Tài liệu Physics exercises_solution: Chapter 35 docx

35.1: Measuring with a ruler from both
21
and SS
to there different points in the
antinodal line labeled
m
= 3, we find that the difference in path length is three times the
wavelength of the wave, as measured from one crest to the next on the diagram.
35.2: a) At
,λ4,
121


rrS
and this path difference stays the same all along the
axis,-y
so
,rrSm λ4,At.4
122

and the path difference below this point, along
the negative
y
-axis, stays the same, so
.4


m
b)
c) The maximum and minimum m-values are determined by the largest integer less
than or equal to

.
λ
d
d) If
,77λ
2
1
7
 md
so there will be a total of 15 antinodes between the
sources. (Another antinode cannot be squeezed in until the separation becomes six times
the wavelength.)
35.3: a) For constructive interference the path diference is
,2,1,0,λ



nm
The
separation between sources is 5.00 m, so for points between the sources the largest
possible path difference is 5.00 m. Thus only the path difference of zero is possible. This
occurs midway between the two sources, 2.50 m from A.
b) For destructive interference the path difference is
,2,1,0,λ)(
2
1
 mm
A path difference of
00.32λ



/
m is possible but a path difference as large as
00.92/λ3

m is not possible. For a point a distance x from A and
Bx from00.5

the
path difference is
m00.1givesm00.3)m00.5(
m00.4givesm00.3)m00.5(
).m00.5(




xxx
xxx
xx
35.4:
a) The path difference is 120 m, so for destructive interference:

.m024λm120
2
λ

b) The longest wavelength for constructive interference is
.m120λ


35.5: For constructive interference, we need
λ)m00.9(λ
12
mxxmrr







,10,1,2,3,Form.8.25m,7.00m,5.75m,4.50m,3.25m,2.00m,75.0
).m25.1(m5.4
Hz)102(120
)sm1000.3(
m5.4
2
m5.4
2
λ
m5.4
6
8





mx
m

m
f
mcm
x
.3,2


(Don’t confuse this m with the unit meters, also represented by an “m”).
35.6:
a) The brightest wavelengths are when constructive interference occurs:

nm.408
5
nm2040
λ
andnm510
4
nm2040
λ,nm680
3
nm2040
λλλ
43


s
m
d
md
b) The path-length difference is the same, so the wavelengths are the same as part

(a).
35.7:
Destructive interference occurs for:

.nm453
5.4
nm2040
λandnm358
5.3
nm2040
λ
21
λ
43



m
d
35.8:
a) For the number of antinodes we have:

,90settingso,,2317.0
Hz)10(1.079m)(12.0
)sm10(3.00
λ
sin
8
8






m
m
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

λ)21(


 m
d
m

So the angles are
.3,2,1,0for2.54,4.35,3.20,65.6











m
35.9:
.m105.90
m2.20
m)10(2.82m)10(4.60
λ
λ
7
34








R
yd
d
R
y
35.10:
For bright fringes:
.mm1.14m101.14
m0.0106
m)10(5.02(20)m)(1.20
λ
3
7





m
y
Rm
d
35.11: Recall
m104.50

m)10(5.00)m(0.750)23(
λ
4
7
2323







d
R
λ
yyy
d
Rm
y
m

.mm0.833m1033.8
4
23


y
35.12:
The width of a bright fringe can be defined to be the distance between its two
adjacent destructive minima. Assuming the small angle formula for destructive

interference
,
λ)(
2
1
d
m
Ry
m


the distance between any two successive minima is
mm.8.00
m)10(0.200
m)10(400
m)00.4(
λ
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 d between
the two slits:
.m103.72
m104.84
m)10(600m)(3.00
λ
so),6.35.Eq(
λ
4
3
9
1
11
1







y
R
d
d
R
y
(R is much greater than d, so Eq.35.6 is valid.)

