their depth and intensity. In addition, the glittering change of hue that accompanies any change of light angle or observer position lends these colours a magic and beauty that is unmatched by any others.

Investigators are studying iridescent plumage to apply the same effects to paint and fabrics upholstery, drapes, clothes and rugs. Since dyes and pigments are not involved, the colours will not fade. Also, more vibrant colours could be possible.

27. Sunglasses or prescription eyeglasses that darken when exposed to the sun were first developed by Corning in the late 1960s and popularized by Transitions in the 1990s. In fact, because of the extreme popularity of the Transitions brand, these lenses are usually referred to as transition lenses. The correct term for these glasses is photochromic or photochromatic, which refers to a specific chemical reaction the lenses have to ultraviolet (UV) radiation.

Photochromic lenses have millions of molecules of substances such as silver chloride or silver halide embedded in them. The molecules are transparent to visible light in the absence of UV light, which is normal for artificial lighting. But when exposed to UV rays, as in direct sunlight, the molecules undergo a chemical process that causes them to change shape. The new molecular structure absorbs portions of the visible light, causing the lenses to darken. The number of molecules that change shape varies with the intensity of the UV rays.

When you go indoors and out of the UV light, a different chemical reaction takes place. The absence of the UV radiation causes the molecules to "snap back" to their original shape, resulting in the loss of their light absorbing properties. In both directions, the entire process happens very rapidly.

In the original PhotoBrown and PhotoGrey products made by Corning, the lenses are made of glass, and the molecules are distributed evenly throughout the entire lens. The problem with this method was apparent in prescription glasses where different parts of the lens were of varying thickness. The thicker parts would appear darker than the thinner areas. But with the increasing popularity of plastic lenses, a new method has been developed. By immersing the lenses in a chemical bath, the photochromatic molecules are actually absorbed to a depth of about 150 microns into the plastic. This is much better than a simple coating, which would only be about 5 microns thick and would not provide enough molecules to make the lenses sufficiently dark.

An important note about photochromic lenses: because they react to UV light and not to visible light, there are circumstances under which the darkening will not occur. A perfect example of this is in your car. Because the windshield blocks out most UV light, photochromic lenses will not darken. For this reason, most sunglasses with photochromic lenses also have a certain amount of tint already applied to them.


28. • The metal body of an aircraft is very good at reflecting radar signals, and this makes it easy to find and track airplanes

with radar equipment. • The goal of stealth technology is to make an airplane invisible to radar. There are two different ways to create radar

invisibility: • The airplane can be shaped so that any radar signals it reflects are reflected away from the radar equipment. • The airplane can be covered in materials that absorb radar signals. • Most conventional aircraft have a rounded shape. This shape makes them aerodynamic, but it also creates a very

efficient radar reflector. The round shape means that no matter where the radar signal hits the plane, some of the signal gets reflected back:

• A stealth aircraft, on the other hand, is made up of completely flat surfaces and very sharp edges. When a radar signal

hits a stealth plane, the signal reflects away an angle, like this:

Extension 29. (a) θ1 = 13.8°

n = 2 θ2 = ?


Since we know the single slit produces a first minimum at 13.8°,

1

sin

1sin

sin13.8

nnw

w

w

λθ

λθ

λ

=

=

° =

Using this relationship, we can find the angle of the second-order minimum:

( )2

2

sin

2sin

2 sin13.828.5

nnw

w

λθ

λθ

θ

=

=

= °= °

The angle for the second-order maximum is at 28.5°. (b) θ3 = ? Using the relationship found in part (a), we can find the angle for the third-order minimum:

( )3

3

sin

3sin

3 sin13.845.7

nnw

w

λθ

λθ

θ

=

=

= °= °

The angle for the third-order maximum is 45.7°. 30. λ1 = 6.00 × 10–7 m

θ1 = 18.5° θ2 = 14.9° λ2 = ?

