Question 76: Ray Optics

Question

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:

Accepted Answer

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:
$$ n_1 \sin(	heta_c) = n_2 \sin(90^{\circ}) $$
Here, $$n_1$$ is the refractive index of the first medium, $$n_2$$ is the refractive index of the second medium, and $$	heta_c$$ is the critical angle. Since $$\sin(90^{\circ}) = 1$$, the equation simplifies to:
$$ n_1 \sin(	heta_c) = n_2 $$
Given that the critical angle $$	heta_c$$ is $$45^{\circ}$$, we have:
$$ \sin(45^{\circ}) = \frac{\sqrt{2}}{2} $$
Substituting this value in the simplified Snell's Law equation, we get:
$$ n_1 \left( \frac{\sqrt{2}}{2} ight) = n_2 $$
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
$$\sqrt{2}: 1$$

Answer

$$1: 2$$

Answer

$$1: \sqrt{2}$$

Answer

$$2: 1$$

Answer

$$\sqrt{2}: 1$$

Ray Optics

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:

Choose the correct answer: