In Young’s double slit experiment when light of wavelength 600 nm is used the fringe width is 0.1 mm. Calculate the fringe width when the entire apparatus is immersed in a medium of refractive index $\frac{4}{3}$ by keeping the source of light in air

0.075 mm

The correct answer is option (3) : 0.075 mm

Fringe width in air, $\beta =\frac{\lambda D}{d}$

Fringe width in liquid , $\beta ' =\frac{\lambda D}{\mu d}=\frac{\beta }{\mu }$

$⇒\beta '=\frac{0.1}{\frac{4}{3}}=0.075\, mm$