In a Young's double slit experiment, the fringe width is $β$. If the entire apparatus is immersed in a liquid of refractive index $μ$, the new fringe width is

$β\, μ^{-1}$

The correct answer is Option (3) → $β\, μ^{-1}$

$\beta = \frac{\lambda D}{d}$

$\lambda_{\text{medium}} = \frac{\lambda}{\mu}$

$\beta' = \frac{\lambda_{\text{medium}} D}{d}$

$\beta' = \frac{\lambda D}{\mu d}$

$\beta' = \frac{\beta}{\mu}$

The new fringe width is $\frac{\beta}{\mu}$.