752
15 Optics
∴ 2α sin α(α cos α − sin α) = 0
∴ tan α = α
15.43 (a) The half-width is the angle between the two points in the pattern where
the intensity is one-half the centre of the pattern:
I θ
I m
=
1
2
=
sin α x
α x
2
(1)
The solution of (1) found by numerical method is
α x = 1.40 rad
(2)
α x =
πa
λ
sin θ x = 1.40
(3)
where θ x =
1
2
sin θ x =
1.4λ
πa
or θ x = sin
−1
1.4λ
πa
or θ = 2θ x = 2 sin
−1
1.4λ
πa
(b) θ = 2 sin −1
1.4
4π
= 12.8 ◦
15.44 a sin θ = mλ
a sin θ 1 = 1 · λ 1
a sin θ 2 = 2λ 2
But θ 1 = θ 2
∴
λ 1 = 2λ 2
15.45 sin θ =
x
D
=
nλ
a
∴ a =
nλD
x
=
2 × 5.6 × 10 −5 × 200
1.6
= 0.014 cm
= 0.14 mm.
15.46 (a) Let m be the order of interference. If the slit width a is maintained constant and the separation of the slits d is varied, the scale of the interference pattern varies, but that of the diffraction pattern remains unchanged.
If the diffraction angle corresponds to the minimum given by (1) then a
particular order of interference maxima may be absent. This is called a
15 Optics
∴ 2α sin α(α cos α − sin α) = 0
∴ tan α = α
15.43 (a) The half-width is the angle between the two points in the pattern where
the intensity is one-half the centre of the pattern:
I θ
I m
=
1
2
=
sin α x
α x
2
(1)
The solution of (1) found by numerical method is
α x = 1.40 rad
(2)
α x =
πa
λ
sin θ x = 1.40
(3)
where θ x =
1
2
sin θ x =
1.4λ
πa
or θ x = sin
−1
1.4λ
πa
or θ = 2θ x = 2 sin
−1
1.4λ
πa
(b) θ = 2 sin −1
1.4
4π
= 12.8 ◦
15.44 a sin θ = mλ
a sin θ 1 = 1 · λ 1
a sin θ 2 = 2λ 2
But θ 1 = θ 2
∴
λ 1 = 2λ 2
15.45 sin θ =
x
D
=
nλ
a
∴ a =
nλD
x
=
2 × 5.6 × 10 −5 × 200
1.6
= 0.014 cm
= 0.14 mm.
15.46 (a) Let m be the order of interference. If the slit width a is maintained constant and the separation of the slits d is varied, the scale of the interference pattern varies, but that of the diffraction pattern remains unchanged.
If the diffraction angle corresponds to the minimum given by (1) then a
particular order of interference maxima may be absent. This is called a