Zemansky Capitulo 35 solucionario (farfismat)

8/14/2019 Zemansky Capitulo 35 solucionario (farfismat)

1/18


2/18

35.4: a) The path difference is 120 m, so for destructive interference:

.m024m1202

==

b) The longest wavelength for constructive interference is .m120=

35.5: For constructive interference, we need )m00.9(12 mxxmrr ==

,10,1,2,3,Form.8.25m,7.00m,5.75m,4.50m,3.25m,2.00m,75.0

).m25.1(m5.4Hz)102(120

)sm1000.3(m5.4

2m5.4

2

m5.4

6

8

==

=

===

mx

mm

f

mcmx

.3,2 (Dont confuse this m with the unit meters, also represented by an m).

35.6: a) The brightest wavelengths are when constructive interference occurs:

nm.4085

nm2040

andnm5104

nm2040,nm680

3

nm2040 43

==

======

s

m

dmd

b) The path-length difference is the same, so the wavelengths are the same as part

(a).

35.7: Destructive interference occurs for:

.nm4535.4

nm2040andnm358

5.3

nm2040

21 43 ====+=m

d

35.8: a) For the number of antinodes we have:

,90settingso,,2317.0Hz)10(1.079m)(12.0

)sm10(3.00sin 8

8

===== mm

df

mc

d

m

the maximum integer value is four. The angles are 9.67and,0.44,6.27,4.13

for .4,3,2,1,0 =m

b) The nodes are given by sin ).21(2317.0)21(

+=+

= md

m So the angles are

.3,2,1,0for2.54,4.35,3.20,65.6 = m

35.9: .m105.90m2.20

m)10(2.82m)10(4.60

734

=

=

==R

yd

d

Ry

35.10: For bright fringes:

.mm1.14m101.14m0.0106

m)10(5.02(20)m)(1.20 37

==

==

my

Rmd


3/18

35.11: Recallm104.50

m)10(5.00)m(0.750)23(4

7

2323

=

===d

Ryyy

d

Rmym

.mm0.833m1033.8 423 ==y

35.12:The width of a bright fringe can be defined to be the distance between its twoadjacent destructive minima. Assuming the small angle formula for destructive

interference

,)( 2

1

d

mRym

+=

the distance between any two successive minima is

mm.8.00m)10(0.200

m)10(400m)00.4(

3

9

1 =

==

+d

Ryy nn

Thus, the answer to both part (a) and part (b) is that the width is 8.00 mm.

35.13: Use the information given about the bright fringe to find the distance dbetween

the two slits: .m103.72m104.84

m)10(600m)(3.00so),6.35.Eq(

43

9

1

111

=

===

y

Rd

d

Ry

(R is much greater than d, so Eq.35.6 is valid.)

The dark fringes are located by ...,2,1,0,)(sin21 =+= mmd The first order dark

fringe is located by sin 22 where,2 d= is the wavelength we are seeking.

d

RRRy

2

sintan 2==

We want2 such that .1yy = This gives 12002and

2

12

21 ===d

R

d

Rnm.

35.14: Using Eq.35.6 for small angles,

,

d

mRym =

we see that the distance between corresponding bright fringes is

.mm3.17m)(10470)(660m)10(0.300

(1)m)(5.00 9

3=

== d

Rmy


4/18

35.15: We need to find the positions of the first and second dark lines:

.m0.0541)tan(8.79m)0.350(tan

79.8m102(1.80

m105.50arcsin

2

arcsin

11

6

7

1

===

=

=

=

Ry

d

Also =

=

=

3.27m)102(1.80m)103(5.50arcsin

23arcsin

6

7

2d

m.0.1805tan(27.30)m)035.0(tan 22 == Ry The fringe separation is then .m0.1264m0.0541m1805.012 === yyy

35.16: (a) Dark fringe implies destructive interference.

2

1sin =d

m1064.10.11sin2

m10624

sin2

69

=

==

d

(b) Bright fringes: sin maxmax md =

The largest that can be is mdm Since6.2/so,90m10624

m101.64

max 9

6

===

is an integer, its

maximum value is 2. There are 5 bright fringes, the central spot and 2 on each side of it.