First we must find the relationship between sin θ1 and sin θ2:

11

22

sin

sin

sin

nnd

nd

nd

λθ

λθ

λθ

=

=

=

1

1

22

1 1

2 2

sinsin

sinsin

d

d

λθ

λθ

θ λθ λ

=

=


Using this relationship, we can substitute our known values and solve for λ2:

( )( )

1 22

1

7

72

sinsin

6.00 10 m sin14.9

sin18.54.86 10 m

λ θλθ

λ

−

−

=

× °=

°= ×

To calculate the highest order observable:

1

1

sin 1

1md

θλ

≤

≤

Substituting in wavelength, we can arrange the equation to isolate d: ( )76.00 10 mm d−× ≤

But,

11

1

17

6

sin

sin

6.00 10 msin18.5

1.89 10 m

d

d

d

λθ

λθ

−

−

=

=

×=°

= ×

Finally, we can use the relationship between m and d to solve for m:

16

71.89 10 m6.00 10 m3.15

dm

m

m

λ−

−

≤

×≤×

≤

The wavelength of the second source is 4.86 × 10–7 m. The highest order observable is 3. 31. If the chamber has a length L and is traversed twice, then the change in path length with and without air is 2Lna – 2L, or 2L(na – 1). We can calculate the number of fringes changed:

( )

( )( )

a

a

2

7

2 1number of fringes

2 8.0 10 m 1.000 29 1

6.38 10 mnumber of fringes 73

L nλ

−

−

−=

× −=

×=

The number of fringe-pairs that shift in the process is 73.

Copyright © 2003 Nelson Unit 4 Performance Task 619

UNIT 4 PERFORMANCE TASK PHYSICAL OPTICS PHENOMENA (Pages 548–549) The Unit 4 Performance Task is research-based and open-ended. It provides students with an opportunity to examine optical phenomena and their applications. The student may choose to photograph physical optical phenomena that exhibit effects discussed in Unit 4, including wave properties in ripple tanks, refraction and dispersion effects, applications of polarization, diffraction effects in slits and diffraction gratings, thin film phenomena, etc. The second option is to create a three-dimensional holographic image. The student is required to conduct and evaluate the work, and present the findings in a suitable form (chosen by the student).

Option 1: Photography of Physical Optical Phenomena In this option, students take four photographs of two different optical phenomena they have studied in the unit. Students may also choose optical phenomena that they find interesting that are not discussed in the textbook. A report will be submitted for assessment/evaluation that contains three of the four images taken, a description of the apparatus used to take the images, and a concise explanation of the underlying physical concepts. The report will also contain historical background information regarding the phenomenon under study. Students are encouraged to use a digital camera in order to simplify the reproduction of the images; however, if a digital camera is not available, they may use a simple SLR camera with a manually-controlled focus, aperture, and exposure settings. Information may be obtained from other physics or optics textbooks, encyclopaedias, and science magazines such as Scientific American and Popular Science. Useful Web sites include: http://www.nikon.co.jp/main/eng/photo_world/kumon/12e.htm http://www.weather-photography.com/Atmospheric_Optics/photography.html

Assessment The report will be assessed and/or evaluated on the basis of the following criteria: • adequacy and accuracy of information • level to which the applicable physics principles seem to have been understood • referencing • photographic quality • quality of the written communication

Option 2: Holography In this option, students will create a simple hologram of a small object found in their home or classroom. They will require a “holography kit” containing the necessary equipment, and a suitable light-tight room for conducting the activity. Students will begin the activity by conducting library and/or Internet research on the semiconductor laser. Their final report should contain an explanation of its operation, including well-labelled diagrams. The report should also contain an explanation of the silver halide process for creating a photographic image, a copy of the hologram, a description of the procedure used to create the hologram, and an evaluation of the process including suggested improvements. Students must be warned about the dangers of allowing direct or reflected laser beams entering the eye well in advance of conducting the activity. Teachers who have little experience in holography are encouraged to purchase a commercially available “holography kit” for this activity. These kits significantly simplify the process, and ensure relatively good quality results. Kits are available from most science supply companies. Internet Web sites provide much useful information on holography. Relevant Web sites include: http://www.cem.msu.edu/~cem181h/projects/96/memory/holoprimer.html http://www.holo.com/holo/book/book1.html http://members.aol.com/gakall/holopg.html#setup

http://www.nikon.co.jp/main/eng/photo_world/kumon/12e.htm

http://www.weather-photography.com/Atmospheric_Optics/photography.html

http://www.cem.msu.edu/~cem181h/projects/96/memory/holoprimer.html

http://www.holo.com/holo/book/book1.html

http://members.aol.com/gakall/holopg.html#setup


Assessment The report will be assessed and/or evaluated on the basis of the following criteria: • adequacy and accuracy of information • level to which the applicable physics principles seem to have been understood • referencing • demonstrated ability to carefully follow instructions • quality of the hologram • quality of the written communication

Copyright © 2003 Nelson Unit 4 Self Quiz 621

UNIT 4 SELF QUIZ (Pages 550–551)

True/False 1. T 2. T 3. F All wave properties can be explained using the wave model except transmission through a vacuum. 4. T 5. T 6. F The fringe pattern would become smaller if the red light source is replaced by a blue light source since blue light has a

shorter wavelength than red light. 7. T 8. F All waves do not have the same properties as electromagnetic waves. Examples include water waves, sound waves, and

earthquake (seismic) waves.

