NTA Abhyas JEE Main2020PhysicsWave OpticsPractice
In Young's double-slit experiment, both the slits produce equal intensities on a screen. A 100 % transparent thin film of refractive index μ = 1 . 5 is kept in front of one of the slits, due to which the intensity at the point O on the screen becomes 75 % of its initial value. If the wavelength of monochromatic light is 720 nm , then what is the minimum thickness (in nm ) of the film?
Correct answer
240
Step-by-step solution
The path difference at the point O is Δ x = μ - 1 t I = 4 l 0 cos 2 ⁡ ϕ 2 0 . 75 4 I 0 = 4 I 0 cos 2 ⁡ ϕ 2 ⇒ cos ϕ 2 = ± 3 2 ⇒ ϕ min = 60 ° ϕ min = 2 π λ Δx = 2 π λ μ - 1 t = π 3 t = λ 6 μ - 1 = λ 0 6 1 . 5 - 1 = 240   nm