CHAPTER 1

PHYSICAL OPTICS: INTERFERENCE

• Introduction

• Waves

• Principle of superposition• Wave packets• Phasors• Interference• Reflection of waves

• Young’s double-slit experiment

• Interference in thin films and air gaps

What is “physical optics”?

The methods of physical optics are used when the wavelength of light and dimensions of the system are of a comparable order of magnitude, when the simple ray approximation of geometric optics is not valid. So, it is intermediate between geometric optics, which ignores wave effects, and full wave electromagnetism, which is a precise theory.

In General Physics II you studied some aspects of geometrical optics. Geometrical optics rests on the assumption that light propagates along straight lines and is reflected and refracted according to definite laws, such as Fermat’s principle and Snell’s Law. As a result the positions of images in mirrors and through lenses, etc. can be determined by scaled drawings. For example, the production of an image in a concave mirror.

•

image

object

1

2

C F

s

! s

y

! y •

Or the use of a convex lens as a magnifying lens:

f f

Object

Image

s ! s

But many optical phenomena cannot be adequately explained by geometrical optics. For example, the iridescence that makes the colors of a hummingbird so brilliant are not due to pigment but to an interference effect caused by structures in the feathers.

The colors you see in a soap bubble are also due to an interference effect between light rays reflected from the front and back surfaces of the thin film of soap making the bubble. The color depends on the thickness of film, ranging from black, where the film is thinnest, to magenta, where the film is thickest.

Likewise, the colors you see in a thin film of oil floating on water are due to an interference effect between light rays reflected from the front and back surfaces of the oil film.

Another example is the “spectrum of colors” you see when you reflect light from the active side of a CD. The effect that produces these colors is closely related to interference; it is called diffraction.

Here is an example of diffraction caused by the edges of the razor blade when viewed from behind with monochromatic blue light.

The pattern of “fringes” is easily observable with monochromatic light. With white light, fringes due to the different wavelengths overlap making them more difficult to observe.

Diffraction and interference cannot be expained by geometrical optics; instead, light has to be treated as waves.

Waves

In physics,we come across three main types of waves:

Mechanical waves: water waves, sound waves, seismic waves, waves on a string ...

Electromagnetic waves: visible and ultraviolet light, radio and television waves, microwaves, x-rays and radar waves. All e-m waves travel in vacuum with a speed of

c = 2.99792458 !108 m s.

Matter waves: the waves associated with electrons, protons and other fundamental particles, atoms and molecules.

However, all waves have features in common.

Vertical and horizontal illustrations of the wave

To describe the “disturbance” of such a wave we need two variables, t and x.

x

Take the ripples (waves) in a pond caused by a small stone being dropped into the water ...

! A

"A

x

y

Shown is a snapshot of the wave at some time t. It is described by the general equation:

y(t,x) = A cos(!t " kx)At a position, x! say, the disturbance y(t) varies

sinusoidally with time (i.e., simple harmonic motion). Similarly, at a time t, as shown here, the disturbance varies sinusoidally with position, x. The parameter

k = 2#$ is called the wavevector, where $ is the

wavelength. The product kx is called the phase angle (= %). A negative (positive) sign indicates the wave is traveling to the right (left).

Also, ! is the angular frequency of the wave given by

! = 2#f , and T = 1f is the periodic time of the wave.

$ A

"A

x

y

x!

Note that the disturbance at some fixed time t!,

y(t! ,x) = y(t! ,x ± n!), i.e., the wave is reproduced at displacements of n! , where n is an integer.

Similarly, the disturbance at some fixed position x!,

y(t,x!) = y(t ± T,x!) = y(t ± m2"# ,x!),

i.e., the wave is reproduced after time intervals of

m2"

# , where m is an integer.

T = 1f = 2"

#

! A

$A

x

y

T A

$A

t

y

To find the speed of a wave, we take two snapshots at a time interval !t apart. If the wave (i.e., the red dot) travels a distance !x in that time, then the speed of the wave is

v = !x!t .

Since the disturbances (y) are equal

y(t,x) = y(t + !t, x + !x),so "t # kx = "(t + !t ) # k(x + !x),

i.e., "!t = k!x,

$v = "k.

Note also v = "

k = 2%f2% & = f& .