Multiple Choice 9. (e) In the deep water the speed is determined using the universal wave equation.

= (12 Hz)(2.0 cm) 24 cm/s

v f

v

λ=

=

10. (c) When the ray passes from medium1 to medium 2 the ray refracts toward the normal. Thus, v2 > v1. When the ray passes from medium 2 to medium 3 it bends away from the normal. Thus, v2 > v3. But the emerging ray is not parallel to the incident ray and is refracted more. Thus, v3 > v1, and therefore, v3 > v1 > v2.

11. (e) When a wave passes form a heavy string to a lighter string, it is moving form a slower medium to a faster medium. Thus, there is no inversion (phase change) for the reflected wave or the transmitted wave.

12. (d) path difference = 12

n − λ for minima

The path difference is 37.0 cm – 28.0 cm = 9.0 cm. Since it is the second nodal line, n = 2

19.0 cm 22

6.0 cm

λ

λ

= − =

13. (d) If the separation between slits A and B is increased, the number of nodal line would increase. The width of the slits B and C is decreased, there will more diffraction through each slit and the pattern will spread out. If the distance from the metal sheet to the screen is changed, the pattern will be less spread out if closer, more spread

out if further away. If the wavelength of the incident light is decreased, the nodal pattern will be closer together. If the distance between the cardboard and the metal sheet is increased, there will be no affect on the pattern. 14. (b)

15. (b) In soap films (air-water-air) there is destructive interference for transmission where the thickness is 3 5, , , etc.4 4 4λ λ λ

Therefore, the second area of destructive interference will have a thickness of 34λ .

16. (e) When dealing with transmission rays 1 and 2 can be ignored and we are only concerned with rays 3 and 4, eliminating (a) and (c). - The degree of transparency has no effect, only the speed in each of the coating and the lens. - To reduce reflected light the velocity of the light in the coating must be greater than in the lens. - The total light energy reflected is a minimum.


Completion 17. period 18. source

19. wλ

20. nodal 21. out of phase, wavelength 22. medium 23. transverse 24. greater

25. sin nnwλθ =

26. greater 27. larger

Matching 28. Scientist Discovery or Innovation

Gabor holography Grimaldi diffraction of light at two successive slits Hertz creation and detection of radio waves Huygens wavelet model for propagation of wave fronts Land commercially viable polarizing filters Maxwell mathematical theory of electromagnetic waves Marconi transmission of radio signals Michelson interferometer Newton particle theory of light Poisson diffraction of light around a small disk Young two-slit interference

Copyright © 2003 Nelson Unit 4 Review 623

UNIT 4 REVIEW (Pages 552–555)

Understanding Concepts 1. (a) reflection, rectilinear propagation, refraction, dispersion (b) diffraction, partial reflection/partial refraction, interference, polarization (c) transmission through a vacuum 2. ng = 1.60

λa = 415 nm λg = ?

a

g

ag

g

415 nm1.60

259 nm

n

n

λλλλ

λ

=

=

=

=

The wavelength of the violet light in glass is 259 nm. 3. The wavelengths of sound are very much longer than light waves. There is little diffraction, if the wavelength is smaller

than the opening. For sound waves, the wavelengths are typically larger than opening and are easily attracted through an opening or around a corner.

4. T = 0.10 s distance between the first and sixth crests is 5 wavelengths. (a) λ = ?

5 12.0 cm12.0 cm

52.4 cm

λ

λ

λ

=

=

=

The wavelength is 2.4 cm. (b) v = ?

2.4 cm0.10 s

24 cm/s

vT

v

λ=

=

=

The speed of the wave is 24 cm/s. 5. n = 2

v = 7.5 cm/s path difference = 29.5 cm – 25.0 cm = 4.5 cm

(a) λ = ?

1 path difference212 4.5 cm21.5 4.5 cm

3.0 cm

n λ

λ

λλ

− = − =

==

The wavelength is 3.0 cm.


(b) f = ?

