Here, d = 2 mm, D = 1.2 m,
λ1 = 650 nm = 650 × 10–9 m, λ2 = 520 nm = 520 × 10–9 m
(a) Distance of third bright fringe from the central maximum for the wavelength 650 nm.
y3 =

3λD

d

=

3(650×10−9)1.2

2×10−3

= 1.17 mm
(b) Let at linear distance ‘y’ from center of screen the bright fringes due to both wavelength coincides. Let n1 number of bright fringe with wavelength λ1 coincides with n2 number of bright fringe with wavelength λ2.
We can write, y = n1β1 = n2β2
n1

λ1D

d

= n2

Dλ2

d

or n1λ1 = n2λ2 ... (i)
Also at first position of coincide, the nth bright fringe of one will coincide with (n + 1)th bright fringe of other.
If λ2 < λ1
So, then n2 > n1
then n2 = n1+1 ... (ii)
Using equation (ii) in equation (i)
n1λ1 = (n1+1)λ2 ⇒ n1 (650) × 10–9 = (n1+1)520×10–9
65 n1 = 52 n1 + 52 or 12 n1 = 52 or n1 = 4
Thus, y = n1β1 = 4 [

(6.5×10−7(1.2)

2×10−3

] = 1.56 × 10−3 m = 1.56 mm
So, the fourth bright fringe of wavelength 520 nm coincides with 5th bright fringe of wavelength 650 nm.