zemansky capitulo 33 solucionario (farfismat)

8/14/2019 Zemansky Capitulo 33 solucionario (farfismat)

1/16

33.1: a) .sm1004.247.1

sm1000.3 88

=

==

n

c

v

b) .m1042.447.1

)m1050.6(

77

0

=

==n

33.2: a)

.m1017.5Hz105.80

sm1000.3

7

14

8

vacuum

=

==f

c

b) .m1040.3Hz)(1.52)1080.5(

sm1000.3

7

14

8

glass

=

==

fn

c

33.3: a) .54.1m/s101.94

m/s1000.38

8

=

==v

cn

b) .m1047.5)m1055.3()54.1(77

0

=== n

33.4: 1.501

m)(1.333)1038.4(

7

Benzene

waterwaterBenzeneBenzenewaterwater 2

===

n

nnn CS

.5.47equalalwaysareanglesreflectedandIncidenta) == ar :33.5

b) .0.665.42sin66.1

00.1arcsin

2sinarcsin

22=

=

==

a

b

abb

n

n

33.6: sm1017.2

s105.11

m50.2 89

=

== t

dv

38.1m/s102.17

m/s1000.38

8

=

==v

cn

33.7: bbaann sinsin =

sm1051.2194.1/)m/s1000.3(so

194.11.48sin

7.62sin00.1

sin

sin

88 ====

=

=

=

ncvvcn

nnb

aab


2/16

33.8 (a)

Apply Snells law at both interfaces.

33.9: a) Let the light initially be in the material with refractive index na and let the thirdand final slab have refractive index nb Let the middle slab have refractive index n1

11 sinsin:interface1st nn aa =

bbnn sinsin:interface2nd 11 =

.sinsingivesequationstwotheCombining bbaa nn =

b) ForNslabs, where the first slab has refractive index na and the final slab has

.sinsin,,sinsin,sinsin,indexrefractive 22221111 bbNNaab nnnnnnn === ofangleon thedependstravelofdirectionfinalThe.sinsingivesThis bbaa nn =

incidence in the first slab and the indicies of the first and last slabs.

33.10: a) .5.250.35sin33.100.1arcsinsinarcsin air

water

airwater =

=

= nn

b) This calculation has no dependence on the glass because we can omit that step in the

.sinsinsin:chain waterwateglassglassairair rnnn ==

33.11: As shown below, the angle between the beams and the prism is A/2 and the anglebetween the beams and the vertical is A, so the total angle between the two beams is 2A.


3/16

33.12: Rotating a mirror by an angle while keeping the incoming beam constant leads

rotation.mirror

thefromarose2ofdeflectionadditionalanwhere22becomesbeamsoutgoingand

incomingbetweenangletheTherefore.byangleincidentin theincreaseanto

+

33.13: .71.862.0sin1.58

1.70arcsinsinarcsin =

=

= a

b

ab

n

n

33.14: 38.2.45.0sin1.52


=

= a

b

ab

n

n But this is the angle

.53.2isangletheThereforesurface.theoftiltthe

ofbecause15additionalanisverticalthefromanglethesosurface,thetonormalthefrom

33.15: a) Going from the liquid into air:

1.48.42.5sin

1.00sin crit =

== aa

b nn

n

.58.135.0sin1.00

1.48arcsinsinarcsin:So b =

=

= a

b

a

n

n

b) Going from air into the liquid:

.22.835.0sin

1.48


=

= a

b

ab

n

n

33.16:

c=

>circle,largestfor theso

escapes,lightnoangle,criticalIf

222

c

1

w

1

c

aircw

m401m)(13.3

m11.3648tan)m0.10(m0.10/tan

48.61.333

1sin)/1(sin

00.1)00.1()00.1(90sinsin

===

===

===

===

RA

.RR

n

nn


4/16

33.17: = 7.48water,glassFor crit

1.77sin48.7

1.333

sinso,90sinsin

crit

crit ====

baba

nnnn

33.18:(a)

00.190sin)00.1(sin:ACatoccursreflectioninternalTotal ==n

==

41.1

1.00sin(1.52)

===+ 48.941.19090 theisanswerthissoangle,criticalthan thelessthusandsmallerislarger,isIf

largestthat can be.

(b) Same approach as in (a), except AC is now a glass-water boundary.