Dark fringes: ( ) .sin21+= md This equation has solutions for ;9.34;0.11 = and

.6.72 Therefore, there are 6 dark fringes.

35.17: Bright fringes for wavelength are located by .sin md = First-order )1( =m

is closest to the central bright line, so ./sin d=

======

0.401andm)10m)/(0.10010(700singivesnm700

0.229andm)10m)/(0.10010(400singivesnm400

39

39

The angular width of the visible spectrum is thus .0.1720.2290.401 =

35.18: m.0.193m101.93m104.20

m)10(4.50m)1.80( 43

7

=

===

y

Rd

d

Ry

35.19: The phase difference is given by )Eq.35.13(sin)/2( d=

rad16700.23sin]m)10(500m)10340.0(2[ 93 ==


5/18

35.20:

differencePath

2=

radians119cm2

cm486cm5242 =

=

35.21: a) Eq.(35.10): 02

0

2

0 750.0)0.30(cos)2(cos IIII === b) rad)3/(0.60 =

Eq. so),)(/2(:)11.35( 12 rr = [ ] nm806/2/)3/()2/()( 12 ==== rr

35.22: a) The source separation is 9.00 m, and the wavelength of the wave is

m.0.20Hz1050.1

m/s1000.3

7

8

=

==f

cSo there is only one antinode between the sources

),0( =m and it is a perpendicular bisector of the line connecting the sources.

b)

=

=

=

sinm)(20.0

m)00.9(

cossincos2cos

2

0

2

0

2

0 I

d

III

)sin)41.1((cos 20 I= ;295045580030forSo, 00 I.I,;I.I, ====

.026090117060 00 I.I,;I.I, ====

35.23: a) The distance from the central maximum to the first minimum is half thedistance to the first maximum, so:

m.1088.8

m)102(2.60

m)10(6.60m)700.0(

2

44

7

=

==

d

Ry

b) The intensity is half that of the maximum intensity when you are halfway to the

first minimum, which is m.1044.44

Remember, all angles are .small


6/18

35.24: a) m,50.2Hz1020.1

m/s1000.3

8

8

=

==f

cand we have:

rad.4.52m)8.1(m50.2

2)(

221 ===

rr

b) .404.02

rad52.4cos2

cos 02

02

0 IIII =

=

=

35.25: a) To the first maximum: m.1081.3m101.30

m)10(5.50m)900.0( 34

7

1

=

==

d

Ry

So the distance to the first minimum is one half this, 1.91 mm.

b) The first maximum and minimum are where the waves have phase differences ofzero and pi, respectively. Halfway between these points, the phase difference between the

waves is :So.

2

.W/m1000.224

cos2

cos 260202

0

==

=

=

IIII

35.26: From Eq. (35.14), .sin

cos20

= d

II So the intensity goes to zero when the

cosines argument becomes an odd integer of +=

)2/1(sin

:isThat.2

md

),2/1(sin += md which is Eq. (35.5).

35.27: By placing the paper between the pieces of glass, the space forms a cavity whose

height varies along the length. If twice the height at any given point is one wavelength

(recall it has to make a return trip), constructive interference occurs. The distancebetween the maxima (i.e., the # of meters per fringe) will be

0.02rad10095.4m))1500/1((2

m105.46arctan

2

arctan

tan2

2

47

==

=

===

xh

lx

35.28: The distance between maxima is

cm.0369.0m)102(8.00

cm)(9.00m)1056.6(

2

5

7

=

==

h

lx

So the number of fringes per centimeter is 1.271

=x

fringes/cm.