7.5 cm/s3.0 cm

2.5 Hz

v fvf

f

λ

λ

=

=

=

=

The frequency is 2.5 Hz. 6. Double-slit interference requires a null result (destructive interference), which is definitive, whereas diffraction could be

open to interpretation at least before wave analysis was possible. 7. 6∆x = 6.0 cm

∆x = 6.0 cm6

= 1.0 cm = 1.0 × 10–2 m

L = 3.00 m d = 220 µm = 2.0 × 10–4 m (a) ∆x = ?

( )

2 4

7

(1.0 10 m)(2.0 10 m)3.00 m

6.7 10 m

x Ldx dL

λ

λ

λ

− −

−

∆ = ∆

=

× ×=

= ×

The wavelength is 6.7 × 10–7 m. (b) The colour of the light is red. 8. θ = 5.2°

n = 2 dλ

= ?

1sin212

sin122

sin 5.2

16.6

n

n

nd

nd

d

λθ

λ θ

λ

= − − =

− =°

=

The ratio of the slit separation d to the wavelength λ of the light is 16.6:1. 9. d = 0.018 mm = 1.8 × 10–5 m

θ = 8.2° n = 5 λ = ?


( )( )5

7

sin12

1.8 10 m sin 8.2152

5.70 10 m

d

n

θλ

λ

−

−

=−

× °=

−

= ×

The wavelength is 5.70 × 10–7 m. 10. λ = 638 nm = 6.38 × 10–7 m

n = 3 θ = 8.0° d = ?

( )7

5

12

sin13 6.38 10 m2

sin 8.01.15 10 m

nd

d

λ

θ−

−

− =

− × =°

= ×

The distance between the slits is 1.15 × 10-5 m. 11. λ = 633 nm = 6.33 10–7 m

d = 0.100 mm = 1.00 10–4 m L = 2.10 m x1 = ?

( )

1

7

4

31

12

1 6.33 10 m1 2.10 m2 1.00 10 m

6.65 10 m

x n Ld

x

λ

−

−

−

= − × = − ×

= ×

The distance of the first dark fringe is 6.65 10–3 m, or 6.65 mm. 12. d = 0.42 mm = 4.2 10–5 m

L = 4.00 m ∆x = 5.5 cm = 5.5 10–2 m λ = ? f = ?

To calculate wavelength:

( )( )2 5

7 7

(minima)

5.5 10 m 4.2 10 m

4.00 m5.78 10 m, or 5.8 10 m

dxL

λ

λ

− −

− −

= ∆ × ×

=

= × ×

To calculate frequency:

8

7

14

3.00 10 m/s5.78 10 m5.2 10 Hz

cf

f

λ

−

=

×=×

= ×

The wavelength is 5.8 10–7 m. The frequency is 5.2 1014 Hz.


13. λ = 639 nm = 6.39 10–7 m d = 0.048 mm = 4.8 10–5 m L = 2.80 m x1 = ?

( )( )7

1 5

2 21

1 (minima)2

6.39 10 m 2.80 m112 4.8 10 m

1.86 10 m, or 1.9 10 m

nx nL d

x

x

λ

−

−

− −

= − × = − ×

= × ×

The first dark fringe is 1.9 10–2 m, or 1.9 cm away from the centre of the pattern. 14. λ = 656 nm = 6.56 10–7 m

L = 1.50 m n = 4 x = 48.0 mm = 4.80 10–3 m d = ?

( )( )7

3

4

sin (maxima)

4 6.56 10 m 1.50 m

4.80 10 m8.20 10 m

nd

x nL d

d

d

λθ

λ

−

−

−

=

=

×=

×= ×

The separation of the two slits is 8.20 10–4 m. 15. λ1 = 4.80 102 nm = 4.80 10–7 m

λ2 = 6.20 102 nm = 6.20 10–7 m d = 0.68 mm = 6.8 10–4 m L = 1.6 m n = 2 ∆x = ?

1 1 2 2 and nL nLx xd d

λ λ= =

( ) ( )

( ) ( )( )

2 1 2 1

7 74

42 1

2 1.60 m6.20 10 m 4.80 10 m

6.8 10 m6.6 10 m

nLx xd

x x

λ λ

− −−

−

− = −

= × − ××

− = ×

The second order maxima are 6.6 10–4 m, or 0.66 mm apart. 16. Let the subscript a represent air, and w represent water.

λa = 4.00 102 nm = 4.00 10–7 m d = 5.00 10–4 m nw = 1.33 L = 50.0 cm = 5.00 10–1 m ∆x = ?