The dark fringes are located by
,2,1,0,λ)(sin
2
1
 mmd

The first order dark
fringe is located by sin
22
λ where,2λ d

is the wavelength we are seeking.
d
R
RRy
2
λ
sintan
2


We want
2
λ
such that
.
1
yy 
This gives
1200λ2λand

2
λλ
12
21

d
R
d
R
nm.
35.14:
Using Eq.35.6 for small angles,
,
λ
d
m
Ry
m

we see that the distance between corresponding bright fringes is
.mm3.17m)(10470)(660
m)10(0.300
(1)m)(5.00
λ
9
3






d
Rm
y
35.15:
We need to find the positions of the first and second dark lines:

.m0.0541)tan(8.79m)0.350(tan
79.8
m102(1.80
m105.50
arcsin
2
λ
arcsin
11
6
7
1

























Ry
d
Also






















3.27
m)102(1.80
m)103(5.50
arcsin
2
λ3
arcsin
6
7
2
d


m.0.1805tan(27.30)m)035.0(tan
22



θRy
The fringe separation is then
.m0.1264m0.0541m1805.0
12
 yyy

35.16:
(a) Dark fringe implies destructive interference.
λ
2
1
sin


d
m1064.1
0.11sin2
m10624
sin2
λ
6
9







d
(b) Bright fringes:
λsin
maxmax
mθd 
The largest that
θ

can be is
mdm Since6.2λ/so,90
m10624
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:


.λsin
2
1
 mθd
This equation has solutions for
;9.34;0.11





θ
and

.6.72


Therefore, there are 6 dark fringes.
35.17: Bright fringes for wavelength
λ
are located by
.λsin md


First-order
)1(

m
is closest to the central bright line, so
./λsin d






0.401andm)10m)/(0.10010(700singivesnm700λ
0.229andm)10m)/(0.10010(400singivesnm400λ
39
39


The angular width of the visible spectrum is thus
.0.1720.2290.401






35.18:
m.0.193m101.93
m104.20
m)10(4.50m)1.80(
λλ
4
3
7







y
R
d
d
R
y
35.19: The phase difference

is given by
)Eq.35.13(sin)λ/2(




d

rad16700.23sin]m)10(500m)10340.0(2[
93



35.20:
λ
differencePath
2




radians119
cm2
cm486cm524
2













35.21: a) Eq.(35.10):
0
2
0
2
0
750.0)0.30(cos)2(cos IIII 

b)
rad)3/(0.60



Eq.
so),)(λ/2(:)11.35(
12
rr 



nm806/λλ2/)3/(λ)2/()(
12


rr
35.22:

a) The source separation is 9.00 m, and the wavelength of the wave is
m.0.20
Hz1050.1
m/s1000.3
λ
7
8




f
c
So there is only one antinode between the sources
),0(

m
and it is a perpendicular bisector of the line connecting the sources.
b)




























sin
m)(20.0
m)00.9(
cossincos
2
cos
2
0
2
0
2
0
I
λ

d
III

)sin)41.1((cos
2
0

I
;295045580030forSo,
00
I.I,;θI.I,θ 

.026090117060
00
I.I,;θI.I,θ 
35.23:
a) The distance from the central maximum to the first minimum is half the
distance to the first maximum, so:
m.1088.8
m)102(2.60
m)10(6.60m)700.0(
2
λ
4
4
7








d
R
y
b) The intensity is half that of the maximum intensity when you are halfway to the
first minimum, which is
m.1044.4
4

Remember, all angles are
.small
35.24: a)
m,50.2
Hz1020.1
m/s1000.3
λ
8
8




f
c
and we have:
rad.4.52m)8.1(
m50.2
2

)(
λ
2
21




rr
b)
.404.0
2
rad52.4
cos
2
cos
0
2
0
2
0
IIII 
















35.25: a) To the first maximum:
m.1081.3
m101.30
m)10(5.50m)900.0(
λ
3
4
7
1







d
R
y
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 of
zero and pi, respectively. Halfway between these poi
nts, the phase difference between the

waves is
:So.
2

.W/m1000.2
24
cos
2
cos
26
0
2
0
2
0

















I
III

35.26: From Eq. (35.14),
.sin
λ
cos
2
0









d
II
So the intensity goes to zero when the
cosine’s argument becomes an odd integer of




)2/1(sin
λ
:isThat.

2
m
d
),2/1(λsin


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 distance
between the maxima (i.e., the # of meters per fringe) will be
0.0235
rad10095.4
m))1500/1((2
m105.46
arctan
2
λ
arctan
tan2
λ
2
λ
4
7






















xh
l
x


35.28:
The distance between maxima is
cm.0369.0
m)102(8.00

cm)(9.00m)1056.6(
2
λ
5
7






h
l
x
So the number of fringes per centimeter is
1.27
1


x
fringes/cm.
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
λ
4

λ
7
7
0





n
t
35.30:
There is a half-cycle phase shift at both interfaces, so for destructive
interference:
nm.5.80
4(1.49)
nm480
4
λ
4
λ
0

n
t
35.31: Destructive interference for
800λ
1

nm incident light. Let

n
be the refractive
index of the oil. There is a
2/λ
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/λ
phase difference due to the reflections, and the condition for destructive interference is
)./λ(2 nmt

