Physics - Geometrical Optics Question with Solution | TestHub

Relative refractive index can be found as,

$\frac{{\mu }_2}{{\mu }_1}=\frac{v_1}{v_2}=\frac{1.5}{2}=\frac{3}{4}$.

When the angle of incident is greater than the critical angle total internal reflection occurs. If the critical angle is $i_c$ then,

$\sin i_c=\frac{{\mu }_2}{{\mu }_1}=\frac{3}{4}$

Now $\tan i_c=\frac{3}{\sqrt{4^2-3^2}}=\frac{3}{\sqrt{7}}$

Therefore, $i_c={\tan }^{-1}\left(\frac{3}{\sqrt{7}}\right)$