The velocity of a fixed point on a wave (such as the red dot) is called the phase velocity.

x

y !x

wave at t wave at t + !t

Principle of superposition

“If two or more waves are traveling through a medium, the resultant disturbance at any point is the algebraic sum of the individual disturbances.”

Waves that obey this principle are called linear waves. One consequence is that two waves can “pass” through each other!

(Waves that do not obey this principle are called non-linear waves.)

y1 y2

y1 + y2

y1 + y2

y1 y2

y1 + y2

y1 y2

y1

y2

y1 = 6sin x

y2 = 5sin 2x

y3 = 4sin 3x

y4 = 3sin 4x

y5 = 2sin 5x

y6 = sin 6x

y = y1 + y2 + y3 + y4 + y5 + y6

So, the superposition of several waves of differing wavelengths and amplitudes produces complex waveforms. For example, to produce a square wave ...

WBX06VD1.MOV

A square wave can be expressed as a so-called Fourier series:

f (x) =

1nn=1,3,5…

!" sin n#

Lx

$ % &

' ( ) ,

where L = *2, i.e., one-half of the wavelength.

A Fourier series decomposes a periodic function into a sum of simple oscillating functions, i.e., sines and/or cosines.

Consider the superposition of two waves of equal amplitude but slightly different frequencies and wavelengths. Then, the resultant is

y(x, t) = Asin !1t " k1x( ) + Asin !2t " k2x( )

= 2Acos

#!2

t "#k2

x$ % &

' ( ) sin ! t " k x( ),

where #! = !1 " !2, #k = k1 " k2, ! = !1 + !2( )2

and k = k1 + k2( )2 . When plotted, at some time t, we

get

i.e., the waves are separated into “groups”.

y(x)

y(x)

x

x

A

2A

y(x)

x

2A

x1 x2

!x

y = (x,t) = 2Acos

!"2

t #!k2

x$ % &

' ( ) sin " t # k x( )

The first term is the envelope, i.e., the green curve. The second term is the wave within the envelope. Both the envelope and wave within the envelope are traveling waves.

The envelope moves with velocity vg = !"!k, called

the group velocity. The wave inside the envelope moves

with velocity vp = " k , called the phase velocity.

The amplitude of the envelope is zero when

!"2

t #!k2

xn$ % &

' ( ) = 2n +1( )*

2,

where n = 0, ±1, ±2 !.

y(x)

x

2A

x1 x2

!x

So, successive minima (at time t) occur at x2 and x1 where

!"2

t #!k2

x2$ % &

' ( ) #

!"2

t #!k2

x1$ % &

' ( ) = * ,

i.e, x2 # x1 = !x =

2*!k

.

Thus, the spatial extent of the group is !x + !k#1. If we

plot the resultant as a function of t at a fixed point, we get

In the case of sound waves, this waveform produces the phenomenon of “beats”.

y(t)

t

2A

y(t)

t

2A

t2 t1

!t

Successive minima (at point x) occur at t2 and t1 where

!"2

t2 #!k2

x$ % &

' ( ) #

!"2

t1 #!k2

x$ % &

' ( ) = * ,

i.e, t2 # t1 = !t =

2*!"

.

Hence, the temporal extent !t + !"#1.

So, we find that

!x.!k , constant 2*( )and

!t.!" , constant 2*( ).

The phase velocity of the individual harmonic waves is

vp = "k ," = vpk.

The group velocity (the velocity of the envelope) is

vg = d!

dk[ ]k =

d kvp( )dk

"

# $ $

%

& ' ' k

= vp + kdvp

dk" # $

% & '

k .

If the phase velocity is the same at all frequencies and wavelengths, i.e., there is no dispersion then

dvp

dk = 0 i.e., vg = vp.

Case A.mpg

A medium in which dvp

dk = 0 is said to be non-

dispersive. (An example is an electromagnetic waves in vacuum.) Glass, for instance, is a dispersive medium. Shown here is a plot of the phase velocity in flint glass.

In this case, dvp

dk < 0, so vp > vg.

If dvp

dk > 0, then vg > vp.

1.95

1.96

1.97

1.0 1.2 1.4 1.6

1.98 vp !108 m/s( )

k !107 m"1( )#

0

dvpdk

# "39.3 m2 s

Red

Violet

Wave Packets

A “wave packet” can be created by superposing many

waves spanning a wavevector range k! ±!k

2 .