Smallest nonzero thickness means
.λ2so,1
1
 tnm
The condition for constructive interference with incident wavelength
λ
is
onsoandnm,320λ2,for
nm533
λ,1for
nm1600
λ,0for
nm.800λwhere),/(λλso,λ2But
.
λ)(2and)/λ)((2
1
2
1
11

2
1
2
1





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
.5.36
m106.48
(1.35)m)1076.8(2
λ
2
λ
7
6
0







tnx
b) The phase difference for the two parts of the light is zero because the path
difference 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 flat
region between the pits, undergo the same phase change due to reflection. The condition
for destructive interference is
),/λ)((2
2
1
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/(λ






nt
35.34:
A half-cycle phase change occurs, so for destructive interference
nm.180
2(1.33)
nm480
2
λ
2
λ
0

n
t
35.35:
a) To have a strong reflection, constructive interference is desired. One part of
the light undergoes a half-cycle phase shift, so:
.
2
1
nm771
2
1
(1.33)nm)290(2
2
1
2
λ
λ
2

1
2
































mmm
dn
n
md
For an integer
value of zero, the wavelength is not visible (infrared) but for
1

m
, the wavelength is
514 nm, which is green.
b) When the wall thickness is 340 nm, the first visible constructive interference
occurs again for








2
1
nm904
λyieldsand1
m
m

= 603 nm, which is orange.
35.36:
a) Since there is a half-cycle phase shift at just one of the interfaces, the
minimum thickness for constructive interference is:
nm.3.74
4(1.85)
nm550
4
λ
4
λ
0

n
t
b) The next smallest thickness for constructive interference is with another half
wavelength thickness added:


nm.223
)85.1(4
nm5503
4
λ3
4
λ3
0

n
t

35.37:
mm.0.570m1070.5
2
)m1033.6(1800
2
λ
4
7





m
x
35.38: a) For Jan, the total shift was
m.1048.2
2
m)1006.6(818
2
1λ
4
7
1





m

x
For Linda, the total shift was
m.1005.2
2
m)1002.5(818
2
λ
4
72
2





m
x
b) The net displacement of the mirror is the difference of the above values:
mm.0.043mm0.205mm248.0
21
 xxx
35.39:
Immersion in water just changes the wavelength of the light from Exercise
35.11, so:
mm,626.0
1.33
mm833.0λ
vacuum

n

y
dn
R
y
using the solution from Exercise
35.11.
35.40:
Destructive interference occurs 1.7 m from the centerline.
22
1
m)2.6(m)0.12( r
=13.51 m
22
2
m)8.2(m)0.12( r
=12.32 m
For destructive interference,
m19.12/λ
21
 rr
and
m.4.2λ

The wavelength we
have 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
at
that position.
b) By moving the speakers toward you by 0.398 m, a ma
ximum is heard, which means
that you moved the speakers one-half wavelength from the min and the signals are back
in phase. Therefore the wavelength of the signals is 0.796 m, and the frequency is
m0.796
m/s340
λ

v
f
=427 Hz.
c) To reach the next maximum, one must move an additional distance of one
wavelength, a distance of 0.796 m.
35.42: To find destructive interference,
λ
2
1
)200(
22
12








mxxmrrd
.λ
2
1
2
1
λ
2
1
m000,20
λ
2
1
2
λ
2
1
)m200(
2
2
222





































m
m
x
mxmxx
The wavelength is calculated by
.m7.51
Hz1080.5
sm1000.3
λ
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
θ
puts speaker
A
ahead of
speaker
B
in phase. Constructive interference occurs when
   
,2,1,0,2381.0
3
2
λ
3
2
sin
,2,1,0,
λ
2
1
6

λsin























mmdm
mmd



,4;8.60,3;4.39,2;4.23,1;13.9,0










mmmmm
no solution
At points on the other side of the centerline, the path from
A
is longer than the path from
B
, and the path difference d sin
θ
puts speaker
A
behind speaker
B
in phase.
Constructive interference occurs when
   
, 2,1,0,2381.0
3
1
λ
3
2

sin
,2,1,0,
λ
2
1
6
λsin
























mmdm
mmd


,4;5.52,3;7.33,2;5.18,1;55.4,0









mmmmm
no solution
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





f
T
s

t
,250.0


t
so the speakers are
41
period out of phase. Let A be ahead of B in
phase.
m210.0
Hz1570
sm330
λ  fv
centerlineofsidesAOn '
: Since A is ahead by
41
period, the path difference must
retard B’s phase enough so the waves are in phase.






















