QJEEM24Ph1413: JEE Main 2024 Physics - Class 12 JEEM24Ph1413 Solution: Critical angle of incidence for a pair of optical media is .…

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:

$n_1 \sin(\theta_c) = n_2 \sin(90^{\circ})$

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:

$n_1 \sin(\theta_c) = n_2$

Given that the critical angle $\theta_c$ is $45^{\circ}$ , we have:

$\sin(45^{\circ}) = \frac{\sqrt{2}}{2}$

Substituting this value in the simplified Snell's Law equation, we get:

$n_1 \left( \frac{\sqrt{2}}{2} \right) = n_2$

Therefore,

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

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

Hence, the ratio of the refractive indices of the first and second media is:

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

The correct answer is therefore:

Option D

$\sqrt{2}: 1$