Critical angle of incidence for a pair of optical media is $$45°$$. The refractive indices of first and second media are in the ratio:

The critical angle is $$45°$$. Find the ratio of refractive indices $$n_1 : n_2$$.

Total internal reflection occurs when light travels from a denser to a rarer medium. The critical angle $$C$$ is given by:

$$\sin C = \frac{n_2}{n_1}$$

where $$n_1$$ is the refractive index of the denser (first) medium and $$n_2$$ is that of the rarer (second) medium.

$$\sin 45° = \frac{n_2}{n_1} \implies \frac{1}{\sqrt{2}} = \frac{n_2}{n_1}$$

$$\frac{n_1}{n_2} = \sqrt{2}$$

$$n_1 : n_2 = \sqrt{2} : 1$$

The correct answer is Option (2): $$\sqrt{2} : 1$$.