In Young's double-slit experiment, the intensity at a point P on the screen is half the maximum intensity in the interference pattern. If the wavelength of light used is λ and d is the distance between the slits, the angular separation between point P and the center of the screen is

Step-by-step solution

If δ is the phase difference between the interfering waves at point P , then the intensity at point P is given by I = I m a x cos 2 ⁡ δ 2 Given I = I m a x 2 . Hence c o s 2 δ 2 = 1 2 , which gives δ 2 = π 4 Or δ = π 2 The angular separation θ between points P and O is given by tan ⁡ θ = y D . Since θ ≃ sin ⁡ θ . Hence sin ⁡ θ = y D ...... (i) If β is the fringe width, then y β = π / 2 2 π = 1 4 ...... (ii) This is because the phase difference δ between two consecutive maxima is 2 π . Now β = λ D d . using this in