==

==

61.3

1.333sin1.52

1.33390sinsin w

nn

== 28.761.390

33.19: a) The slower the speed of the wave, the larger the index of refractionso air has a larger indexof refraction than water.

.15.1sm1320

sm344arcsinarcsinarcsinb)

water

aircrit =

=

=

=

v

v

n

n

a

b

c) Air. For total internal reflection, the wave must go from higher to lower index of refractionin this

case, from air to water.

.24.42.42

1.00arcsinarcsincrit =

=

=

a

b

n

n:33.20

.40.140.154.5tantana) ==== ba

bp n

n

n:33.21

.35.654.5sin1.40

1.00arcsinsinarcsinb) =

=

= ab

ab

n

n


5/16

:soand,0.37page,nexton thepicturetheFrom =r:33.22

1.77.37sin

53sin1.33

sin

sin=

==b

aab nn

1.65.tan31.2

1.00

tantana) =

===

p

ba

a

bp

nn

n

n

:33.23

.58.731.2sin1.00

1.65arcsinsinarcsinb) =

=

= a

b

ab

n

n

.58.91.001.66arctanarctanairIn)a =

=

=

a

bp

nn:33.24

.51.31.33

1.66arctanarctanIn waterb) =

=

=

a

bp

n

n

.2

1:filterfirsteThrough tha) 01 II =:33.25

.285.0)0.41(cos2

1:filtersecondThe 0

2

02 III == b) The light is linearly polarized.

=== max2

max

2

max 0.854)(22.5coscosa) IIIII :33.26

=== max2

max

2

max 500.0)(45.0coscosb) IIIII

=== max2

max

2

max 146.0)5.67(coscosc) IIIII

polarizedislighttheandmW10.0isintensityfilter thefirstAfter the 2021 =I:33.27

where,cosisfiltersecondafter theintensityThefilter.firsttheofaxisthealong 20 II= .37.025.062.0andmW10.0 20 === I .mW6.38Thus,

2=I


6/16

33.28: Let the intensity of the light that exits the first polarizer be I1, then, according to repeatedapplication of Malus law, the intensity of light that exits the third polarizer is

).0.230.62(cos)0.23(coscmW0.75 2212 = I

incidentintensitythealsoiswhich,

)0.230.62(cos)0.23(cos

cmW0.75thatseeweSo

22

2

1

=I

on the third polarizer after the second polarizer is removed. Thus, the intensity that exits the third polarizer

after the second polarizer is removed is

.cmW32.3)23.0(62.0cos)(23.0cos

)(62.0coscmW75.0 222

22

=

.125.0)(45.0cos,250.0)0.45(cos2

1,

2

1a) 0

2

230

2

0201 IIIIIIII =====:33.29

0.)(90.0cos2

1,

2

1b) 20201 === IIII

33.30: a) All the electric field is in the plane perpendicular to the propagation direction,and maximum intensity through the filters is at 90 to the filter orientation for the case ofminimum intensity. Therefore rotating the second filter by 90 when the situationoriginally showed the maximum intensity means one ends with a dark cell.

b) If filterP1 is rotated by 90, then the electric field oscillates in the direction pointingtoward theP2 filter, and hence no intensity passes through the second filter: see a darkcell.

c) Even ifP2 is rotated back to its original position, the new plane of oscillation of theelectric field, determined by the first filter, allows zero intensity to pass through the

second filter.

33.31: Consider three mirrors, M1 in the (x,y)-plane, M2 in the (y,z)-plane, and M3 in the(x,z)-plane. A light ray bouncing from M1 changes the sign of thez-component of the

velocity, bouncing from M2 changes thex-component, and from M3 changes they-

component. Thus the velocity, and hence also the path, of the light beam flips by 180

.46.69.73sin344

1480arcsinsinarcsinsinarcsina) =

=

=

= a

a

ba

b

ab

v

v

n

n:33.32

b) .13.41480

344arcsinarcsincrit =

=

=

b

a

v

v

33.33: a)331133222211

sinsinso,sinsinandsinsin nnnnnn ===

withmaterialin thenoraltherespect towithanglesamethemakeslighttheandsin

sinso,sinsinandsinsinb)/)sin(sin

33

11112222333113

n

nnnnnnn ====

1n as it did in part (a).

c) For reflection, .ar = These angles are still equal if r becomes the incidentangle; reflected rays are also reversible.


