A Young's double slit experiment is first performed in air using light of wavelength 500 nm . The slit separation is 2.0 mm and the screen is placed at a distance of 1.2 m from the slits. The entire setup is then immersed in a liquid of refractive index 1.25 . What is the minimum non-zero distance from the central maximum where a bright fringe of the interference pattern in air would exactly coincide with a dark frin

Step-by-step solution

The position of the n^ th bright fringe in air is given by: y₁ = n D d When immersed in the liquid, the wavelength becomes . The position of the m^ th dark fringe in the liquid is given by: y₂ = (m - 0.5) D d For the fringes to coincide, y₁ = y₂ : n D d = (m - 0.5) D d Simplifying, we get: n = m - 0.5 Substitute = 1.25 = 5 4 : n = 4(m - 0.5) 5 5n = 4m - 2 We need the smallest positive integers n and m that satisfy this equation. Testing values for m ( m = 1, 2, 3, ): If m = 1 , 5n = 2 (no integer solution) If m = 2