7/18

35.29: Both parts of the light undergo half-cycle phase shifts when they reflect, so for

destructive interference nm.114m1014.1)42.1(4

m1050.6

4

4

77

0 ==

===

nt

35.30: There is a half-cycle phase shift at both interfaces, so for destructive

interference:

nm.5.804(1.49)

nm480

4

4

0 ====n

t

35.31: Destructive interference for 8001 = nm incident light. Let nbe the refractive

index of the oil. There is a 2/ phase shift for the reflection at the air-oil interface but no

phase shift for the reflection at the oil-water interface. Therefore, there is a net 2/

phase difference due to the reflections, and the condition for destructive interference is

)./(2 nmt= Smallest nonzero thickness means .2so,1 1== tnm The condition for constructive interference with incident wavelength is

onsoandnm,3202,for

nm533,1for

nm1600,0for

nm.800where),/(so,2But

.)(2and)/)((2

121

11

21

21

======

=+==+=+=

m

m

m

mtn

mtnnmt

The visible wavelength for which there is constructive interference is 533 nm.

35.32: a) The number of wavelengths is given by the total extra distance traveled,divided by the wavelength, so the number is

.5.36m106.48(1.35)m)1076.8(2

2

7

6

0===

tnx

b) The phase difference for the two parts of the light is zero because the pathdifference is a half-integer multiple of the wavelength and the top surface reflection has a

half-cycle phase shift, while the bottom surface does not.

35.33: Both rays, the one reflected from the pit and the one reflected from the flatregion between the pits, undergo the same phase change due to reflection. The condition

for destructive interference is ),/)((221 nmt += where n is the refractive index of the

plastic substrate. The minimum thickness is for ,0=m and equals

m.0.11nm1108)]nm/[(4)(1.790)4/( ==== nt

35.34: A half-cycle phase change occurs, so for destructive interference

nm.1802(1.33)

nm480

2

2

0 ====n

t


8/18

35.35: a) To have a strong reflection, constructive interference is desired. One part ofthe light undergoes a half-cycle phase shift, so:

.

2

1

nm771

2

1

(1.33)nm)290(2

2

1

2

2

12

+

=

+

=

+

=

+=

mmm

dn

nmd For an integer

value of zero, the wavelength is not visible (infrared) but for 1=m , the wavelength is514 nm, which is green.

b) When the wall thickness is 340 nm, the first visible constructive interference

occurs again for

+

==

2

1

nm904yieldsand1

m

m = 603 nm, which is orange.

35.36: a) Since there is a half-cycle phase shift at just one of the interfaces, theminimum thickness for constructive interference is:

nm.3.744(1.85)

nm5504

4 0 ====

nt

b) The next smallest thickness for constructive interference is with another half

wavelength thickness added:( )

nm.223)85.1(4

nm5503

4

3

4

3 0 ====n

t

35.37: mm.0.570m1070.52

)m1033.6(1800

2

47

==

== m

x

35.38: a) For Jan, the total shift was m.1048.22

m)1006.6(8182

1 471

=== mx

For Linda, the total shift was m.1005.22

m)1002.5(818

2

472

2

=

==m

x

b) The net displacement of the mirror is the difference of the above values:

mm.0.043mm0.205mm248.021 ==== xxx

35.39: Immersion in water just changes the wavelength of the light from Exercise

35.11, so: mm,626.01.33

mm833.0 vacuum ====n

y

dn

Ry using the solution from Exercise

35.11.


9/18

35.40: Destructive interference occurs 1.7 m from the centerline.

22

1 m)2.6(m)0.12( +=r =13.51 m22

2 m)8.2(m)0.12( +=r =12.32 m

For destructive interference, m19.12/21 == rr and m.4.2= The wavelength wehave calculated is the distance between the wave crests.Note: The distance of the person from the gaps is not large compared to the separation of

the gaps, so the path length is not accurately given by .sin d

35.41: a) Hearing minimum intensity sound means that the path lengths from the

individual speakers to you differ by a half-cycle, and are hence out of phase by 180 atthat position.

b) By moving the speakers toward you by 0.398 m, a maximum is heard, which means

that you moved the speakers one-half wavelength from the min and the signals are backin phase. Therefore the wavelength of the signals is 0.796 m, and the frequency is

m0.796

m/s340

==

vf =427 Hz.

c) To reach the next maximum, one must move an additional distance of one

wavelength, a distance of 0.796 m.


