Light is incident on an interface between water \( (\mu = 4/3) \) and glass \( (\mu = 3/2) \). For total internal reflection, light should be travelling from :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → glass to water and \( \angle i \gt \angle i_c \)
Total internal reflection needs two conditions: the light must go from the optically denser to the rarer medium, and the angle of incidence must be greater than the critical angle.
Glass \( (\mu = 3/2) \) is denser than water \( (\mu = 4/3) \), so the light must travel from glass to water, with \( \angle i \gt \angle i_c \). For this pair,
\( \displaystyle \sin i_c = \frac{\mu_{\mathrm{water}}}{\mu_{\mathrm{glass}}} = \frac{4/3}{3/2} = \frac{8}{9} \;\Rightarrow\; i_c \approx 62.7^\circ \)
Option (1) water to glass and \( \angle i \gt \angle i_c \) — Incorrect. Going from water to glass is from rarer to denser; the light bends towards the normal and is always partly refracted.
Option (2) water to glass and \( \angle i \lt \angle i_c \) — Incorrect. Wrong direction of travel, and an angle below the critical angle gives refraction anyway.
Option (3) glass to water and \( \angle i \lt \angle i_c \) — Incorrect. The direction is right, but for \( \angle i \lt \angle i_c \) the light is refracted into water.
Option (4) glass to water and \( \angle i \gt \angle i_c \) — Correct. Denser to rarer, with the angle of incidence above the critical angle.
NCERT (2026-27):
Summary – “… The critical angle of incidence ic for a ray incident from a denser to rarer medium, is that angle for which the angle of refraction is 90°. For i > ic, total internal reflection occurs.”