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8/14/2019 Zemansky Capitulo 33 solucionario (farfismat)
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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
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8/14/2019 Zemansky Capitulo 33 solucionario (farfismat)
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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.
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8/14/2019 Zemansky Capitulo 33 solucionario (farfismat)
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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
1.33arcsinsinarcsin =
=
= 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
1.00arcsinsinarcsin =
=
= 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
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8/14/2019 Zemansky Capitulo 33 solucionario (farfismat)
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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
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8/14/2019 Zemansky Capitulo 33 solucionario (farfismat)
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: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
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8/14/2019 Zemansky Capitulo 33 solucionario (farfismat)
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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.
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8/14/2019 Zemansky Capitulo 33 solucionario (farfismat)
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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=
=
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8/14/2019 Zemansky Capitulo 33 solucionario (farfismat)
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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
1.00arcsinsinarcsin =
=
= 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
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8/14/2019 Zemansky Capitulo 33 solucionario (farfismat)
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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
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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
=+
=
+==
+
+=
++
+==
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8/14/2019 Zemansky Capitulo 33 solucionario (farfismat)
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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
===
=
=
=
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8/14/2019 Zemansky Capitulo 33 solucionario (farfismat)
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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
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8/14/2019 Zemansky Capitulo 33 solucionario (farfismat)
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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.
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8/14/2019 Zemansky Capitulo 33 solucionario (farfismat)
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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.
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8/14/2019 Zemansky Capitulo 33 solucionario (farfismat)
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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.
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8/14/2019 Zemansky Capitulo 33 solucionario (farfismat)
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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.