10/18

35.42: To find destructive interference, 2

1)200( 2212

+=+== mxxmrrd

.2

1

2

1

2

1

m000,20

2

12

2

1)m200(

2

2

222

+

+

=

++

++=+

m

m

x

mxmxx

The wavelength is calculated by .m7.51Hz1080.5

sm1000.3

6

8

=

==f

c

.m0.20;3andm,1.90:2andm,219:1andm,761:0 ======== xmxmxmxm

35.43: At points on the same side of the centerline as point ,A the path from B is

longer than the path from ,A and the path difference d sin puts speakerA ahead of

speakerB in phase. Constructive interference occurs when

( ) ( ) ...,2,1,0,2381.03

2

3

2sin

,2,1,0,2

16sin

=

+=

+=

=

+=

mmdm

mmd

,4;8.60,3;4.39,2;4.23,1;13.9,0 ===== mmmmm no solutionAt points on the other side of the centerline, the path from A is longer than the path from

B , and the path difference dsin puts speakerA behind speakerB in phase.

Constructive interference occurs when

( ) ( ) ,...2,1,0,2381.03

1

3

2sin

...,2,1,0,2

1

6sin

=

+=

+=

=

+=+

mmdm

mmd

,4;5.52,3;7.33,2;5.18,1;55.4,0 ===== mmmmm no solution


11/18

35.44: First find out what fraction the 0.159 ms time lag is of the period.

)Hz(1570s)10159.0()s10159.0(10159.0 33

3

==

= fT

st

,250.0=t so the speakers are 41 period out of phase. Let A be ahead of B in

phase.

m210.0Hz1570

sm330 === fv

centerlineofsidesAOn ' : Since A is ahead by 41 period, the path difference mustretard Bs phase enough so the waves are in phase.

=

=

=

=

=

6.60m422.0

m210.0

4

7sin

9.21m0.422

m0.210

4

3sin

,...4

7,

4

3sin

22

11

d

centerlineofsidesBOn ' : The path difference must now retard As sound by

,,45

41

= 5.38,2.7gives,4

5,

4

1sin d

35.45: a) If the two sources are out of phase by one half-cycle, we must add an extrahalf a wavelength to the path difference equations Eq. (35.1) and Eq. (35.2).

This exactly changes one for the other, for ,and21

21 mmmm ++ since m

in any integer.

b) If one source leads the other by a phase angle, the fraction of a cycle difference is

.2

Thus the path length difference for the two sources must be adjusted for both

destructive and constructive interference, by this amount. So for constructive

inference: ,)2(21 += mrr and for destructive interference,.)221(21 ++= mrr


12/18

35.46: a) The electric field is the sum of the two wave functions, and can be written:

2)./cos(2)/cos(2)()cos()cos()()(( 12 +=++=+= tEtEtEtEtEtEt)E pp

b) ),2/cos()( += tAtEp so comparing with part (a), we see that the amplitude ofthe wave (which is always positive) must be .|)2/cos(|2 EA =

c) To have an interference maximum, m 22 = . So, for example, using ,1=m the

relative phases are

22

:;4:;0: 12 ===== pEEE , and all waves are in

phase.

d) To have an interference minimum, .2

1

2

+= m

So, for example using

,0=m relative phases are ,22:;:;0: 12 ===== pEEE and the resultingwave is out of phase by a quarter of a cycle from both of the original waves.

e) The instantaneous magnitude of the Poynting vector is:

)).2(cos)2(cos4()(|| 222020 +== tEctcEpS

For a time average, ).2(cos2||so,2

1)2(cos 220

2 cESt av ==+

35.47: a) mr=

.)()(So

.)(

.)(

2222

22

2

22

1

mdyxdyxr

dyxr

dyxr

=+++=

++=

+=

b) The definition of hyperbola is the locus of points such that the difference between

12 toandto SPSP is a constant. So, for a given andm we get a hyperbola. Or, in the

case of all m for a given , a family of hyperbola.

