Q 12-10-047JEE MainJEE Main 2025 (24 Jan, Shift 2)Easy
Young's double slit interference apparatus is immersed in a liquid of refractive index $1.44$. It has slit separation of $1.5\ \text{mm}$. The slits are illuminated by a parallel beam of light whose wavelength in air is $690\ \text{nm}$. The fringe-width on a screen placed behind the plane of slits at a distance of $0.72\ \text{m}$ will be
Answer: (A) $0.23\ \text{mm}$
In the liquid the wavelength becomes $\lambda/\mu$:
$$\beta = \frac{\lambda D}{\mu d} = \frac{690\times10^{-9}\times0.72}{1.44\times1.5\times10^{-3}} = \frac{496.8\times10^{-9}}{2.16\times10^{-3}} = 2.3\times10^{-4}\ \text{m} = 0.23\ \text{mm}$$
Solution by Sreeraj P, M.Sc Physics