One face of a prism of refractive index 1.5 and angle 75° is covered with a liquid of refractive index \( \displaystyle\frac{3\sqrt{2}}{4} \). 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?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \sin^{-1}\left(\displaystyle\frac{3}{4}\right) \)
Light is just totally reflected at the liquid-covered face when its angle of incidence there equals the critical angle C for the glass-liquid boundary. Then the angle of refraction at the first face follows from \( r_1 + r_2 = A \), and Snell’s law gives the angle of incidence.
Critical angle at the second face
\( \displaystyle \sin C = \frac{\mu_{\mathrm{liquid}}}{\mu_{\mathrm{glass}}} = \frac{3\sqrt{2}/4}{1.5} = \frac{\sqrt{2}}{2} \Rightarrow C = 45^\circ \)
So \( r_2 = 45^\circ \).
Refraction at the clear face
\( r_1 = A - r_2 = 75^\circ - 45^\circ = 30^\circ \)
\( \sin i = \mu\sin r_1 = 1.5\times\sin30^\circ = 1.5\times0.5 = \displaystyle\frac{3}{4} \)
\( i = \sin^{-1}\left(\displaystyle\frac{3}{4}\right) \)