In Young's double slit experiment, monochromatic light of wavelength 5000 A ∘ is used. The slits are 1 . 0 mm apart and screen is placed at 1 . 0 m away from slits. The distance from the centre of the screen where intensity becomes half of the maximum intensity for the first time is ______ × 10 - 6 m .

Step-by-step solution

Let intensity of light on screen due to each slit is I 0 . So internity at centre of screen is 4 I 0 (as cos ϕ = 1 ). Intensity at distance y from centre, I = I 0 + I 0 + 2 I 0 I 0 cos ϕ As we know, I max = 4 I 0 . Therefore, ⇒ I max 2 = 2 I 0 = 2 I 0 + 2 I 0 cos ϕ ⇒ cos ϕ = 0 ⇒ ϕ = π 2 Hence, k Δ x = π 2 ⇒ 2 π λ dsin θ = π 2 ⇒ 2 λ d × y D = 1 2 ⇒ y = λ D 4 d = 5 × 10 - 7 × 1 4 × 10 - 3 = 125 × 10 - 6 = 125