Chapter 35
5
(a) 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
Á
1
=
k
1
L
¡
!t
,whe
re
k
1
(= 2
¼=¸
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
Á
2
=
k
2
L
¡
!t
,whe
re
k
2
(= 2
¼=¸
2
) is the
angular wave number and
¸
2
is the wavelength in plastic. The angular frequencies are the same
since the waves have the same wavelength in air and the frequency of a wave does not change
when the wave enters another medium. The phase difference is
Á
1
¡
Á
2
=(
k
1
¡
k
2
)
L
=2
¼
μ
1
¸
1
¡
1
¸
2
¶
L:
Now
¸
1
=
¸
air
=n
1
, where
¸
air
is the wavelength in air and
n
1
is the index of refraction of the
glass. Similarly,
¸
2
=
¸
air
=n
2
,where
n
2
is the index of refraction of the plastic. This means that
the phase difference is
Á
1
¡
Á
2
=(2
¼=¸
air
)(
n
1
¡
n
2
)
L
.Thev
a
lueo
f
L
that makes this 5
:
65 rad is
L
=
(
Á
1
¡
Á
2
)
¸
air
2
¼
(
n
1
¡
n
2
)
=
5
:
65(400
£
10
¡
9
m)
2
¼
(1
:
60
¡
1
:
50)
=3
:
60
£
10
¡
6
m
:
(b) 5
:
65radi
sle
s
sth
an2
¼
rad (= 6
:
28 rad), the phase difference for completely constructive
interference, and greater than
¼
rad (= 3
:
14 rad), the phase difference for completely destruc-
tive interference. The interference is therefore intermediate, neither completely constructive nor
completely destructive. It is, however, closer to completely constructive than to completely de-
structive.
15
Interference maxima occur at angles
μ
such that
d
sin
μ
=
m¸
,whe
re
d
is the separation of the
sources,
¸
is the wavelength, and
m
is an integer. Since
d
=2
:
0m and
¸
=0
:
50 m, this means
that sin
μ
=0
:
25
m
. You want all values of
m
(positive and negative) for which
j
0
:
25
m
j
∙
1.
These are
¡
4,
¡
3,
¡
2,
¡
1, 0, +1, +2, +3, and +4. For each of these except
¡
4 and +4, there are
two different values for
μ
. A single value of
μ
(
¡
90
±
) is associated with
m
=
¡
4andas
ing
le
value (
¡
90
±
) is associated with
m
= +4. There are sixteen different angles in all and therefore
sixteen maxima.
17
The angular positions of the maxima of a two-slit interference pattern are given by
d
sin
μ
=
m¸
,
where
d
is the slit separation,
¸
is the wavelength, and
m
is an integer. If
μ