7/16

33.34: It takes the light an additional 4.2 ns to travel 0.840 m after the glass slab isinserted into the beam. Thus,

ns.4.2m0.840

)1(m0.840m0.840

==c

ncnc

We can now solve for the index of refraction:

2.50.1m0.840

)sm10(3.00s)10(4.2 89 =+=

n

The wavelength inside of the glass is

nm.200nm1962.50

nm490 ==

33.35: .6.4338.1

00.1arcsin90arcsin90 =

=

=

b

ab

n

n

But .1.7200.1

)6.43sin(38.1arcsin

sinarcsinsinsin =

=

==

a

bb

abbaa n

nnn

33.36:

==2

nnn abbbaa

sinsinsin

.51.72

1.80arccos2)80.1(

2cos2

2sin(1.80)

2cos

2sin2

22sinsin(1.00)

=

==

=

=

=

aa

aaaaa

33.37: The velocity vector maps out the path of the light beam, so the geometry asshown below leads to:

,arccosarccosandyy

yy

ra

r

r

a

a

rara vvv

v

v

vvv =

=

== with the minus

sign chosen by inspection. Similarly, .arcsinarcsinxx

xx

ra

r

r

a

avv

v

v

v

v=

=


8/16

33.38: ( )

+

=+=+= m105.40m0.00250

m105.40

m)0.00250m(0.0180

#)(##

77

glassairglassair n

dd

.103.52(1.40)4=

33.39:.40sin(

10.1arcsinarcsin7.40m00310.0

2/)m00534.0(arctancrit =

=

==

= n

nnna

b

Note: The radius is reduced by a factor of two since the beam must be incident at ,crit the

on the glass-air interface to create the ring.

33.40: =

= 51

m2.1

m5.1arctana

.3651sin1.33


=

= a

b

ab

n

n

So the distance along the bottom of the pool from directly below where the lightenters to where it hits the bottom is:

.m2.936tanm)0.4(tanm)0.4( === bx .m4.4m2.9m5.1m5.1total =+=+= xx

33.41 .14cm16.0

cm4.0arctanand27

cm16.0

cm8.0arctan =

==

= ba

So, .8.114sin

27sin00.1

sin

sinsinsin =

=

==

b

aabbbaa

nnnn

33.42: The beam of light will emerge at the same angle as it entered the fluid as seen byfollowing what happens via Snells Law at each of the interfaces. That is, the emergent

beam is at 5.42 from the normal.

33.43: a) .61.48333.1

000.1arcsin

90sinarcsin =

=

=

w

ai

n

n

The ice does not come into the calculation since .sinsin90sin iceair iwc nnn == b) Same as part (a).

33.44: .9.145sin

90sin33.1

sin

sinsinsin =

=

==

a

bbabbaa

nnnn


9/16

33.45:

==

b

aabbbaa

n

nnn

sinarcsinsinsin

.6.4400.1

)0.25(sin66.1arcsin =

=

So the angle below the horizontal is ,6.190.256.440.25 ==b and thusthe angle between the two emerging beams is .2.39

33.46: .40.190sin

60sin62.1

sin

sinsinsin =

=

==

a

bbabbaa

nnnn

33.47: .28.190sin

2.57sin52.1

sin

sinsinsin =

=

==

a

bbabbaa

nnnn

33.48:a) For light in air incident on a parallel-faced plate, Snells Law yields:.sinsinsinsinsinsin aaaaabba nnnn =====

b) Adding more plates just adds extra steps in the middle of the above equation that

always cancel out. The requirement of parallel faces ensures that the angle nn = andthe chain of equations can continue.

c) The lateral displacement of the beam can be calculated using geometry:

.cos

)sin(

cosand)sin(

b

ba

b

ba

td

tLLd

=

==

d) =

=

= 5.30

80.1

0.66sinarcsin

sinarcsin

n

n ab

.cm62.15.30cos

)5.300.66sin()cm40.2( =

= d

33.49: a) For sunlight entering the earths atmosphere from the sun BELOW the

horizon, we can calculate the angle as follows:

nnnnn bbabbaa === where,sinsin)00.1(sinsin is the atmospheresindex of refraction. But the geometry of the situation tells us:

+

+

==+

=+

=hR

R

hR

nR

hR

nR

hR

Rbaab arcsinarcsinsinsin .

b)

+

+ = m102.0m1064.

m106.4arcsinm)102.0m106.4

m)10(6.4(1.0003)arcsin 46

6

46

6

.22.0 = This is about the same as the angular radius of the sun, .25.0


10/16

33.50: A quarter-wave plate shifts the phase of the light by = 90 . Circularly polarizedlight is out of phase by 90 , so the use of a quarter-wave plate will bring it back intophase, resulting in linearly polarized light.