6.60
m422.0
m210.0
4
7
sin
9.21
m0.422
m0.210
4
3
sin
,
λ
4
7
,
λ
4
3
sin

22
11



d

centerlineofsidesBOn '
: The path difference must now retard A’s sound by
,λ,λ
4
5
4
1

 5.38,2.7gives,λ
4
5
,
λ
4
1
sin


d
35.45:
a) If the two sources are out of phase by one half-cycle, we must add an extra
half a wavelength to the path difference equations Eq. (35.1) and Eq. (35.2).
This exactly changes one for the other, for

,and
2
1
2
1
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:
,λ)2(
21

 mrr
and for destructive interference,
.λ)221(
21

 mrr
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

 ωtEtEtEtEtEtEt)E
pp
b)
),2/cos()(

 tAtE
p
so comparing with part (a), we see that the amplitude of
the wave (which is always positive) must be
.|)2/cos(|2

EA

c) To have an interference maximum,
m


2
2

. So, for example, using
,1

m
the
relative phases are




2
2
:;4:;0:
12

p
EEE
, 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


p
EEE
and the resulting
wave 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()(||
222
0
2
0

 tEctcE
p
S
For a time average,
).2(cos2||so,
2
1
)2(cos
22
0
2

cESt
av


35.47: a)
mλr



.λ)()(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
λandm
we get a hyperbola. Or, in the
case of all
m
for a given
λ

, a family of hyperbola.
c)
.λ)()()(
2
1
2222
 mdyxdyx
35.48: a)
)cos(2
21
2
2
2
1
2

 EEEEE
p
.
2
9
0
.cos
2
4
2
5
2
1
.cos45cos44

2
00
22
0
2
0
22222
cEI
EEccEI
EEEEE
p



























.cos
9
4
9
5
So
0








II
b)
odd).(whenoccurswhich
9
1
0min
nnII



35.49: For this film on this glass, there is a net
2λ
phase change due to reflection and
the condition for destructive interference is
.750.1where),λ(2  nnmt
Smallest nonzero thickness is given by
.2λ 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



35.50: For constructive interference:
nm.2100nm)700(3sinλsin
1






dmd
For destructive interference:
.
nm2100sin
λλ
2
1
sin
2
1
2
1
22












mm
d
md


So the possible wavelengths are
.4fornm,467λand,3fornm,600λ
22
 mm
Both

andd
drop out of the calculation since their combination is just the path
difference, 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.
d
d
m
d
m
m
d

d
I
d
II
mmm
m











2
λ
4
λ3
:1;
4
λ
:0
.
2
)21(
λ
sin

λ2
sin
λ
cos
0
2
0







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:
.λ)(λ)(2
2
2
1
21
2
1
1
 mmt

But the two different wavelengths differ by
just one
.1value,-
12
 mmm
nm.1334
4(1.52)
nm)0.477(17
λ
2
1
82
.8
nm)477.0nm2(540.6
nm540.6nm0.477
)(2
λλ
2
λλ
)λλ(λ
2
1
λ
2
1
1
1
12
21
1

21
1212111
































t
n
t
m
λλ
mmmm
35.53:
a) There is a half-cycle phase change at the glass, so for constructive
interference:
.λ
2
1
4
λ
2
1
2
22
22
2
2






















mxhx
mx
x
hxd
Similarly for destructive interference:
.λ4
22
mxhx 
b) The longest wavelength for constructive interference is when
:0

m
cm.72
21
cm14cm)24(4cm)14(

4
22
2
1
22






m
xhx
λ
35.54:
a) At the water (or cytoplasm) to guanine interface, is a half-cycle phase shift for
the 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(533λ
)(
nm266
)(
(1.80)nm)74(2
)(
2
λ
λ
)
2

1
(2
2
1
2
1
2
1







m
mmm
nt
n
mt
gg
g
g
For the cytoplasm layers:
).0nm(533λ
)(
nm267
)(
(1.333)nm)100(2
)(

