Explanation

(c) Velocity of light water in material is
$v = v\lambda\ \ \ \ \ \ \ldots(i)$
Refractive index of material is
$\mu = \frac{c}{v}\ \ \ \ \ \ \ \ldots(ii)$
Where $c$ is speed of light in vacuum or air
or $\mu = \frac{c}{v\lambda}\ \ \ \ \ \ldots(iii)$
Given, $v = 2 \times 10^{1}⁴Hz,$
$\lambda = 5000Å = 5000 \times 10^{- 10}\ m$
$c = 3 \times 10^{8}{ms}^{- 1}$
Hence, from Eq (iii), we get
$\mu = \frac{3 \times 10^{8}}{2 \times 10^{14} \times 5000 \times 10^{- 10}} = 3.00$