33.51: a) .2sin

8

1)sin(cos

2

1)90(coscos

2

1 20

2

0

22

0 IIII ===

b) For maximum transmission, we need .45so,902 ==

33.52: a) The distance traveled by the light ray is the sum of the two diagonal segments:

( ) ( ) .)( 21222212

1

2 yxlyxd +++=

Then the time taken to travel that distance is just:

( )c

yxlyx

c

dt

212

2

2212

1

2 )()( +++==

b) Taking the derivative with respect tox of the time and setting it to zero yields:

( )[ ]

( )[ ]

.sinsin)(

)(

0)()()(1

)()(1

21212

2

22

1

2

212

2

2212

1

2

212

2

2212

1

2

==+

=

+

=++=

+++=

yxl

xl

yx

x

yxlxlyxxcdx

dt

yxlyxdt

d

cdx

dt

33.53: a) The time taken to travel from point A to point B is just:

.)(

2

22

2

1

22

1

2

2

1

1

v

xlh

v

xh

v

d

v

dt

+

+

+

=+=

Taking the derivative with respect tox of the time and setting it to zero yields:

.sinsin)(

)(andBut

.)(

)()(0

221122

2

2

22

1

1

2

2

1

1

22

22

22

112

22

2

1

22

1

nnxlh

xln

xh

xn

n

cv

n

cv

xlhv

xl

xhv

x

v

xlh

v

xh

dt

d

dx

dt

=+

=

+==

+

+=

++

+==


11/16

33.54: a) n decreases with increasing , so n is smaller for red than for blue. So beam ais the red one.

b) The separation of the emerging beams is given by some elementary geometry.

,tantan

tantanvr

vrvr

xdddxxx

=== wherex is the vertical beam

separation as they emerge from the glass .mm2.9220sin

mm00.1 =

=x From the ray

geometry, we also have

.cm95.34tan7.35tan

mm92.2

tantan

:so,5.3466.1

70sinarcsinand7.35

61.1

70sinarcsin

=

=

=

=

==

=

vr

vr

xd

33.55: a) .2

sinsinsinsinA

nnn babbaa ==

But .2

sin2

2sin

2sin

2

An

AAAa =

+=

++=

At each face of the prism the deviation is .2

sin2

sin2so,A

nA

=+

=

b) From part (a), AA

n

=

2sinarcsin2

.9.380.602

0.60sin)52.1(arcsin2 =

=

c) If two colors have different indices of refraction for the glass, then the deflection

angles for them will differ:

.0.52.472.522.520.602

0.60sin)66.1(arcsin2

2.470.602

0.60sin)61.1(arcsin2

violet

red

===

=

=

=


12/16

33.56:

Direction of ray A: by law of reflection.Direction of ray B:

At upper surface: sinsin 21 nn = The lower surface reflects at . Ray B returns to upper surface at angle of

incidence sinsin: 12 nn = Thus

sinsin 11 nn = =

Therefore rays A and B are parallel.

33.57: Both l-leucine and d-glutamic acid exhibit linear relationships betweenconcentration and rotation angle. The dependence forl-leucine is:

Rotation angle ( ) :isacidglutamic-forandml),(g/100)gml10011.0( dC= Rotation angle ( ) ml).(g/100)gml100124.0( C=

33.58: a) A birefringent material has different speeds (or equivalently, wavelengths) intwo different directions, so:

.)(4

41

41

and

21

0

0

2

0

1

212

02

1

01

nnDDnDnDD

nn =+=+===

b) .m106.14)635.1875.1(4

m10895

)(4

77

21

0

=

=

=

.

nnD


13/16

33.59: a) The maximum intensity from the table is at ,35= so the polarized

component of the wave is in that direction (or else we would not have maximum intensityat that angle).

b) At )3540(cos2

1mW8.24:40 20

2 +=== pIII

pII 996.0500.0mW8.24 02 += (1).

At )35120(cos2

1mW2.5:120 20

2 +=== pIII

pII3

0

21060.7500.0mW2.5 += (2).

