754
15 Optics
The angular locations of the bright interference fringes are given by
d sin θ = mλ
(m = 0, 1, 2 . . .)
(2)
We can locate the first diffraction minimum within the double-slit fringe pattern by dividing (2) by (1)
n =
d
a
n =
0.2
0.012
= 16.66
Therefore 16 interference fringes will lie within the central maximum.
15.48 m 1 λ 1 = m 2 λ 2
λ 1 =
m 2 λ 2
m 1
=
3
4
× 5460 = 4095 Å
15.49 d sin θ = mλ
For m = m max , θ = θ max = 90
◦
, Given
1
d
= 5000
m max =
d
λ
=
1
5000
×
1
6 × 10 −5 = 3.33
∴ m max = 3
15.50
d
a
=
m
n
(condition for missing orders)
0.3
0.1
=
3
1
=
m
n
where m and n are integers. The above relation is satisfied for
m = 3, 6, 9
n = 1, 2, 3
Thus the maxima will be missing in the third, sixth, ninth, etc. orders.
15.51
d
a
=
a + b
a
=
a + 2a
a
= 3 =
m
n
The above relation is satisfied for
m = 3, 6, 9, . . .
n = 1, 2, 3, . . .
15 Optics
The angular locations of the bright interference fringes are given by
d sin θ = mλ
(m = 0, 1, 2 . . .)
(2)
We can locate the first diffraction minimum within the double-slit fringe pattern by dividing (2) by (1)
n =
d
a
n =
0.2
0.012
= 16.66
Therefore 16 interference fringes will lie within the central maximum.
15.48 m 1 λ 1 = m 2 λ 2
λ 1 =
m 2 λ 2
m 1
=
3
4
× 5460 = 4095 Å
15.49 d sin θ = mλ
For m = m max , θ = θ max = 90
◦
, Given
1
d
= 5000
m max =
d
λ
=
1
5000
×
1
6 × 10 −5 = 3.33
∴ m max = 3
15.50
d
a
=
m
n
(condition for missing orders)
0.3
0.1
=
3
1
=
m
n
where m and n are integers. The above relation is satisfied for
m = 3, 6, 9
n = 1, 2, 3
Thus the maxima will be missing in the third, sixth, ninth, etc. orders.
15.51
d
a
=
a + b
a
=
a + 2a
a
= 3 =
m
n
The above relation is satisfied for
m = 3, 6, 9, . . .
n = 1, 2, 3, . . .