Physics - Geometrical Optics Question with Solution | TestHub

$\therefore \text{sin }{\theta }_{\text{c}}=\frac{1}{\mu }$

$=\frac{3}{4}$

$\therefore \frac{\text{r}}{\sqrt{{\text{h}}^2+{\text{r}}^2}}=\frac{3}{4}$

$\therefore \frac{{\text{r}}^2}{{\text{h}}^2+{\text{r}}^2}=\frac{9}{16}$

$\therefore 16{\text{r}}^2=9{\text{h}}^2+9{\text{r}}^2$

$7{\text{r}}^2=9{\text{h}}^2$

$\text{r}=\text{h}\sqrt{\frac{9}{7}}$

$=\frac{3\text{h}}{\sqrt{7}}$

$=\frac{3}{\sqrt{7}}\times 15\text{ cm}$