To generate the wave packet shown above, we put

y = a(kn )cos knx( )

n="25

25#

where kn = k! + n!k50( ) and the amplitudes a(kn ) are

a Gaussian distribution, i.e.,

a(kn ) = e"n2 $2.

We set k! = 5, !k = 3 and $ = 10. Thus, there are 51

equally spaced values of kn in the sum.

x

y

kn k!

k! +!k

2 k! "!k

2

a(kn)

!k

In constructing a wave packet in this way we find the interesting result that the spatial extent or “length” of the wave packet, !x, is inversely proportional to !k, viz:

Although it is difficult to locate exactly the start and end points of the wave packets, it is clear that

!x " !k # constant,a result we found earlier. The actual value of the

constant depends on the amplitude function a(kn ) and

the precise definitions of !x and !k.

!x

!x

!x

!x

!k = 2

!k = 3

!k = 4

!k = 1 x

x

x

x

!k !x !x " !k

1 $ 45±1 $ 45

2 $ 22 ±1 $ 44

3 $ 15±1 $ 45

4 $ 11±1 $ 44

arbitrary units

For example, if we use the full-width at half maximum (FWHM) for !x and !k we find that !x " !k # 5.54.

The important point is that there is a relationship between !x and !k, i.e.,

the range of k values and the width of the packet. We will see the significance of this result in chapter 6.

k k!

a(k)

!k

x

!x

!k = 0.333

x

!x

!k = 0.666

x

!x

!k = 1.00

x

!x

!k = 1.33

!k !x !x " !k

0.333 16.65 5.54

0.666 8.33 5.55

1.00 5.52 5.52

1.33 4.16 5.53

arbitrary units

Again, if we had we summed originally over a frequency range, !f , instead of !k, we would have found similar condition also occurs between the frequency range and the temporal extent, !t ,

i.e., !f " !t # !$ " !t # constant.

Furthermore, a wave packet is composed of many individual component waves of differing wavevector (hence wavelength and frequency). So, a wave packet traveling through a dispersive medium, i.e., one in which the phase velocity depends on frequency (and wavevector) - such as glass, water, etc. - will broaden because the component waves travel with different velocities.

x

vg

Question 1:

For the wave packet shown, (a) estimate the range in wavevectors !k. (b) If !t is the time it takes the wave packet to pass a point in space, what is the range of frequencies !f in the wave packet?

!x

(a) First, we determine how many oscillations there are in the wave packet; but it is difficult to say definitively where the waves start and stop. We find there are between 12 and 13 oscillations, so, let us take N = 12 but with an uncertainty !N = 1.

"# = !xN = 16

12 = 1.33 units.

But k =

2$#

=2$N!x

.

"!k =

2$!x

!N =2$16

%1 = 0.393 inverse units.

Note that !x & !k = 2$ , i.e., a constant whose value depends on the uncertainty !N.

!x

(b) The frequency is given by the number of oscillations passing the point in time !t , i.e.,

f = N!t .

"!f = !N!t =

1!t .

But # = 2$f , i.e., !# = 2$!f ,

"!#

2$ = 1!t ,

i.e., !# % !t = 2$ (a constant).

Earlier we found that !x % !k = 2$ = !#.!t( )

"!#

!k = !x!t ,

which the speed of the wave packet; it is called the

group velocity ( vg). We note that

!x % !k = 2$ and !# % !t = 2$ represent the minimum values, with !N = 1, i.e., the uncertainty in the number of oscillations.

A monchromatic source is defined as one with a constant wavelength (or frequency). However, if a source produces a single wave train of finite length, i.e., for a finite length of time, the disturbance cannot be monchromatic.

As the wave train shown above is not a simple harmonic wave - it has a start and an end - and it is non-repeating, it must be represented as an infinite sum of harmonic waves of a range of different frequencies, centered on the principal frequency. Thus, strictly speaking, unless the wave train produced by a source is infinitely long, it cannot be truly monchromatic.

However, in what follows, we will ignore such effects.

y

!t

t

Phasors

Returning to single frequency (i.e., monochromatic) waves. Suppose we add two sine waves. What is the resultant?

Consider two separated sources, S1 and S2, with

identical frequency, wavelength and amplitude. Such sources are described as

coherent. Then at P we have:

y1 = A cos(!t " kr1) and y2 = A cos(!t " kr2)Thus, the resultant disturbance at P is:

y1 + y2 = Acos(!t " kr1) + Acos(!t " kr2)

How can we work this out ?? ... remember phasors ??

