Explanation

(b) In the case of refraction if $CD$ is the refracted wave front and $v_{1}$ and $v_{2}$ are the speed of light in the two media, then in the time the wavelets from $B$ reaches $C$, the wavelet from $A$ will reach $D$, such that
$t = \frac{BC}{v_{a}} = \frac{AD}{v_{g}} \Rightarrow \frac{BC}{AD} = \frac{v_{a}}{v_{g}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
But in $\Delta ACB,\ BC = AC\sin\theta$ …(ii)
While in $\Delta ACD,\ AD = AC\sin{\phi'}$ …(iii)
From equations (i), (ii) and (iii) $\frac{v_{a}}{v_{g}} = \frac{\sin\theta}{\sin{\phi'}}$
Also $\mu \propto \frac{1}{v} \Rightarrow \frac{v_{a}}{v_{g}} = \frac{\mu_{g}}{\mu_{a}} = \frac{\sin\theta}{\sin\phi^{'}} \Rightarrow \mu_{g} = \frac{\sin\theta}{\sin{\phi'}}$