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Question:

In the Young's double slit experiment using a monochromatic light of wavelength λ, the path difference (in terms of an integer n) corresponding to any point having half peak intensity is

(2n+1)λ8

(2n+1)λ16

(2n+1)λ2

(2n+1)λ4

Solution:

Sol (B)
Imax/2=Imcos²(φ/2) ⇒cos(φ/2)=1/√2 ⇒φ/2=π/4 ⇒φ =π/2(2n+1) ⇒Δx=λ/2πφ =λ/2π × π/2(2n+1)=λ/4(2n+1)