2
λ
λ
2
1
2
2
1
2
1
2
1














m
mmm
nt
n

mt
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 percentage
reflected.
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 wavelengths
that will interfere constructively upon reflection.
35.55:
a) Intensified reflected light means we have constructive interference. There is
one half-cycle phase shift, so:
).3(nm424and2),nm(593λ
.
)(
nm1484
)(
(1.53)nm)485(2
)(
2
λ
λ
2
1
2
2
1
2
1

2
1














mλm
mmm
tn
n
mt
b) Intensified transmitted light means we have destructive interference at the upper
surface. There is still a one half-cycle phase shift, so:
.
nm1484(1.53)nm)485(22
λ
λ
2
m
m

m
tn
n
m
t


)3(nm495λ



m
is the only wavelength of visible light that is intensified. We could also think of this as
the result of internal reflections interfering with the outgoing ray without any extra phase
shifts.
35.56:
a) There is one half-cycle phase shift, so for constructive interference:
.
)(
nm1102
)(
(1.45)nm)380(2
)(
2
λ
λ
2
1
2
2

1
2
1
2
1
0













mmm
tn
n
mt
Therefore, we have constructive interference at
),2(nm441


mλ
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
λ
λ
2
0
0
m
m
m
tn
n
m
t

Therefore, the strongest transmitted wavelength (as measured in air) is
),2(nm551λ


m
which corresponds to green.
35.57:
For maximum intensity, with a half-cycle phase shift,
.λfor,
2
λ)12(
4
λ)12(

2
λ)12(
2
λ)12(
4
λ)12(
4
λ)12(
4
λ)12(
and
λ
2
1
2
2
2
22222
2222




































R
Rm
r
mRm
r
Rmm

RrR
m
RrR
rRR
m
rRRtmt
The second bright ring is when
:1

m
mm.0.910m1010.9
2
m)(0.952m)1080.5()1)1(2(
4
7





r
So the diameter of the third bright ring is 1.82 mm.
35.58: As found in Problem (35.51), the radius of the
thm
bright ring is in general:
,
2
λ)12( Rm
r



for
.λ

R
Introducing a liquid between the lens and the plate just changes the
wavelength from
.
λ
λ
n

So:
mm.737.0
1.33
mm850.0
2
λ)12(
)(



n
r
n
Rm
nr
35.59:
a) Adding glass over the top slit increases the effective path length from that slit
to the screen. The interference pattern will therefore change, with the central maximum

shifting downwards.
b) Normally the phase shift is
,sin
2



λ
d

but now there is an added shift from the
glass, so the total phase shift is now
)).1(sin(
λ
2
λ
)1(2
sin
λ
2
λ
2
λ
2
sin
λ
2











nLd
nLdLLn
πd





So the intensity becomes
.))1(sin(
λ
cos
2
cos
2
0
2
0








nLdIII


c) The maxima occur at
)1(λsin))1(sin(
λ
 nLmdmnLd


35.60:
The passage of fringes indicates an effective change in path length, since the
wavelength of the light is getting shorter as more gas enters the tube.
L
m
nn
LL
n
L
m
2
λ
)1()1(
λ
2
λ
2
λ
2



.
So here:
.1062.2
m)2(0.0500
m)1046.5(48
)1(
4
7





n
35.61:
There are two effects to be considered: first, the expansion of the rod, and second,
the change in the rod’s 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
λ
)1(2
λ

2
λ
2 TLn
Ln
Ln
N









.22.1
m1089.5
)C00.5()C10(5.00m)030.0()148.1(2
7
6
1






N
The change in the number of wavelengths due to the change in refractive index of
the rod is:

.73.12
m105.89
m)(0.0300min)00.1()minC00.5()C1050.2(2
λ
2
7
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,

Snell’s Law tells us that an incident angle of
θ
on the flat side of the prism enters the prism
at an angle of
,nθ
where
n
is the index of refraction of the prism. Similarly on leaving the
prism, the in-going angle is
Anθ 
from the normal, and the outgoing, relative to the prism,
is
).( Anθn

So the beam leaving the prism is at an angle of
AAnθnθ




)(
from the
optical axis. So
.)1( Anθθ




At the plane of the source
,

0
S
we can calculate the height of one image above the source:
).1(2)1()()tan(
2





naAdAanaa
d

b) To find the spacing of fringes on a screen, we use:
m.1057.1
1.00)(1.50rad)10(3.50m)2(0.200
m)10(5.00m)0.200m00.2(
)1(2
λλ
3
3
7










naA
R
d
R
y