In a Young's double slit experiment, the slits are illuminated by a light beam containing two wavelengths, 400 nm and 560 nm . A transparent glass slab of refractive index 1.5 is introduced in the path of light from one of the slits. If the minimum thickness of the slab required such that the geometrical center of the screen remains a bright fringe for both the wavelengths simultaneously is X 10⁻¹ m , then the value

Step-by-step solution

When a glass slab of thickness t and refractive index is introduced in front of one slit, the optical path difference at the geometrical center of the screen becomes: x = ( - 1)t For the center to be a bright fringe for a particular wavelength, the path difference must be an integer multiple of that wavelength. For it to be a bright fringe for both wavelengths simultaneously, x must be a common multiple of both ₁ and ₂ . To find the minimum thickness, x should be the Least Common Multiple (LCM) of ₁ and ₂ . ₁ = 400