c) .)()()(212222 +=+++ mdyxdyx


13/18

35.48: a) )cos(2 212

2

2

1

2 += EEEEEp

.2

90

.cos2

4

2

5

2

1

.cos45cos44

2

00

22

0

2

0

22222

cEI

EEccEI

EEEEE

p

==

+

==

+=++=

.cos9

4

9

5So 0

+= II

b) odd).(whenoccurswhich9

10min nnII ==

35.49: For this film on this glass, there is a net 2 phase change due to reflection and

the condition for destructive interference is .750.1where),(2 == nnmt Smallest nonzero thickness is given by .2 nt=

15

00

0

0

0

)C(108.6)](150Cnm)[(166.4nm)7.1()()(

so)1(

nm.168.1(1.750)][(2)nm)5.588(,C170At

nm.166.4(1.750)][(2)nm)(582.4,C20.0At

===

+====

Tttt

Ttt

t

t


14/18

35.50: For constructive interference: nm.2100nm)700(3sinsin 1 === dmd

For destructive interference: .nm2100sin

2

1sin

21

2122 +

=+

=

+=

mm

dmd

So the possible wavelengths are .4fornm,467and,3fornm,600 22 ==== mm

Both andd drop out of the calculation since their combination is just the pathdifference, which is the same for both types of light.

35.51: First we need to find the angles at which the intensity drops by one-half from the

value of the m th bright fringe.

ddm

dm

mddId

II

mmm

m

=======

+==

=

+

2

4

3:1;

4

:0

.2

)21(

sin2

sin

cos 020

so there is no dependence on the -m value of the fringe.

35.52: There is just one half-cycle phase change upon reflection, so for constructive

interference: .)()(2 221

2121

1 +=+= mmt But the two different wavelengths differ byust one .1value,- 12 = mmm

nm.13344(1.52)

nm)0.477(17

2

182

.8nm)477.0nm2(540.6

nm540.6nm0.477

)(2

2

)(

2

1

2

1

1

1

12

211

211212111

==

+=

=+

=

+

=+

=

=

+

tn

t

m

mmmm


15/18

35.53: a) There is a half-cycle phase change at the glass, so for constructiveinterference:

.214

2

1

222

22

2

2

+=+

+=

+=

mxhx

mxx

hxd

Similarly for destructive interference:

.4 22 mxhx =+

b) The longest wavelength for constructive interference is when :0=m

cm.7221

cm14cm)24(4cm)14(422

21

22

=+

=+

+=

m

xhx

35.54: a) At the water (or cytoplasm) to guanine interface, is a half-cycle phase shift forthe reflected light, but there is not one at the guanine to cytoplasm interface. Therefore

there will always be one half-cycle phase difference between two neighboring reflected

beams. For the guanine layers:

).0nm(533)(

nm266

)(

(1.80)nm)74(2

)(

2

)

2

1(2

21

21

21

==+

=+

=+

=+= mmmm

nt

nmt

gg

g

g

For the cytoplasm layers:

).0nm(533)(

nm267

)(

(1.333)nm)100(2

)(

2

2

12

21

21

21

==+

=+

=+

=

+= m

mmm

nt

nmt cc

c

c

b) By having many layers the reflection is strengthened, because at each interface

some more of the transmitted light gets reflected back, increasing the total percentagereflected.

c) At different angles, the path length in the layers change (always to a larger value

than the normal incidence case). If the path length changes, then so do the wavelengthsthat will interfere constructively upon reflection.


16/18

35.55: a) Intensified reflected light means we have constructive interference. There isone half-cycle phase shift, so:

).3(nm424and2),nm(593

.)(

nm1484

)(

(1.53)nm)485(2

)(

2

2

12

21

21

21

====+

=+

=+

=

+=

mm

mmm

tn

nmt

b) Intensified transmitted light means we have destructive interference at the uppersurface. There is still a one half-cycle phase shift, so:

.nm1484(1.53)nm)485(22

2mmm

tn

n

mt ====

)3(nm495 == m

is the only wavelength of visible light that is intensified. We could also think of this asthe result of internal reflections interfering with the outgoing ray withoutany extra phase

shifts.

