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EP 225.3 WAVES, FIELDS AND OPTICSFINAL EXAMINATION

April 13, 2009Solutions

1. A thin lens made of glass with ng = 1:5 has a focal length of 10 cm in air. If it is

placed in a liquid, it becomes a divergent lens with f = 50 cm. Find the index of refraction of the liquid.

Solution

In air,1f

= 110

= ( ng 1) 1R 1

1R 2

= 0 :5 1R 1

1R2

1R1

1R2

= 15

In liquid,1f l =

150 =

ng

n l 1 1R1

1R 2 =

15

ng

n l 1

n l == ng

0:9 = 1 :67

2. Rays from a lens converge at P: Determine the thickness of glass plate (ng = 1:5) tobe inserted so that the image is moved to P 0 (3 mm from P ):

P

P'

t

glass3 mm

Assuming small angles,(sin i sin r ) t = sin i

t = ng

ng 1 = 3 = 9 mm

1



Matrix

1 i0 1

1 0(1:5 1) 1

5 1:5 1 100 1

1 0 12 :5 1

1 100 1

1 0 11 :5 1 1

5 1=1:5 1 100 1

= 1: 6667 0:13333i 30:0 3:0i0:13333 3:0

From B = 0, the image location is i = 10 cm (at the mirror). Magni…cation is

m = A = 1:666 + 1:333 = 0:333 =13

4. A microscope consists of two convergent lenses, objective lens with f o = 2:0 cm andeyepiece with f e = 3:0 cm. The tube length is standard, L = 16 cm. The lenses areseparated by 21 cm.

(a) When the image is formed at in…nity, where is the object location from the objective lens?

(b) Find the magni…cation.

(c) When the image is formed at 25 cm, what is the new magni…cation? What changehas to be made to the object location?

Solution

(a) The object distance1o +

118 =

12 ! o = 2:25 cm

(b) The magni…cation is

m =25 162 3

= 66:7

(c) New magni…cation is

m 0 =25 16

213

+ 125

= 74:7

1i

125 =

13 ! i = 2:679 = 3:0 0:321

The new object location o0 can be found from

1o0

+ 1

18:321 =

12

o0 = 1

:5 1=18:321 = 2:2451

3



0 = 1

! s 2I cs v

= 12 103 r 2 10 7

1450 103

= 5 : 9 10 11 m

The RMS displacement is

rms = 1p 2 0 = 4:2 10 11 m

(b) The purpose is for improving impedance matching. Sound wave incident on wateris almost totally re‡ected. Elastic (bulk) modulus of gel is much smaller than thatof water and the gel impedance is in between the air and water.

7. Unpolarized light falls on a glass surface at the Brewster’s angle. The index of refractionof the glass is n g = 1 :65:

(a) Explain what is meant by unpolarized light wave.(b) What is the incident angle?(c) How much light power is re‡ected at the surface? The re‡ection coe¢cient for

the electric …eld tangential to the surface is

r = sin( 0

B )sin( 0 + B )

where 0 is the refraction angle.

Solution

(a) In unpolarized wave, the electric …eld is randomly oriented without preferreddirection. If the wave is propagating in z direction, the electric …eld consists of xand y components,

E x e x + E y e y

In unpolarized wave, the …elds E x and E y are uncorrelated but the intensities areequal, I x = I y : If the total intensity is I 0 ; I x = I y = 0:5I 0 :

(b) The Brewster’s angle is

B = tan 1 (n

g) = tan 1 (1:65) = 58:78

(c) The re‡ection coe¢cient of the electric …eld parallel to the boundary is

r = sin( 0

B )sin( 0 + B )

= sin(31:22 58:78 )

sin (90 )= 0:4627

5



Then power re‡ection is12

r 2 = 0 :107 = 10:7%

8. (a) In double slit arrangement, the slit separation distance d is 3 times the slit openinga; d = 3a: Sketch expected interference-di¤raction pattern as a function of theangle :

(b) (Unrelated to (a)). One of double slits is wider than the other by a factor of 1.5The slit separation is d: Show that the interference pattern is described by

I ( ) = I 014

+ 6 cos2 d sin

where I 0 is the peak intensity due to the narrower slit. The maximum intensityis 6:25I 0 : Is it reasonable?

Solution

(a) The interference/di¤raction pattern is

I 0 cos2 d sin sin2

where =

a sin

If d = 3a; the light intensity is proportional to

cos2 ( x )sin( x= 3)

x= 3

2

wherex =

d sin

-5 -4 -3 -2 -1 0 1 2 3 4 5

0.2

0.4

0.6

0.8

1.0

x

y

The third (m = 3) interference peaks are suppressed.

6



(b) The electric …elds due to the unequal slits are E 0 and 1:5E 0 : The phase di¤erence

is still given by = 2 d sin

: The total …eld is

E t = q E 20 + (1 :5E 0 )2 + 2 1:5E 20 cos

=

q 3:25E 20 + 3E 20 cos

and the light intensity is proportional to

I = I 0 (3:25 + 3 cos )= I 0 3:25 + 3 2cos2 ( = 2) 1= I 0 0:25 + 6 cos2 ( = 2)

= I 014

+ 6 cos2 d sin2

The peak intensity is (2:5)2 I 0 = 6:25I 0 as expected.