First we must find the wavelength of the light in water:

aw

w7

7w

4.00 10 m1.33

3.00 10 m

nλλ

λ

−

−

=

×=

= ×


To calculate the distance between the fringes:

( )( )1 7

4

4

5.00 10 m 3.00 10 m

5.00 10 m3.00 10 m

Lxd

x

λ

− −

−

−

∆ =

× ×=

×∆ = ×

The fringes are 3.00 10–4 m apart on the screen. 17. Dispersion requires double refraction to be visible. Since the sides of window glass are parallel, little refraction occurs. 18. Polarization was important because it showed the waves of light were transverse. 19. Light travels in all directions until it is reflected off a horizontal surface (such as the hood of a car, or a puddle on the

road). This reflected light become horizontally polarized—the light is only vibrating in one plane. Polaroid sunglasses have polarizing filters that are arranged in the vertical plane to absorb these horizontally polarized light waves. As a result, Polaroid sunglasses reduce the effect of glare.

20. w = 5.60 10–4 m L = 3.00 m y1 = 3.5 mm = 3.5 10–3 m λ = ?

( )( )( )( )

11

3 4

7 71

1 (minima)2

12

3.5 10 m 5.60 10 m

1 3.00 m

6.53 10 m, or 6.5 10 m

ny nL d

y d

n L

λ

λ

λ

− −

− −

= −

= −

× ×=

= × ×

The wavelength of the light is 6.5 10–7 m. 21. λ = 675 nm = 6.75 10–7 m

θ = ? (a) w = 1.80 10–4 m

( )( )7

1 4

1

sin

1 6.75 10 msin

1.80 10 m0.21

nnwλθ

θ

θ

−

−

=

×=

×= °

The angle that locates the first dark fringe is 0.21°. (b) w = 1.80 10–6 m

( )( )7

1 6

1

sin

1 6.75 10 msin

1.80 10 m22

nnwλθ

θ

θ

−

−

=

×=

×= °

The angle that locates the first dark fringe is 22°.


22. λ = 638 nm = 6.38 10–7 m w = 4.40 10–4 m L = 1.45 m y1 = ?

( )( )

1

1

7

4

31

(minima)

1.45 m 6.38 10 m

4.40 10 m2.10 10 m

yL w

Lyw

y

λ

λ

−

−

−

=

=

×=

×= ×

( )1

3

3

central maxima 2

2 2.10 10 m

central maxima 4.20 10 m

y−

−

=

= ×

= ×

The width of the central fringe is 4.20 10–3 m. 23. (a) λ = 589 nm = 5.89 10–7 m

w = 1.08 10–6 m n = 1 θ = ?

( )( )7

6

sin (minima)

1 5.89 10 m

1.08 10 m33

nnwλθ

θ

−

−

=

×=

×= °

The first minimum is at an angle of 33°.

(b) Since 22sin 1.09wλθ = = , sin θ > 1, and θ is undefined. Therefore, there is no second minimum.

24. λ = 1.15 10–7 m θ = 8.4° w = ?

( )( )7

6 6

sin (minima)

sin

8 1.15 10 m

sin 8.46.29 10 m, or 6.3 10 m

n

n

nwnw

w

λθ

λθ

−

− −

=

=

×=

°= × ×

The width of the slit is 6.3 10–6 m. 25. λ = 451 nm = 4.51 10–7 m

w = 0.10 mm = 1.0 10–4 m L = 3.50 m y1 = ?

( )( )

1

7

4

21

4.5 10 m 3.50 m

1.0 10 m1.575 10 m

Lyw

y

λ

−

−

−

=

×=

×= ×


( )1

2

2 2

central maximum 2

2 1.575 10 m

central maximum 3.15 10 m, or 3.2 10 m

y−

− −

=

= ×

= × ×

The width of the central maximum is 3.2 10–7 m, or 32 mm. 26. λ = 639 nm = 6.39 10–7 m

w = 4.2 10–4 m L = 3.50 m ∆y = ?

( )( )7

4

3

3.50 m 6.39 10 m

4.2 10 m5.32 10 m

Lyw

y

λ

−

−

−

∆ =

×=

×∆ = ×

The maxima are 5.32 10–3 m, or 5.3 mm far apart.

27. ?wλ =

Since 12y L= , therefore:

1

1

1212

yL w

yy w

w

λ

λ

λ

=

=

=

The ratio wλ is 1

2, or 0.5.