Solving equations (1) and (2) we find:

.mW8.19989.0mW6.1922

== pp II

Then if one subs this back into equation (1), we find:

5.049 = .mW1.10500.02

00 = II

33.60: a) To let the most light possible throughNpolarizers, with a total rotation of ,90

we need as little shift from one polarizer to the next. That is, the angle between

successive polarizers should be constant and equal to .2N

Then:

==

=

=

NIII

NII

NII NN

2cos,

2cos,

2cos 20

4

02

2

01

.

b) If +=

+=>> 2

2

21

21cos,1

nn

n

n

,14122

)2(

12cos

22

2

=

NNN

N

N for largeN.


14/16

33.61: a) Multiplying Eq. (1) by sin and Eq. (2) by sin yields:

(1): sinsincossincossinsin tta

x=

(2): sinsincossincossinsin tta

y=

Subtracting yields: ).sincossin(cossinsinsin

= ta

yx

b) Multiplying Eq. (1) by cos and Eq. (2) by cos yields:

cossincoscoscossincos:)2(

cossincoscoscossincos:)1(

tta

y

tta

x

=

=

Subtracting yields: ).cossincos(sincoscoscos

=

ta

yx

(c) Squaring and adding the results of parts (a) and (b) yields: 2222 )cossincos(sin)coscos()sinsin( =+ ayxyx (d) Expanding the left-hand side, we have:

).cos(2)coscossin(sin2

)coscossin(sin2)cos(sin)cos(sin2222

222222

+=++=++++

xyyxxyyx

xyyx

The right-hand side can be rewritten:

).(sin)cossincos(sin 2222 = aa Therefore: ).(sin)cos(2 2222 =+ axyyx Or: .where,sincos2

2222 ==+ axyyx (e) ,0)(2:0 222 yxyxxyyx

===+= which is a straight diagonal line.

circle.aiswhich,:2

.ellipseaniswhich,2

2:4

222

222

ayx

axyyx

=+=

=+=

This pattern repeats for the remaining phase differences.


15/16

33.62: a) By the symmetry of the triangles, .and, AbB

a

B

r

C

a

B

a

A

b ====

Therefore, .sinsinsinsin AaC

b

A

a

A

b

C

a

C

b nn ===== b) The total angular deflection of the ray is:

.422 +=++= AbA

a

C

a

C

b

B

a

A

b

A

a

c) From Snells Law, sin

== Aa

A

b

A

b

A

an

n sin1arcsinsin

.sin1

arcsin4242 +

=+= Aa

A

a

A

b

A

an

d)

=

==

nnnd

d

d

d AaA

a

A

a

1

2

1

2

cos

sin1

420sin

1arcsin420

).1(3

1cos1cos3

cos1cos4cos16sin

14

2

1

22

1

2

1

22

1

2

2

1

2

2

1

2

==

+=

=

nn

nnn

e) For violet: =

=

= 89.58)1342.1(

3

1arccos)1(

3

1arccos 221 n

.8.402.139 violetviolet ==

For red: =

=

= 58.59)1330.1(

3

1arccos)1(

3

1arccos

22

1 n

.5.425.137 redred ==

Therefore the color that appears higher is red.


16/16

33.63: a) For the secondary rainbow, we will follow similar steps to Pr. (34-51). Thetotal angular deflection of the ray is:

,26222 +=+++= AbA

a

A

b

A

a

A

b

A

b

A

b

A

a where we have used

the fact from the previous problem that all the internal angles are equal and the two

external equals are equal. Also using the Snells Law relationship, we have:

.sin1arcsin

= Aa

A

bn

.2sin1

arcsin62262 +

=+= Aa

A

a

A

b

A

an

b)

=

==

nnnd

d

d

d AaA

a

A

a

2

2

2

2

cos.

sin1

620sin

1arcsin620

).1(8

1coscos9)cos1()sin1( 22

2

2

2

2

222

2

22 ==+= nnnn

c) For violet: =

=

= 55.71)1342.1(

8

1arccos)1(

8

1arccos 222 n

.2.532.233 violetviolet ==

For red: .94.71)1330.1(8

1arccos)1(

8

1arccos

22

2 =

=

= n

.1.501.230 redred == Therefore the color that appears higher is violet.

zemansky capitulo 33 solucionario (farfismat)

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