JEE Physics 2026 Question with Solution

Given: An equilateral prism with refractive index $\mu = \sqrt{2}$.

Find: The angle of refraction at the incident surface, $r_1$.

For an equilateral prism, the prism angle is

$$A = 60^\circ$$

When the emergent ray grazes along the surface, the angle of emergence is

$$e = 90^\circ$$

So the angle of refraction at the second surface equals the critical angle:

$$r_2 = C$$

Using the critical angle condition,

$$\sin C = \frac{1}{\mu} = \frac{1}{\sqrt{2}}$$

Therefore,

$$C = 45^\circ$$

Hence,

$$r_2 = 45^\circ$$

For a prism,

$$r_1 + r_2 = A$$

So,

$$r_1 = A - r_2 = 60^\circ - 45^\circ = 15^\circ$$

Therefore, the angle of refraction at the incident surface is $15^\circ$. The correct option is A.

Given: The prism is equilateral, so $A = 60^\circ$, and its refractive index is $\sqrt{2}$.

Find: The refracted angle at the first surface.

If the ray emerges grazing the second surface, then at that surface the refracted ray makes $90^\circ$ with the normal outside. That means the internal angle of incidence there is the critical angle.

From

$$\sin C = \frac{1}{\mu}$$

we get

$$\sin C = \frac{1}{\sqrt{2}}$$

and hence

$$C = 45^\circ$$

Thus the internal refraction angle at the second face is $r_2 = 45^\circ$.

Now use prism geometry:

$$r_1 + r_2 = A$$

So,

$$r_1 + 45^\circ = 60^\circ$$

Therefore,

$$r_1 = 15^\circ$$

Thus, the required angle is $15^\circ$.