28. Spacing between fringes is much greater. Bright fringes are much brighter and sharper. 29. Immersed in water, the wavelength of the light will decrease.

sin

sin

m

m

mdλθ

θ λ

=

∝

Since θ will decrease, the smaller the wavelength, the smaller the distances between the bright lines. 30. d = 2.2 10–6 m

λ1 = 412 nm = 4.12 10–7 m λ2 = 661 nm = 6.61 10–7 m L = 3.10 m ∆y = ?

Since sin mmdλθ = and sin m

yL

θ = :

1 21 2 and

mL mLy yd d

λ λ= =

( )

( )( )( )2 1 2 1

7 7

6

12 1

1 3.10 m 6.61 10 m 4.12 10 m

2.2 10 m3.5 10 m

mLy yd

y y

λ λ

− −

−

−

− = −

× − ×=

×− = ×

The first order spectrum is 3.5 10–1 m, or 35 cm wide.


31. diffraction grating = 4 61 cm 2.00 10 cm 2.00 10 m5000 line

− −= × = ×

m = 2 θ = 35.0° λ = ?

( )( )6

7

sin

sin

2.00 10 m sin 35.0

25.74 10 m

m

m

md

dm

λθ

θλ

λ

−

−

=

=

× °=

= ×


32. (a) d = 5 71 cm 3.33 10 cm 3.33 10 m30 000 line

− −= × = ×

The largest wavelength of visible light would be red at approximately 710 nm, and the shortest would be violet at approximately 410 nm.

sin mmdλθ =

However, the largest value would be for n = 1, therefore:

( )

7

1sin

3.33 10 mm

λθ −=

×

Whether you substitute for red or violet light, the value for sin θm > 1, therefore no wavelength of visible light can be diffracted.

(b) Since sin θ ≤ 1, 1mdλ ≤ . For the first nodal to show:

7

1

3.33 10 m

md

d

λ

λ −

=

= = ×

Therefore, the longest wavelength that will show is 3.33 10–7 m.

33. d = 4 61 cm 1.18 10 cm 1.18 10 m8500 line

− −= × = ×

θ1 = 26.6°, 26.8° θ2 = 41.1°, 41.3° λ1 = ? λ2 = ?

1

2

26.6 26.8average 26.72

41.1 41.3average 41.22

θ

θ

° + °= = °

° + °= = °

To calculate wavelength:

( )

11

6

71

sin

1.18 10 m sin 26.7

15.30 10 m

dm

θλ

λ

−

−

=

× °=

= ×

( )

22

6

72

sin

1.18 10 m sin 41.2

17.77 10 m

dm

θλ

λ

−

−

=

× °=

= ×

The wavelengths are 5.30 10–7 m, or 530 nm, and 7.77 10–7 m, or 777 nm.


34. λblue = 482 nm noil = 1.40 t = ?

First we must calculate the wavelength of the light in oil:

airoil

oil

482 nm1.40

362 nm

nλλ

λ

=

=

=

For bright reflection, the path difference must be 2λ .

4362 nm

490.6 nm

t

t

λ=

=

=

The minimum thickness of the oil slick is 90.6 nm. 35. Let the subscript c represent the coating, g represent the glass, and a represent the air.

nc = 1.38 ng = 1.52 λa = 565 nm t = ?

First we must determine the wavelength of the light in the coating:

ac

c

565 nm1.38

409 nm

nλλ

λ

=

=

=

The minimum path difference for cancellation is 2λ .

4409 nm

4102 nm

t

t

λ=

=

=

The minimum thickness of the coating is 102 nm. 36. Let the subscript c represent the coating, g represent the glass, and a represent the air.

nc = 1.61 ng = 1.52 λa = 589 nm t = ?


Since the coating appears black there is no reflection (i.e., there is cancellation). Therefore, the thickness is 0, 2λ , and λ.

First we must determine the wavelength of the light in the coating:

ac

c

589 nm1.61

366 nm

nλλ

λ

=

=

=

Therefore, the two smallest non-zero thicknesses are 183 nm, and 366 nm. 37. Let the subscript f represent the film, g represent the glass, and a represent air.

nf = 1.36 ng = 1.52 λa = 525 nm t = ? First we must calculate the wavelength of the light in film:

af

7

7f

5.25 10 m1.36

3.86 10 m

nλλ

λ

−

−

=

×=

= ×

To calculate the thickness:

7

7

23.86 10 m

21.93 10 m

t

t

λ

−

−

=

×=

= ×

The thickness is 1.93 10–7 m. 38. Let the subscript s represent soap, and a represent air.

t = 112 nm θ = 90° λ = ?