You might think that we could use trigonometric identities but, in general, that is not the best approach.

S1 S2

r2 r1

P

A “phasor” is a rotating vector. Phasors can be added just like ordinary vectors.

y(t) = A cos!t = Acos"(t)

A

y(t)"

y(t)

t

http://www.animations.physics.unsw.edu.au/jw/phasor-addition.html

Let’s solve this problem using phasors ...

The length of the resultant phasor is R = 2Acos ! .

But, 2! + (" + # $%) = "& ! = $ (# $%)2 .

'R = 2Acos $ (# $%)

2( ) * +

, - .

But cos. is an even function, i.e., cos! = cos($! ).

'R = 2Acos (# $ %)

2( ) * +

, - .

Thus, the projection of this phasor onto the x-axis is

y1 + y2 = Rcos(# + ! ),

= 2Acos (# $ %)

2( ) * +

, - cos # + !( ).

amplitude

Put y1 = A cos(/t $ kr1) = Acos#and y2 = A cos(/t $ kr2 ) = Acos% i.e., # = /t $ kr1 and % = /t $ kr2.

#

%

(" + # $%)

!

!

R

A

A

y1 y2

Substituting for !, " and # we get:

y1 + y2 = 2Acos

$2 % $12

& ' (

) * + cos ,t %

$2 + $12

& ' (

) * + ,

where ($2 % $1) = (kr2 % kr1) = - , i.e., the phase difference

between the two waves.

We can also get the result trignometrically, viz:

y1 + y2 = Acos(,t % kr1) + Acos(,t % kr2)

= A cos(,t % $1) + Acos(,t % $2).Using the trignometric relationship:

cosX + cosY = 2cos

X % Y2

& ' (

) * + cos

X + Y2

& ' (

) * + ,

we find, as before, (see top of page),

y1 + y2 = 2Acos

$2 % $12

& ' (

) * + cos ,t %

$2 + $12

& ' (

) * + .

Although this seems like much less work, it is only

applicable if the amplitudes of y1 and y2 are equal. The

method using phasors is more generally applicable ... see the next example ...

Question 2:

Find the resultant of the two waves whose disturbances at a given location vary with time as follows:

y1 = 4cos!t and y2 = 3cos !t + "3( ).

Draw the two waves as phasors.

The length of the resultant phasor is given by

R2 = 4 + 3cos !3( )( )2

+ 3sin !3( )( )2

= 37,

"R = 6.08.The resultant wave (y1 + y2) is given by the projection

of R onto the horizontal axis, i.e.,

Rcos(#t + $),

where $ = tan%1 2.60

(4 + 1.5)& ' (

) * + = 25.3!, 0.44 rad.

"y1 + y2 = 6.08cos(#t + 0.44 rad).

4

3

#t

R 3sin

!3( ) = 2.60

$

!

3

3cos!

3( ) = 1.50

InterferenceEarlier, we found the resultant at P for two identical sources S1 and S2 is:

y1 + y2 = 2Acos

!2 " !12

# $ %

& ' ( cos )t "

!2 + !12

# $ %

& ' ( ,

where (!2 " !1) = k(r2 " r1) = k*r =

2+,

*r. Here,

• (r2 " r1) = *r is called the path difference, and• (!2 " !1) = - is the phase difference

between the two waves. Note, the resultant disturbance

(y1 + y2) is a maximum when

(!2 " !1)

2= n+ =

-2

=k2*r =

+,*r ,

i.e., when *r = n, or - = 2n+.Thus, when the path difference is an integral number (n) of wavelengths or the phase difference is an even number of + , (y1 + y2) is a maximum. This condition is known as constructive interference.

S1 S2

r2 r1

P

EVEN

S1 S2

r2 r1

P

The resultant at P is:

y1 + y2 = 2Acos

!2 " !12

# $ %

& ' ( cos )t "

!2 + !12

# $ %

& ' ( .

But, if (!2 " !1)

2=*2,

3*2,

5*2, etc =

(2n + 1)2

*,

the resultant disturbance (y1 + y2) = 0. That occurs

when

(!2 " !1)

2=+2

=k2,r =

*-,r =

(2n + 1)2

*

i.e., ,r = n +

12

# $ %

& ' ( - or + = (2n + 1)*

So, when the path difference is a half-integer number of wavelengths or the phase difference is an odd number of * , (y1 + y2) = 0. This is destructive interference.

