The CRITICAL ANGLE for total internal reflection at the boundary between a denser medium (refractive index $n$) and a rarer medium (refractive index 1, e.g. air) is

1. TIR occurs when light goes from denser to rarer medium at an angle of incidence GREATER than the critical angle. 2. At the critical angle, the refracted ray grazes the boundary ($\theta_2 = 90^\circ$). 3. Apply Snell's law: $n\sin\theta_c = 1\cdot\sin 90^\circ = 1$. 4. So $\sin\theta_c = 1/n$. 5. Example: for water ($n = 4/3$), $\sin\theta_c = 3/4$, giving $\theta_c \approx 48.6^\circ$. For glass ($n = 1.5$), $\sin\theta_c \approx 0.67$, giving $\theta_c \approx 42^\circ$. 6. Options A, C, D have wrong functional dependence on $n$. _Source: NCERT Class 12 Physics Part 2, Ch 9, §9.4 (Total Internal Reflection, Eq. 9.7), p. 6–8._