The reflected bright path difference is 2λ , therefore:

( )

s

s

444 112 nm448 nm

t

t

λ

λ

λ

=

===

This is the wavelength in the soap. To calculate the wavelength in air:

( )( )

a

s

a s

a

1.33 448 nm596 nm

n

n

λλ

λ λ

λ

=

===

The wavelength of the reflected light is 596 nm, and the colour is yellowish-orange.


39. (a) The film is thin enough that there is destructive interference. The two rays are out-of-phase, as illustrated.

(b) The pattern will change as the soap moves under gravity creating a larger wedge of soap. The top will be a larger,

darker area and the distance between the areas of dark (or bright) will come closer together as the bottom thickens. Eventually, the bubble breaks since the surface tension is not sufficient to hold it in place.

(c) The path differences for bright reflection are 3 5, ,2 2 2λ λ λ , etc. The thickness will be 3,

4 4λ λ , etc. Therefore, the

difference in thickness between adjacent bands would be 2λ .

(d) λa = 588 nm ts= ? First we must calculate the wavelength of the light in the soap film:

a

s

as

s

588 nm1.33

442 nm

n

n

λλλλ

λ

=

=

=

=

The lowest dark band is the second fringe, so the path difference would be 2λ, and the thickness would be λ. Therefore, the thickness is 442 nm.

40. λ = 639 nm = 6.39 10–7 m m = 38 t = ?

Thirty-eight dark fringes with one at each end is equal to 37 spaces. Each dark fringe represents a change in thickness of

2λ , therefore:

7

5

2

6.39 10 m372

1.18 10 m

t m

t

λ

−

−

= ×=

= ×

The thickness of the foil is 1.18 10–5 m.


41. t = 7.62 10–5 m λ = 539 nm = 5.39 10–7 m m = ?

( )7

7

22

2 7.62 10 m

5.39 10 m283

t m

tm

m

λ

λ−

−

=

=

×=

×=

There are 283 bright fringes across the wedge. 42. L = 15.8 cm = 1.58 10–1 m

λ = 548 nm = 5.48 10–7 m ∆x = 1.3 mm = 1.3 10–3 m t = ?

( )( )

( )1 7

3

5

2

21.58 10 5.48 10 m

2 1.3 10 m

3.3 10 m

x Lt

Ltx

t

λ

λ

− −

−

−

∆ =

=∆

× ×=

×

= ×

The thickness of the paper strip is 3.3 10–5 m. 43. λ = 589 nm = 5.89 10–7 m number of fringes = 2000

For each fringe the path difference is λ and the distance the mirror moves is 2λ .

The distance moved is:

( )4

2000 1000 589 nm2

5.89 10 m

λ

−

= = ×

The distance the mirror must be moved is 5.89 10–4 m 44. (a) “Coherent” means same wavelength and same phase. The distance between each slit and the screen is nearly equal, but

the mirror changes the phase 180°. Therefore, the image on the screen will not be coherent with the source. (b) It is a combination of two single slit diffractions overlapping. All we see is a double slit interference, which is the

central maximum of the single slit diffraction patterns. (c) It will be dark, since the sources are essentially 180° out-of-phase. 45.

Type Nature of Source Typical Means of Detection

Nonionizing or Ionizing

radio oscillating charge m radio receiver no infrared electron transitions in

atoms and molecules temperature measurement

no

ultraviolet higher energy electron transitions in atoms

fluorescence no

X ray rapid deceleration of charges

photo sensitive device yes

46. w = 20.0 cm = 2.00 10–1 m θ1 = ?


( )( )1

1

sin

sin

2.0 10 m sin 36

11.17 10 m

n

n

nww

n

λθ

θλ

λ

−

−

=

=

× °=

= ×

The wavelength of the microwaves is 1.17 10–1 m, or 11.7 cm.

Applying Inquiry Skills 47. In both cases, Figures 5(a) and (b) represent a single slit interference pattern. If the apparatus is the same, the wavelength

of the light will be constant. From the single slit relationship, 1sin or, if is constant sin nnw wλθ λ θ= ∝ . Since θ is

smaller in the 5 (b) that in 5(a), then it follows that the slit width must have increased. Also, note that intensity of the central maximum is higher in (b) than in (a), additional evidence that the width has increased. Since more light passes through the slit, the intensity is higher.

48. f = 10.5 GHz = 10.5 109 Hz A = L = 50 cm = 0.5 m B = y1 = ? C = θ = ?


