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. The intensity of the light in front of one of the slits on a screen at a distance D = 10 d is

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. The intensity of the light in front of one of the slits on a screen at a distance D = 10 d is, I 0 2, 3 4 I 0, I 0, I 0 4

MARKS App · Accepted Answer

Path difference at point P is Δ x = d x D = d 2 2 D Δ x = ( 5 λ ) 2 2 × 10 d Δ x = ( 5 λ ) 2 2 × 10 × 5 λ = λ 4 Therefore, phase difference is Δ ϕ = 2 π λ × Δ x = π 2 If I is the intensity due to each slit, then the maximum intensity I 0 is, I 0 = 4 I ⇒ I = I 0 4 Therefore, intensity at point P is I n e t = I + I + 2 I I cos ⁡ π 2 I n e t = 2 I = I 0 2

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