A layer of benzene floats on water. If the angle of incidence of the light entering benzene from air is 60°, then the angle that light makes with normal in water is (Refractive index of water = 4/3)

$\sin^{-1}(\frac{3\sqrt{3}}{8})$

Here one might be tempted to think that since the refractive index of benzene is not given, (D) is correct. But actually we do not need this data.

$μ_1\, \sin i_1 = μ_2\, \sin i_2 = μ_3\, \sin i_3$

$∴ \sin i_3 = \frac{μ_1\, \sin i_1}{μ_3}=\frac{3\sqrt{3}}{8}$