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

<p>The critical angle is the angle of incidence at which light is refracted along the boundary, meaning the angle of refraction is $90^{\circ}$. The relationship between the critical angle and the refractive indices of two media can be understood using Snell's Law, given as:</p> <p>$n_1 \sin(\theta_c) = n_2 \sin(90^{\circ})$</p> <p>Here, $n_1$ is the refractive index of the first medium, $n_2$ is the refractive index of the second medium, and $\theta_c$ is the critical angle. Since $\sin(90^{\circ}) = 1$, the equation simplifies to:</p> <p>$n_1 \sin(\theta_c) = n_2$</p> <p>Given that the critical angle $\theta_c$ is $45^{\circ}$, we have:</p> <p>$\sin(45^{\circ}) = \frac{\sqrt{2}}{2}$</p> <p>Substituting this value in the simplified Snell's Law equation, we get:</p> <p>$n_1 \left( \frac{\sqrt{2}}{2} \right) = n_2$</p> <p>Therefore,</p> <p>$n_1 = n_2 \cdot \frac{2}{\sqrt{2}}$</p> <p>$n_1 = n_2 \cdot \sqrt{2}$</p> <p>Hence, the ratio of the refractive indices of the first and second media is:</p> <p>$n_1 : n_2 = \sqrt{2} : 1$</p> <p>The correct answer is therefore:</p> <p><b>Option D</b></p> <p>$\sqrt{2}: 1$</p>