One face of a prism of refractive index 1.5 and angle 75° is covered with a liquid of refractive index $\frac{3\sqrt{2}}{4}$. What should be the angle of incidence of light on the clear face of prism for which light is just totally reflected at the liquid covered face?

$\sin^{-1}(\frac{3}{4})$

The correct answer is Option (3) → $\sin^{-1}(\frac{3}{4})$

The critical angle is -

$\sin θ_c=\frac{μ_{liquid}}{μ_{prism}}=\frac{\sqrt{4/3}}{1.5}≃0.77$

Using Snell's law

$μ_{air}\sin i=μ_{prism}\sin r$

$⇒\sin i=1.5 \sin r$

$⇒\sin i=1.5×\sin(75°-50°)$

$≃0.631$

$⇒i=\sin^{-1}(\frac{3}{4})$