35.56: a) There is one half-cycle phase shift, so for constructive interference:

.)(

nm1102)((1.45)nm)380(2

)(2

212

21

21

21

0

+=

+=

+=

+=

mmmtn

nmt

Therefore, we have constructive interference at ),2(nm441 == m which corresponds

to blue-violet.b) Beneath the water, looking for maximum intensity means that the reflected

part of the wave at the wavelength must be weak, or have interfered destructively. So:

.nm1102(1.45)nm)380(22

2 00

mmm

tn

n

mt ====

Therefore, the strongest transmitted wavelength (as measured in air) is

),2(nm551 == m which corresponds to green.

35.57: For maximum intensity, with a half-cycle phase shift,

.for,2

)12(

4

)12(

2

)12(

2

)12(

4

)12(

4

)12(

4

)12(and

2

12

2

2

22222

2222

>>+

+

+=

+

++=

+=

=+

=

+=

RRm

rmRm

r

RmmRrR

mRrR

rRRm

rRRtmt

The second bright ring is when :1=m

mm.0.910m1010.92

m)(0.952m)1080.5()1)1(2( 47

==+

r

So the diameter of the third bright ring is 1.82 mm.


17/18

35.58: As found in Problem (35.51), the radius of the thm bright ring is in general:

,2

)12( Rmr

+

for .>>R Introducing a liquid between the lens and the plate just changes the

wavelength from .

n So:

mm.737.01.33

mm850.0

2

)12()( ===

+

n

r

n

Rmnr

35.59: a) Adding glass over the top slit increases the effective path length from that slitto the screen. The interference pattern will therefore change, with the central maximum

shifting downwards.

b) Normally the phase shift is ,sin2

d= but now there is an added shift from the

glass, so the total phase shift is now

)).1(sin(

2

)1(2sin

2

2

2sin

2+=

+=

+= nLd

nLdLLnd

So the intensity becomes .))1(sin(

cos2

cos 202

0

+== nLdIII

c) The maxima occur at )1(sin))1(sin(

==+ nLmdmnLd

35.60: The passage of fringes indicates an effective change in path length, since thewavelength of the light is getting shorter as more gas enters the tube.

L

mnn

LL

n

Lm

2

)1()1(

2

2

2 === .

So here:

.1062.2m)2(0.0500

m)1046.5(48)1( 4

7

=

=n


18/18

35.61: There are two effects to be considered: first, the expansion of the rod, and second,the change in the rods refractive index. The extra length of rod replaces a little of the air

so that the change in the number of wavelengths due to this is given by:

0

0glass

0

air

0

glass

1

)1(2

2

2 TLnLnLnN

=

=

.22.1m1089.5

)C00.5()C10(5.00m)030.0()148.1(27

6

1 ==

N

The change in the number of wavelengths due to the change in refractive index of

the rod is:

.73.12m105.89

m)(0.0300min)00.1()minC00.5()C1050.2(2

27

5

0

0glass

2 =

=

= Ln

N

So the total change in the number of wavelengths as the rod expands is

0.1422.173.12 =+=N fringes/minute.

35.62: a) Since we can approximate the angles of incidence on the prism as being small,

Snells Law tells us that an incident angle of on the flat side of the prism enters the prism

at an angle of ,n where n is the index of refraction of the prism. Similarly on leaving the

prism, the in-going angle is An from the normal, and the outgoing, relative to the prism,is ).( Ann So the beam leaving the prism is at an angle of AAnn += )( from theoptical axis. So .)1( An =

At the plane of the source ,0S we can calculate the height of one image above the source:

).1(2)1()()tan(2

=== naAdAanaad

b) To find the spacing of fringes on a screen, we use:

m.1057.11.00)(1.50rad)10(3.50m)2(0.200

m)10(5.00m)0.200m00.2()1(2

33

7

=

+=

==naA

RdRy

Documents

Zemansky Capitulo 35 solucionario (farfismat)