In a Young's double slit experiment, the fringe width is \( \beta \). If the entire apparatus is immersed in a liquid of refractive index \( \mu \), the new fringe width is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \beta\,\mu^{-1} \)
The fringe width is proportional to the wavelength of light. In a medium of refractive index \( \mu \) the frequency stays the same but the speed falls to \( c/\mu \), so the wavelength becomes \( \lambda/\mu \).
\( \displaystyle \beta = \frac{\lambda D}{d} \)
\( \displaystyle \beta' = \frac{(\lambda/\mu)\,D}{d} = \frac{\beta}{\mu} = \beta\,\mu^{-1} \)