In a Young's double slit experiment with light of wavelength λ, the separation of slits is d and distance of screen is D such that D ≫ d ≫ λ . If the Fringe width is β , the distance from point of maximum intensity to the point where intensity falls to half of the maximum intensity on either side is:

Step-by-step solution

As I m a x = 4 I Then I n e t = 1 2   I m a x = 2 I ⇒ 2 I = 2 I 1 + c o s ϕ ⇒ 1 + c o s ϕ = 1 ⇒ c o s ϕ = 0 ⇒ ϕ = π 2 , 3 π 2 , 5 π 2 ,   …   ϕ = 2 π λ Δ x ⇒ π 2 = 2 π λ Δ x ⇒ Δ x = λ 4 , 3 λ 4 , 5 λ 4 ,   … Also path difference Δ x = d sin ⁡ θ = λ 4 At desired location ⇒ d   y n D = λ 4 ⇒   y n = λ D 4