ODD

Question 3:

If a transparent sheet of plastic of thickness 1.00mm and refractive index µ = 1.25 is placed in the path of a ray,

what is the optical path difference ( !r) it introduces?

1.00mm

Let the overall distance be L, the thickness of the plastic be t and the wavelength of the light be ! . Then, the number of wavelengths in case (a) is

Na =

L!

,

and the number of wavelengths in case (b) is

Nb =

L " t!

+t# !

=L " t!

+µt!

=L + µ "1( )t

!.

So, Nb > Na , and the optical path difference is

$r = Nb " Na( )! =

L + µ "1( )t!

"L!

% & '

( ) * ! = µ "1( )t .

Using t = 1.00 +10"3m and µ = 1.25, we get

$r = 0.25+1.00 +10"3m = 0.25mm.

t

L

(a)

(b)

The general wave equation

Earlier we wrote:

y(t,x) = A cos(!t " kx)to describe the wave. However, this is not the only expression we could have chosen. For example,

y(t,x) = Bsin(!t " kx) and y(t,x) = Ce"i(!t"kx)

are also waves. They may look different but they are all solutions of the “general wave equation”

#2y#x2 =

1v2

#2y#t2 .

Example: if y(t,x) = A cos(!t " kx),

#y

#x = kA sin(!t " kx) and #2y

#x2 = "k2Acos(!t " kx),

#y

#t = "!A sin(!t " kx) and #2y

#t2 = "!2Acos(!t " kx).

#2y#x2 = k

!$ % &

' ( )

2 #2y#t2 , i.e.,

#2y#x2 = 1

v2#2y#t2 ,

where v is the phase velocity of the wave.

http://www.walter-fendt.de/ph14e/stwaverefl.htm

Standing waves

Consider two waves with the same amplitude traveling in opposite directions along the x-axis,

i.e., yR = a cos(!t " kx) and yL = a cos(!t + kx).Then applying the superposition principle, the resultant is

yR + yL = a cos(!t " kx) + a cos(!t + kx)Using the trignometric relationship:

cosA + cosB = 2cos

A + B2

# $ %

& ' ( cos

A " B2

# $ %

& ' ( ,

we get

y(x, t) = yR + yL = 2a cos(!t)cos "kx( )

= 2a cos(!t)cos kx( ).

This is a standing wave; because the crests and troughs “stand in-place” but the amplitude varies sinusiodally with a maximum of 2a!

See animation ...

Preliminary observations prior to discussing interference and diffraction effects.

Look at what happens when plane water waves, traveling from the left, encounter a barrier with a small opening. The waves

on the right are circular and centered on the opening just as if there was a point source of waves at the opening.

If we vary the wavelength and the size of the opening we find that when ! << d, as in (a), the ray approximation is valid but when ! " d the waves spread out. The spreading effect in (b) and (c) is called diffraction.

(a) ! << d (b) ! " d (c) ! >> d

d

Two-slit interference pattern (Young, 1801)

If the path difference to P between the two rays is

!r = (r2 " r1) = n# , where n is an integer, we have constructive interference, i.e., maximum disturbance at the point P on the screen. However, if

!r = (r2 " r1) = n +

12

$ % &

' ( ) # ,

we have destructive interference, i.e., zero disturbance at the point P.

r1

r2

S2

S1

L

P

yn

**

d sin*

d

L >> d

Thus, we have interference maxima at angle !, when

"r = dsin !n = n# ,

and interference minima when

"r = dsin !n = n +

12

$ % &

' ( ) # .

The phase difference at point P is

* = k"r =

2+#

d sin !.

If L >> d, yn = L tan!n. For small angles, tan! , sin !,

so the condition for maximum disturbance at P is

yn = n

#Ld

$ % &

' ( ) . ( n = 0, ± 1, ... ).

r1

r2

S2

S1

L

P

yn

!!

d sin!

d

L >> d

For zero disturbance at P,

yn = n +

12

! " #

$ % & 'Ld

. ( n = 0, ± 1, ... ).

So, providing ( ) small, the positions of the maxima and minima are separated by a distance

*y =

12'Ld

,

i.e., they are equally spaced.

