The maximum intensity in Young's double-slit experiment is I 0 . Distance between the slits is d = 5 λ , where λ is the wavelength of monochromatic light used in the experiment. What will be the

Question

The maximum intensity in Young's double-slit experiment is I 0 . Distance between the slits is d = 5 λ , where λ is the wavelength of monochromatic light used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance D = 10 d ?

TestHub · Accepted Answer

Δ x at P = d x D = d 2 2 D = 5 λ 2 2 × 10 × d Δ x = 5 λ 2 2 × 10 × 5 λ = λ 4 Δ ϕ = 2 π λ × Δ x = π 2 I 0 = 4 I ⇒ I = I 0 4 Given, I 0 = 10 , d = 5 λ and D = 10 d . Path difference = Δ x at P = x d D At x = d 2 Δ x = d 2 2 D Δ x = 5 λ 2 2 × 10 d Δ x = 5 λ 2 2 × 10 × 5 λ = λ 4 Δ ϕ = 2 π λ Δ x Δ ϕ = π 2 I n e t = I + I + 2 I cos ⁡ π 2 = 2 I = 2 I 0 4 = I 0 2 as I m a x = I 0 = 4 I ⇒ I = I 0 4

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