15.3 Solutions
749
The intensity of the transmitted beam at the first face will be
I
1 =
3
4
I
The corresponding amplitude will be
a
1 =
√
3
2
a
The reflected beam at the second face will have amplitude
a
1 =
1
2
×
√
3
2
a =
√
3
4
a
The emerging beam from the first face will have amplitude
a 2 =
√
3
2
a
1 =
√
3
2
√
3
4
a =
3
8
a
The two beams reflected from the first face will interfere:
I min
I max
=
(a 1 − a 2 ) 2
(a 1 + a 2 ) 2 =
a
2
−
3
8
a
2
a
2
+
3
8
a
2 =
1
49
15.33 For constructive interference
2 μt =
m +
1
2
λ
∴ λ =
2μt
m +
1
2
=
2 × 1.5 × 4 × 10 −5
m +
1
2
=
12 × 10 −5
m +
1
2
cm
Only for m = 2, we get λ = 4.8 × 10 −5 cm or 4800 Å, corresponding to blue
colour in the visible region.
15.34 2μt cos r = mλ (minima)
For smallest thickness, m = 1
t =
mλ
2μ cos r
=
1 × 5890
2 × 1.5 × cos 60 ◦ = 3927 Å