Question

In Young’s double slit experiment, the screen is kept at a distance of 1.2 m from the plane of the slits. The two slits are separated by 5 mm and illuminated with monochromatic light having wavelength 600 nm. Calculate:

1. Fringe width i.e. fringe separation of the interference pattern.
2. Distance of 10^th bright fringe from the centre of the pattern.

Answer 1

Given: D = 1.2 m

d = 5 × 10⁻³ m

λ = 600 × 10⁻⁹ m

i. β = `(D lambda)/d`

= `(1.2 xx 600 xx 10^-9)/(5 xx 10^-3)`

= 0.144 × 10⁻³ m

= 0.144 mm

ii. Distance of n^th bright fringe from the centre of the pattern:

x_n = `(D n lambda)/d`

Here, n = 10

∴ x₁₀ = `10((D lambda)/d)`

= 10 × 0.144 mm

= 1.44 mm