In Young's double slit experiment, the width of the fringes obtained with light of wavelength 6000 Å is 3.0 mm. If the entire apparatus is immersed in a liquid medium of refractive index 1.50, then the fringe width would be

2.0 mm

The correct answer is Option (2) → 2.0 mm

Given:

$\lambda = 6000 \, \text{Å} = 6 \times 10^{-7} \, \text{m}$

$\text{Fringe width in air}, \, \beta = 3.0 \, \text{mm}$

Refractive index of medium, $\mu = 1.5$

Fringe width in a medium is given by:

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

$\beta' = \frac{3.0}{1.5} = 2.0 \, \text{mm}$

∴ The new fringe width = 2.0 mm