8

9

2

3.00 10 m/s10.5 10 Hz2.8 10 m

cf

λ

λ −

=

×=×

= ×

To calculate the value of C:

1

2

1 2

1

sin

12

1 2.8 10 msin 12 2.0 10 m

sin 7.00 1044

nyL

nd

θ

λ

θ

θθ

−

−

−

=

= − × = − ×

= ×= °

To calculate the value of B:

( )

1

1

1

sin

sin0.50 m sin 44

0.35 m

yLy L

y

θ

θ

=

== °=

Therefore, the value of C is 44°, and the value of B is 0.35 m, or 35 cm.

Making Connections 49. No, in fact it could be worse depending on the path difference between the two signals. Only when the path difference is

zero will you get constructive interference. At any other path difference there will be some destructive interference with a

maximum with a path difference 2λ . (Assuming no reflected radiations.)


50.

The path of the reflected wave is twice the hypotenuse of the triangle, therefore:

( ) ( )2 222 2.50 10 m 60.0 m 514 m× + =

Therefore, the path difference is 514 m – 500 m = 14 m. Since this is a phase inversion at Earth, a wavelength of 14 m will interfere destructively. Therefore, the largest possible wavelength to interfere constructively will be 13 m.

51. f1 = 3.0 104 Hz f2 = 4.5 104 Hz (a) λ1 = ? λ2 = ?

1

1

8

4

41

3.00 10 m/s3.0 10 Hz

1.0 10 m

c fcf

λ

λ

λ

=

=

×=×

= ×

22

8

4

32

3.00 10 m/s4.5 10 Hz

6.7 10 m

cf

λ

λ

=

×=×

= ×

The typical wavelengths would be 1.0 104 m, and 6.7 103 m. (b) v = 3.4 102 m/s λ1 = ? λ2 = ?

1

2

4

2 21

3.4 10 m/s3.0 10 Hz

1.13 10 m, or 1.1 10 m

v fvf

λ

λ

λ − −

=

=

×=×

= × ×

2

2

4

3 32

3.4 10 m/s4.5 10 Hz7.56 10 m, or 7.6 10 m

vf

λ

λ − −

=

×=×

= × ×

The actual wavelengths would be 1.1 10–2 m, and 7.6 10–3 m.

Extension 52. f = 14.0 kHz = 1.40 104 Hz

v = 1.40 102 m/s w = ?


2

4

2

1.40 10 m1.40 10 Hz1.00 10 m

vf

λ

λ −

=

×=×

= ×


For n = 1, and θ1 = 30.0°, to calculate the width of the slit for single slit maxima:

( )2

2

1sin2

11 1.00 10 m2sin 30.0

3.00 10 m

m mw

ww

λθ

−

−

= + + × ° =

= ×

The width of the slit is 3.00 10–2 m, or 3.00 cm. 53. The white light range is 400 nm to 750 nm (red).

λR = 7.50 10–7 m number of lines = ? sin θm ≤ 1

( )7

8

sin

2 7.50 10 m

1.50 10 m

mmd

d m

d

λθ

λ−

−

=

=

= ×

= ×

Since d is equal to 1.50 10–8 m, or 1.50 10–6 cm:

6

3

1number of lines1.50 10 cm

number of lines 6.67 10 lines/cm

−

−

=×

= ×

The maximum possible number of lines per centimetre is 6.67 103. 54. Let the subscript a represent air, and s represent soap.

n = 1.33

d = 6500 line 2.00 10 mmm

−= ×

m = 1 θ = 18° t = ?

First we must calculate the wavelength in air:

( )

a

a

6

7a

sin

sin

sin18 2.00 10 m

16.18 10 m

m

m

md

dm

λθ

θλ

λ

−

−

=

=

° ×=

= ×

Next we must calculate the wavelength in soap:

as

7

7s

6.18 10 m1.33

4.65 10 m

nλλ

λ

−

−

=

×=

= ×


The path difference is 2λ and the thickness is

4λ , therefore:

7

7

44.65 10 m

41.2 10 m

t

t

λ

−

−

=

×=

= ×

The minimum possible thickness of the soap film is 1.2 10–7 m. 55. number of fringes = 31

λ = 589 nm = 5.89 10–7 m t = ?

A dark fringe occurs at the edges where t = 0. The next dark fringe occurs when the path difference is λ, and the thickness

is 2λ , therefore:

( )( )5

4

number of fringes2

31 5.89 10 m

29.1 10 m

t

t

λ

−

−

= ×

=

= ×

The centre is 9.1 10–4 m thicker than the edges.

physics 12 unit 4 solns

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