Question:medium

\(\text{ Light is incident on an interface between water } (\mu = \frac{4}{3}) \text{ and glass } (\mu = \frac{3}{2}).\\\) \(\text{For total internal reflection, light should be traveling from:}\)

Updated On: Mar 27, 2026
  • \(\quad \text{water to glass and } \angle i > \angle i_c \\\)
  • \(\quad \text{water to glass and } \angle i < \angle i_c \\\)
  • \(\quad \text{glass to water and } \angle i < \angle i_c \\\)
  • \(\quad \text{glass to water and } \angle i > \angle i_c\)
Show Solution

The Correct Option is D

Solution and Explanation

Total internal reflection at an interface between two media occurs when light travels from a higher refractive index medium to a lower refractive index medium, and the angle of incidence exceeds the critical angle. The critical angle is determined by the refractive indices of the two media.

Given:

1. Refractive index of water, \(\mu_{\text{water}}=\frac{4}{3}\).

2. Refractive index of glass, \(\mu_{\text{glass}}=\frac{3}{2}\).

Since \(\mu_{\text{glass}} > \mu_{\text{water}}\), total internal reflection can only happen when light moves from glass to water.

The condition for total internal reflection is \(\angle i > \angle i_c\). The critical angle \(\angle i_c\) is calculated as:

\[\sin(\angle i_c)=\frac{\mu_{\text{water}}}{\mu_{\text{glass}}}=\frac{4}{3}/\frac{3}{2}=\frac{8}{9}\]

Thus, \(\angle i_c\) is the angle whose sine is \(\frac{8}{9}\). Total internal reflection requires light to travel from glass to water with an angle of incidence \(\angle i\) greater than \(\angle i_c\).

\(\quad \text{Condition: light travels from glass to water and } \angle i > \angle i_c\)

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