In a Young's double slit setup operated completely in air, the fringe width is exactly $0.6 \mathrm{

Question

In a Young's double slit setup operated completely in air, the fringe width is exactly $0.6 \mathrm{~mm}$. If the entire apparatus is gently immersed in a transparent liquid of refractive index $1.5$ without altering its geometry, the new fringe width becomes

Professor PC Thomas Classes · Answer

(a) : When an optical setup is immersed in a medium of refractive index $\mu$, the speed of light decreases, causing the wavelength to decrease to $\lambda' = \lambda/\mu$. Since fringe width $\beta$ is directly proportional to wavelength ($\beta = \frac{\lambda D}{d}$), it scales identically: $\beta' = \frac{\beta}{\mu}$. Given $\beta = 0.6 \mathrm{~mm}$ and $\mu = 1.5$, the new fringe width is $\beta' = \frac{0.6}{1.5} = 0.4 \mathrm{~mm}$.