Chapter 35
1. Comparing the light speeds in sapphire and diamond, we obtain
v
v
v
c
n
n
s
d
s
d
F
H
G
I
K
J
F
H
G
I
K
J
1
1
2 998
10
1
177
1
2 42
4 55
10
8
7
.
.
.
.
.
m s
m s
c
h
2. (a) The frequency of yellow sodium light is
f
c
2 998
10
589
10
509
10
8
9
14
.
.
m s
m
Hz.
(b) When traveling through the glass, its wavelength is
n
n
589
152
388
nm
nm.
.
(c) The light speed when traveling through the glass is
14
9
8
5.09 10 Hz 388 10 m
1.97 10 m s.
n
v
f
l
3. The index of refraction is found from Eq. 353:
n
c
v
2 998
10
192
10
156
8
8
.
.
.
.
m s
m s
4. Note that Snell’s Law (the law of refraction) leads to
1
=
2
when
n
1
=
n
2
.
The graph
indicates that
2
= 30° (which is what the problem gives as the value of
1
) occurs at
n
2
=
1.5.
Thus,
n
1
= 1.5, and the speed with which light propagates in that medium is
v
=
= 2.0
×
10
8
m/s.
5. The fact that wave
W
2
reflects two additional times has no substantive effect on the
calculations, since two reflections amount to a 2(
±
/2) =
±
phase difference, which is
effectively not a phase difference at all. The substantive difference between
W
2
and
W
1
is
the extra distance 2
L
traveled by
W
2
.
1407
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View Full DocumentCHAPTER 35
(a) For wave
W
2
to be a halfwavelength “behind” wave
W
1
, we require 2
L
=
/2, or
L
=
/4 = 155 nm using the wavelength value given in the problem.
(b) Destructive interference will again appear if
W
2
is
3
2
“behind” the other wave. In
this case,
2
3
2
L
, and the difference is
L
L
3
4
4
2
310
nm .
6. In contrast to the initial conditions of problem 30, we now consider waves
W
2
and
W
1
with an initial effective phase difference (in wavelengths) equal to
1
2
, and seek positions
of the sliver which cause the wave to constructively interfere (which corresponds to an
integervalued phase difference in wavelengths). Thus, the extra distance 2
L
traveled by
W
2
must amount to
1
2
3
2
,
, and so on. We may write this requirement succinctly as
2
1
where
0,1, 2,
.
4
m
L
m
K
(a) Thus, the smallest value of
/
L
that results in the final waves being exactly in phase
is when
m
=0, which gives
/
1/ 4 0.25
L
.
(b) The second smallest value of
/
L
that results in the final waves being exactly in
phase is when
m
=1, which gives
/
3/ 4 0.75
L
.
(c) The third smallest value of
/
L
that results in the final waves being exactly in phase
is when
m
=2, which gives
/
5/ 4 1.25
L
.
7. (a) We take the phases of both waves to be zero at the front surfaces of the layers. The
phase of the first wave at the back surface of the glass is given by
1
=
k
1
L –
t
, where
k
1
(= 2
/
1
) is the angular wave number and
1
is the wavelength in glass. Similarly, the
phase of the second wave at the back surface of the plastic is given by
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 Fall '08
 Many
 Physics, Pythagorean Theorem, Light, Wavelength, NM

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