9. (a) Monochromatic light of = 600 nm falls normal on a nonuniform plastic …lmof index of refraction n = 1:4:When 10 stripes of constructive interference areobserved in re‡ected light, what is the di¤erence = b a between the thicknessat both edges?

a bn = 1.4

light

The condition for constructive interference is

2n f d = m + 12

Then2n f = 10 600 10 9

= 2:14 m

(b) (Unrelated to (a)) Lake with cleat water appears dark blue but lake with conta-minated water appears whitish. Explain in terms of Rayleigh and Mie scattering.Blue water color is due to Rayleigh scattering by water molecules much smallerthan wavelengths in visible range. The intensity of Rayleigh scattering is propor-tional to 1= 4 : Blue color is dominant.Whitish color is caused by Mie scattering by larger molecules. In this case, scat-tered power is insensitive to , all colors scattered evenly.

10. A nonrelativistic electron having an initial energy 10 keV undergoes cyclotron motionin a magnetic …eld of 1.5 T.

7



(a) What is the initial electron acceleration?(b) Show that the radiation power at time t is given by

P = e2 ! 2

c

3 " 0 mc3

12

mv 2 (t)

where ! c = eB

m is the electron cyclotron frequency and v (t) is the electron veloc-

ity.(c) Derive a di¤erential equation for the electron energy and …nd the time by which

99% of the initial electron energy is lost to radiation.

Solution

(a) The initial electron velocity is from12

mv 2 = 10 4 1:6 10 19 J

v = 5:93 107 m/sand the initial acceleration is

a = evB

m = 1 :56 1019 m/s 2

(b) The radiation power is

P = e2 a2

6 " 0 c3 = e2 ! 2

c

3 " 0 mc3

12

mv 2

(c) The di¤erential equation is

dE

dt = P =

e2 ! 2c

3 " 0 mc3 E = E

Solution isE (t) = E 0 e t

where E 0 is the initial energy (10 keV) and

= e2 ! 2

c

3 " 0 mc3

=(1:6 10 19 )

2 1 :6 10 19 1 :5

9 :1 10 31

2

3 8:85 10 12 9:1 10 31 (3 108 )3

= 0 :869=s

Time by which 99% energy is lost

e t = 0 :01t = ln(0:01) = 4:505

t = 4:505

0:869 = 5 :18 s.

8



Solution EP225 Final Exam April 2008

1. (a) What is the focal length of a convergent lens of f = 15 cm in air if it is placed inwater? Assume ng = 1:5 and nw = 1:33:

(b) Explain why goggles help viewing in water.

Solution

(a) The focal length in air is

1f a

= ( ng 1) 1R 1

1R 2

= 115

(1)

and the focal length in water is

1f w

=ng

nw1

1R 1

1R 2

From (1), 1R 1

1R 2

= 10:5

115

= 17:5

Then

1f w

= 1:51:33 1

1R1

1R2

= (1 :128 1) 17:5

f w = 58:68 cm

(b) The liquid in human eye lens has an index of refraction close to that of water andin water lens action of human eye is weakened (no focusing). Goggle restores airmedium in front of the eye as in usual viewing.

2. A microscope has an objective lens with f o = 15 mm and eyepiece with f e = 20 mm.The tube length of microscope (the distance between the inner focal points F1 and F2as shown) is 16 cm.

(a) Determine the magni…cation when the …nal image is formed at 25 cm.

(b) For the image to be projected on a screen 5 cm from the eyepiece, what adjustment

has to be made?

F1 F1 F2 F2

L

objective eyepiece

Solution

1



(a) The magni…cation when the image is formed at in…nity is

m =161:5

252

= 133:3

When formed at 25 cm,

m = 161:5 1 + 252 = 144

(b) For viewing, the object distance should be

1ov

+ 1

16 + 160 =

1f o

; ov = 17:60 mm

The object distance for the eyepiece can be found from

1o

1250

= 120

; o = 18:52 mm

For projection (real image at 50 mm), the object distance for the eyepiece is

1o0

+ 150

= 120

; o0 = 33:33 mm

The object distance is to be found from

1o p

+ 1

16 + 160 13:33 =

116

; o p = 17:745

Therefore, the objective lens has to be moved away from the object by 17:745

17:60 = 0:145 mm.3. A Galileo type telescope has an objective lens of f o = 50 cm and eyepiece f e = 8

cm. When used to view an object 5 m away with image formed at in…nity, what is thedistance between the two lenses? What is the magni…cation?

