15.3 Solutions
751
2d
λ 2
−
2d
λ 1
= 1
∴ λ 1 − λ 2 = λ =
λ 1 λ 2
2d
(b) λ =
(5.89 × 10 −5 ) 2
2 × 2.89 × 10 −5 = 6 Å
15.39 λ =
2d
N
=
2 × 2.948 × 10 −3 cm
100
= 5.896 × 10 −5 cm = 5896 Å
15.40 Resolving power
λ
λ
=
gπ D
λ
where D = distance between the plates = 2 mm = 0.2 cm:
λ = 600 nm = 6 × 10
−5 cm
g =
2r
1 − r 2 =
2 × 0.9
1 − (0.9) 2 = 9.4737
R.P =
λ
λ
=
9.4737 × 3.14 × 0.2
6 × 10 −5
= 9.9 × 10
4
15.3.5 Diffraction
15.41 a sin θ = mλ
(1)
∴
sin θ =
mλ
a
= 1 ×
λ
a
=
λ
a
(2)
If D is the slit screen distance and x the distance of the first minimum from
the central maximum, then
tan θ sin θ = x/D
(3)
Combining (2) and (3)
λ =
ax
D
=
0.02 × 0.5
200
= 5 × 10
−5 cm = 5000 Å
15.42 I θ = I m
sin α
α
2
Differentiating I θ with respect to α and setting
dI θ
dα
= 0,
dI θ
dα
=
I m (2α 2 sin α cos α − 2α sin
2
α)
α 4
= 0