Two beams of light are incident normally on water (R.I. = 4/3). If the beam 1 passes through a glass (R.I. = 3/2) slab of height h as shown in the figure, the time difference for both the beams for reaching the bottom is

$\frac{h}{6C}$

The refractive index of glass is greater than that of water. Therefore the speed of light in glass is lesser than that of water. It is given as

$v = \frac{C}{n}$ where $C = 3 × 10^8 m/s$

v = speed of light in a medium of R.I. n

∴ The time difference for the rays

$=t_1-t_2=\frac{h}{v_g}-\frac{h}{v_w}$

$⇒Δt=\frac{h}{(C/n_g)}-\frac{h}{(C/n_w)}=\frac{h}{C}(n_g-n_w)$

$⇒Δt=\frac{h}{C}(\frac{3}{2}-\frac{4}{3})=\frac{h}{6C}$