If light is used as the source, what is the intensity distribution across the screen?

With all e-m radiation (like light) the “disturbance” or “wave function” is the electric field vector, i.e.,

! E =! A " sin+t .

Let E1 be the electric field at P due to the waves from S1

and E2 be the electric field at P due to the waves from S2.

Since both electric fields result from the same single source that illuminates the two slits, they have the same frequency

and amplitude, and the ! E -vectors will be parallel. When

they reach P, they will have a phase difference ! = k"r. So, we can represent the individual wave functions at P as

E1 = A" sin#t and E2 = A" sin(#t + !).The resultant is

E = E1 + E2 = A" sin#t + A" sin(#t + !).Using the identity

sin$ + sin % = 2cos

($ &%)2

sin($ + %)

2,

we find

E = 2A" cos

!2.sin #t +

!2

' ( )

* + , .

Thus the amplitude of the wave is 2A" cos

!2

.

The intensity of e-m radiation is proportional to the square of the amplitude, i.e.,

I = 4I! cos

2 !2

,

where I! is the intensity of the light on the screen from

either slit separately and ! = k"r =

2#$

d sin %.

An alternating intensity pattern is observed on the screen, called an interference pattern. Maximum intensity occurs

where y = n

$Ld

& ' (

) * + , with n = 0, ± 1, ± 2, etc.

n = ,4 ,3 ,2 ,1 0 1 2 3 4

Intensity 4I!

Iav = 2I!

y

Question 4:

Using a conventional two-slit apparatus and light that has a 589nm wavelength, 28 bright fringes per centimeter are observed near the center of a screen 3.00m away. What is the slit separation?

From earlier, when ! is small, the spacing between bright

and dark fringes is "y =

12

#Ld

$ % &

' ( ) . Since bright and dark

fringes are equally spaced, the spacing between neighboring bright fringes is

"yn,n+1 =

#Ld

=1*10+2 m/cm28 fringes/cm

= 3.57 *10+4 m .

,d =

#L"yn,n+1

=(589 *10+9 m)(3 m)

3.57 *10+4 m= 4.95 *10+3m

= 4.95mm.

Class discussion problem: At the dark fringes there is zero light intensity and so no energy is arriving. When light waves interfere and produce an interference pattern what happens to the energy in the light waves?

At the dark fringes there is zero light intensity and so no energy is arriving. When light waves interfere and produce an interference pattern what happens to the energy in the light waves?

The energy is re-distributed non-uniformly. The energy in the dark regions is less than the average; the energy in the bright regions is above the average.

So, the energy “missing” in the dark fringes is “transferred” to the bright fringes.

Intensity 4I!

Iav = 2I!

y

Conditions for interference

To observe interference,

the sources must be coherent, i.e., they must maintain a constant phase difference with respect to each other.

Class discussion problem: Which of the following are coherent sources:

(a) two candles,(b) a point source and its image in a plane mirror,(c) two pinholes illuminated by the same source,(d) two headlights of a car,(e) two images of a point source reflecting from the

top and bottom surfaces of a glass block.

Class discussion problem: What would the interference pattern look like if the light was composed of two different wavelengths; say, red light with !R = 750nm

and blue light with !B = 430nm?

Class discussion problem: If you were to blow smoke into the region between the slits and the viewing screen, would the smoke show evidence of the interference?

r1

r2

S2

S1

P

Question 5:

Two radio antennas, separated by 300m, simultaneously broadcast identical coherent signals. A radio in a car

traveling north receives the signals. (a) If the car is at the position of the second maximum, what is the wavelength of the signals?

(b) How much farther north does the car have to travel to reach the next minimum?

1000m 300m

400m

Note: we cannot use the small angle approximation since d ! L ! y.

(a) From earlier the condition for a maximum is

"r = dsin # = n$ . At the n = 2 maximum,

# = tan%1 400 m 1000 m( ) = 21.8!.

&$ =

dsin #n

=(300 m) sin21.8!

2= 55.7 m.

(b) After the n = 2 maximum, the next minimum occurs when the path length is one-half wavelength greater, i.e., when

dsin # = n +

12

' ( )

* + , $ =

52$ .

&# = sin%1 5$

2d' ( )

* + , = sin%1 5 - 55.7 m

2 - 300 m' ( )

* + , = 27.7!.