SolutionThe objective forms image at i

1500

+ 1i =

150

; i = 55:56 cm

To form …nal image at in…nity the eye piece should be at 55:56 8 = 47:56 cm fromthe objective. The magni…cation is

m = +55:56

8 = 6:94

4. An object is placed 15 cm in front of the surface of the lens shown. Find the location,magni…cation, and nature of the …nal image.

2



n = 1.5

10 cm5 cm

SolutionFirst refraction

115

+ 1:5

i = (1 :5 1)

110

; i = 90:0 cm

Second refraction1:595

+ 1i0

= 0:515

; i0 = 8:636 cm

Magni…cation is

m =90=1:515

8:63695=1:5 = 0:545

The image is virtual and erect.The matrix is

1 i00 1

1 0(1:5 1) =5 1:5

1 50 1

1 0(1=1:5 1) =10 1=1:5

1 150 1

= 0:83333 + 3: 3333 10 2 i0 15: 833 + 1: 8333i03: 3333 10 2 1: 8333

From B = 0 ;

i0 = 8:636 cmThe magni…cation is

m = A = 0:83333 + 3: 3333 10 2 i0= 0 :545

5. A rod of length l = 0:5 m and mass m = 0:3 kg is freely pivoted at one end. The otherend is attached to a mass-less spring (spring constant ks = 20 N/m ) as shown.

(a) If the spring elongates vertically a distance y ; what is the total potential energyin the system, one due to the spring elasticity and another due to gravity?

3



(b) What is the kinetic energy of the rod if the displacement y is varying with time?

(c) Then, using the energy conservation principle, show that the oscillation frequencyis given by

! = r 3ks

m ;

independent of both l and g:

Solution

The potential energy is12

ks y2 + 12

mgy

The kinetic energy is12

I 1l

dydt

2

= 12

13

mdydt

2

Energy conservation is

12

ks y2 + 12

mgy + 16

mdydt

2

= const.

Di¤erentiate13

mdydt

d2 ydt 2 + ks

dydt

y + 12

mgdydt

= 0

d2 ydt 2 +

3ks

m y +

32

g = 0

Let~y = y +

mg2ks

(Notice thaty =

mg2ks

is the equilibrium position when the rod is attached.) ~y satis…es

d2 ~ydt2 +

3ks

m ~y = 0

The oscillation frequency is

! = r 3ks

m = 12 r

3 200:3 = 14:1 rad/sec, f = 2:25 Hz

6. Analyze how the voltage wave develops after the switch S is closed. Assume V = 5 V,Rg = Z 1 = RL = 50 ; and Z 2 = 30 : Check whether the …nal voltage is consistentwith that expected from dc theory.

4



(b) The maximum re‡ection occurs when the …lm is f =4 because both re‡ected wavesare additive. The e¤ective impedance of the quarter wavelength thick …lm is

Z =Z 2f Z a

= 1

p 2 3772

377 =

12 377 :

Then the power re‡ection is

2 =188:5 377188:5 + 377

2

= 11 :1%

8. (a) A grating in a spectrometer has 30000 grooves equally spaced over a width of a = 8 cm. Red light of = 650 nm is incident normal to the grating. Determinethe angular locations of intensity maxima. What is the maximum number of order m?

(b) A distant binary star consists of two stars with an angular separation of = 0:200.

(100 = 1

3600 degrees.) What should the mirror diameter be in order for a telescopeto recognize the two stars as separate objects? Assume = 550 nm.

Solution

(a)

d = 0:08 m30000

= 2:67 10 6 m

d sin = m = m 650 10 9

sin = m0:65 10 6

2:67 10 6 = 0 :2434m

m = 1 : sin 1 (0:2434) = 0:246 radm = 2 : sin 1 (2 0:2434) = 0:508m = 3 : sin 1 (3 0:2434) = 0:8186m = 4 : sin 1 (4 0:2434) = 1:3405

(b)

0:2" = 0 :21

3600 180

= 9 :7 107

rad

= 1:22 D

= 9 :7 10 7

D = 1:229:7 10 7 =

1:22 550 10 9

9:7 10 7 = 69 cm

9. An electron is placed in an electromagnetic wave with an rms intensity of I = 10 W/m 2

and frequency = 10 15 Hz.

6



(c) In wiggler radiation, what is the spatial period w of magnets if 500 eV photonis to be produced?

Solution

(a)

= E mc2 =

2:9 109 eV512 keV = 5664

(b) The frequency of 100 eV photon is

h = 100 eV = 100 1:6 10 19 J

= 100 1:6 10 19

6:63 10 34 = 2 : 4133 1016 Hz

From! = 2 eB

m

B = !m

e 2 = 2 2: 4133 1016 9:1 10 31

1:6 10 19 (5664)2

= 2 : 69 10 2 T

(c) The frequency of 500 eV photon is

= 500 1:6 10 19

6:63 10 34 = 1 : 21 1017 Hz

and the wavelength is

= c =

3 108

1: 21 1017 = 2: 4793 10 9 m

From =

12 2 w

we …nd

w = 2 2

= 2 (5664) 2 2: 4793 10 9

= 0 :16 m

8



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