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 o

Question

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:

TestHub · Accepted Answer

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 d = β 4 ∵ β = λ D d ⇒ Separation between central maxima and desired y = β 4

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