A ray of light is incident at the glass-water interface at an angle i, it emerges finally parallel to the surface of water, then the value of ${\mathrm{\mu }}_{\mathrm{g}}$ would be

$\begin{array}{l}\mathrm{For}\mathrm{glass}-\mathrm{water}\mathrm{interface}{}_{\mathrm{g}}{\mathrm{\mu }}_{\mathrm{w}}=\frac{\sin \mathrm{i}}{\sin \mathrm{r}}.....\left(\mathrm{i}\right)\\ \mathrm{For}\mathrm{water}-\mathrm{air}\mathrm{interface}{}_{\mathrm{w}}{\mathrm{\mu }}_{\mathrm{\alpha }}=\frac{\sin \mathrm{r}}{\sin {90}^{\mathrm{o}}}....\left(\mathrm{ii}\right)\\ {\Rightarrow }_{\mathrm{g}}{\mathrm{\mu }}_{\mathrm{w}}{\times }_{\mathrm{w}}{\mathrm{\mu }}_{\mathrm{\alpha }}=\frac{\sin \mathrm{i}}{\sin \mathrm{r}}\times \frac{\sin \mathrm{r}}{\sin {90}^{\mathrm{o}}}=\sin \mathrm{i}\\ \Rightarrow \frac{{\mathrm{\mu }}_{\mathrm{w}}}{{\mathrm{\mu }}_{\mathrm{g}}}\times \frac{{\mathrm{\mu }}_{\mathrm{\alpha }}}{{\mathrm{\mu }}_{\mathrm{w}}}=\sin \mathrm{i}\Rightarrow \sin \mathrm{i}\Rightarrow {\mathrm{\mu }}_{\mathrm{g}}=\frac{1}{\sin \mathrm{i}}\end{array}$