15.3 Solutions
745
(b) Now the average intensity of the interference pattern is obtained by averaging over the cos 2 function in (6)
< I >=
λL/d
0
I dx
λL/d
0
dx
=
4A 2
λL/d
0
cos
2
(πdx/λL)dx
λL/d
= 2A
2
(7)
The average intensity is equal to 2A 2 as expected. Although at maximum
the intensity is double the average value, at minima, it becomes zero and
on the whole it averages out to 2 A 2 . Hence there is no violation of energy
conservation.
15.24 u + v = D = 100 cm → v = 100 − u = 100 − 30 = 70 cm
I
O
=
0.7
O
=
v
u
=
70
30
∴ 2d = O = 0.30 cm
(distance between two coherent sources)
β =
.λD
2d
(bandwidth)
∴ λ =
2dβ
D
=
(0.3)(0.0195)
100
= 5.85 × 10
−5 cm = 5850 Å
15.25 D = y 1 + y 2 = 10 + 100 = 110 cm
2d = 2(μ − 1)y 1 α = 2 × (1.5 − 1) ×
2
57.3
× 10
= 0.349 cm
λ =
(2d)β
D
=
0.349 × 0.018
110
= 5.711 × 10
−5 cm
= 5711 Å.
15.26 θ = 20 s =
20
3600
×
1
57.3
rad
= 9.695 × 10
−5 rad
β =
λ
2μθ
=
5.82 × 10 −5 cm
2 × 1.5 × 9.695 × 10 −5 = 0.2 cm
Number of fringes per centimetre=
1
β
=
1
0.2
= 5.
15.27 θ =
λ
2μβ
=
6 × 10 −5 cm
2 × 1.4 × (0.2 cm)
= 1.07 × 10
−4 rad
= 1.07 × 10
−4
× 57.3
◦
= 22
of arc.