JEE Main 2026 — Physics Optics

2026mcqmedium

As shown in the diagram, when the incident ray is parallel to base of the prism, the emergent ray grazes along the second surface.

JEE Main 2026 Physics, Ray Optics — question figure

If refractive index of the material of prism is $\sqrt{2}$ , the angle $\theta$ of prism is.

Official previous-year question

Held on 21 Jan 2026 · Verified 6 Jul 2026.

Options

A
$60^{\circ}$
B
$45^{\circ}$
C
$90^{\circ}$
D
$75^{\circ}$

Did you get this right?

Solution

From the geometry of the prism, the angle of the prism is $A = 180^\circ - (45^\circ + \theta)$ .

The incident ray is parallel to the base, so the angle of incidence $i$ at the first surface is equal to the base angle $45^\circ$ . Thus, $i = 45^\circ$ .

Applying Snell's law at the first surface: $1 \cdot \sin(45^\circ) = \mu \sin(r_1)$ .

Given $\mu = \sqrt{2}$ , we have $\frac{1}{\sqrt{2}} = \sqrt{2} \sin(r_1) \implies \sin(r_1) = \frac{1}{2} \implies r_1 = 30^\circ$ .

At the second surface, the emergent ray grazes the surface, so the angle of emergence $e = 90^\circ$ .

Applying Snell's law at the second surface: $\mu \sin(r_2) = 1 \cdot \sin(90^\circ)$ .

$\sqrt{2} \sin(r_2) = 1 \implies \sin(r_2) = \frac{1}{\sqrt{2}} \implies r_2 = 45^\circ$ .

The relation between prism angle and internal angles is $A = r_1 + r_2$ .

$A = 30^\circ + 45^\circ = 75^\circ$ .

Substituting $A$ in the first equation: $75^\circ = 180^\circ - (45^\circ + \theta)$ .

$75^\circ = 135^\circ - \theta \implies \theta = 135^\circ - 75^\circ = 60^\circ$ .

Your note

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