So at the minimum: y = (1000 m) tan27.7! = 525 m.

Therefore, the car must travel an additional

"y = (525% 400) m = 125 m.

1000m 300m

400m

Reflection of waves

When a traveling wave or pulse encounters a change in the medium, part of the wave (or all of the wave, if it is a hard medium) is reflected. However, the reflected pulse or wave is inverted, i.e., it undergoes a phase change of

! ( 180!).

Incident pulse

Reflected pulse

http://www.kettering.edu/~drussell/Demos/reflect/reflect.html

Interference in thin films and air gaps

Consider a thin film (of water or oil) that is viewed at a small angle to the normal. Some of the light from the source is (directly) reflected from the top surface (#1). Most light enters the film and is refracted. Some of that

light is reflected from the lower surface and refracted at the upper surface (#2). At the upper surface, a change of phase of ! occurs on reflection. At the lower

surface, there is no change of phase on reflection. If the rays are nearly parallel, the path difference is "r = 2t so the phase difference is

# k "r = 2!# $ ( )2t ,

where # $ = $n is the wavelength of light in the medium

(of refractive index n).

#1 #2

t

Because of the phase change on reflection at the upper surface, the total phase difference between #1 and #2 is

! = 2"# $ ( )2t + ".

• if 2t = m # $ , i.e., a whole number of wavelengths, then

! = " + 2"m = (2m + 1)" (= ", 3", 5" …),and we get destructive interference.

• if 2t = (2m + 1) # $ 2, i.e., an odd number of half-

wavelengths, then

! = " + (2m + 1)" = 2(m + 1)" (= 2", 4", 6" …)and we get constructive interference.

Note: there is only a change in phase of " if the reflection occurs at an interface where the incident and reflected rays are in the less dense medium; a so-called hard reflection.

So, if a film of water is on a glass slide, say, rays reflected at the air-water interface and at the water-glass interface both undergo a phase change of ! , so the phase difference between these rays when they emerge in air is determined solely by the thickness of the water film, i.e., 2t.

Air wedge

Consider a wedge between two glass slides. Ray #1 is reflected from the upper glass- air interface, where

there is no change of phase. Ray #2 is reflected from the lower air-glass interface, where there is a change of phase of ! .

n1 = 1

n2 " 1.33

n3 " 1.5

1 2

x

t # = tx

Therefore, the total phase difference between rays #1 and #2 is

! = k"r + # = 2#$( )2t + # .

When 2t = m$ , then

! = 2#m+ # = (2m +1)# (= #, 3#, 5# …),i.e., destructive interference occurs.

But, since % = tx, that condition occurs when

m = 2t$ = 2x%

$ , i.e., x = m$2%.

So, dark fringes appear that are equally spaced in x. The linear density of dark fringes, i.e., the number per unit length is

m

x = 2%$ .

Shown alongside are the fringes produced by two very flat glass plates inclined at a very small angle.

Newton’s rings

They are caused by interference between light reflected from the bottom of the curved surface and the light reflected from an optically flat surface. If the curved surface is spherical the fringes are circular. Since the thickness of the air gap increases away from the point of contact, the fringe spacing decreases.

air gap

optical flat

Question 6:

Light, with wavelength 600nm, is used to illuminate two glass plates at normal incidence. The plates are 22.0cm in length. They touch at one end but are separated at the other end by a wire of radius 0.025mm. How many bright fringes appear along the length of the plates?

From the notes, dark fringes appear when 2t = m! .Therefore, if dark fringes appear when 2t = m! , bright fringes occur when

2t = m +

12

" # $

% & ' ! , ( m = 0, 1, 2, 3 …).

(m =

2t!)

12

.

The maximum value of m occurs when t = 2r. Then

m =

2(2r)!

)12

=4 * 0.025*10)3

600 *10)9 )12

= 166.2.

Hence, mmax = 166,

since m must be a whole number.

Note, the spacing of bright fringes is

Lmmax

=22.0 *10)2

166= 1.33*10)3m (1.33mm)

L

2r

Class discussion question:

Consider the air-wedge shown below, where the air gap increases linearly. Is the first band (fringe) at the right hand edge - where the gap is zero - bright or dark?

Class discussion question:

Consider the thin film we looked at previously. What difference, if any, would it make to the appearance of the fringes if white light was used instead of monochromatic light?

#1 #2

t