The critical angle is maximum when light travels from

water to air

The correct answer is Option (4) → water to air

Critical angle $\theta_c$ is given by:

$\sin \theta_c = \frac{n_2}{n_1}$, where $n_1$ is refractive index of initial medium and $n_2$ of the second medium ($n_1 > n_2$)

The critical angle is maximum when $n_2/n_1$ is maximum, i.e., when the difference between $n_1$ and $n_2$ is minimum.

Note: Among the options, air to water: $n_1 = 1$ (air), $n_2 = 1.33$ (water) → invalid (light must go from denser to rarer medium).

Water to air: $n_1 = 1.33$, $n_2 = 1$ → $\sin \theta_c = 1/1.33 = 0.7519$ → largest possible among the given options.

Critical angle is maximum when light travels from water to air