JEE Main 2026 — Physics Optics

2026mcqmedium

The exit surface of a prism with refractive index $n$ is coated with a material having refractive index $\frac{n}{2}$ . When this prism is set for minimum angle of deviation, it exactly meets the condition of critical angle. The prism angle is $\_\_\_\_$ .

Official previous-year question

Held on 24 Jan 2026 · Verified 6 Jul 2026.

Options

A
$30^{\circ}$
B
$15^{\circ}$
C
$45^{\circ}$
D
$60^{\circ}$

Did you get this right?

Solution

Let the refractive index of the prism be $n_1 = n$ and the refractive index of the coating on the exit surface be $n_2 = \frac{n}{2}$ .

At the condition of minimum deviation, the angle of refraction at the first surface $r_1$ and the angle of incidence at the second surface $r_2$ are equal, i.e., $r_1 = r_2 = \frac{A}{2}$ , where $A$ is the prism angle.

The problem states that at the exit surface, the light meets the condition of critical angle. Therefore, $r_2 = \theta_c$ .

The critical angle $\theta_c$ for the interface between the prism and the coating is given by $\sin \theta_c = \frac{n_2}{n_1}$ .

Substituting the values, $\sin \theta_c = \frac{n/2}{n} = \frac{1}{2}$ .

Thus, $\theta_c = 30^\circ$ .

Since $r_2 = \theta_c$ , we have $r_2 = 30^\circ$ .

Using the relation $r_1 = r_2 = \frac{A}{2}$ for minimum deviation, we get $\frac{A}{2} = 30^\circ$ .

Therefore, $A = 60^\circ$ .

Your note

JEE Main Optics in other years

Optics 2025 Optics 2024 Optics 2023 Optics 2022 Optics 2021 Optics 2020 Optics 2019 Optics 2018 Optics 2017 Optics 2016 Optics 2015 Optics 2014

More JEE Main Optics questions

When an unpolarized light falls at a particular angle on a glass plate (placed in air), it is observed that the reflected beam is linearly polarized. The angle of refracted beam with respect to the normal is $\_\_\_\_$. $\left(\tan ^{-1}(1.52)=57.7^{\circ}\right.$, refractive indices of air and glass are 1.00 and 1.52, respectively.)

2026medium

In Young's double slit experiment, the fringe width is β. If the wavelength of light is doubled and the slit separation is halved, the new fringe width is:

2026medium

A biconvex lens is formed by using two thin planoconvex lenses, as shown in the figure. The refractive index and radius of curved surfaces are also mentioned in figure. When an object is placed on the left side of lens at a distance of 30 cm from the biconvex lens, the magnification of the image will be : ![JEE Main 2026 Physics, Ray Optics — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Ray_Optics/6983c1b28bfdbb24356397fd/question_1__q_6983c1b28bfdbb24356397fd__cdn-question-pool.getmarks.app__28S2_q_43_1__a26fa0afbd_final_ppt_sync.png)

2026hard

A collimated beam of light of diameter 2 mm is propagating along $x$-axis. The beam is required to be expanded in a collimated beam of diameter 14 mm using a system of two convex lenses. If first lens has focal length 40 mm, then the focal length of second lens is $\_\_\_\_$ mm.

2026medium

Light ray incident along a vector $\vec{AO}$ $(\vec{AO} = 2\hat{i}-3\hat{j})$ emerges out along vector $\vec{OB}$ $(\vec{OB} = C\hat{i}-4\hat{j})$ as shown in the figure below. The value of $C$ is ________. ![JEE Main 2026 Physics, Ray Optics — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Ray_Optics/69d8073a81239f6567c4f345/question_1__q_69d8073a81239f6567c4f345__cdn-question-pool.getmarks.app__08APRS2__q42_d1__9c03865d34_final_ppt_sync.jpg)

2026medium

View all Optics questions

Keep practicing Optics

Work through every JEE Main Optics PYQ, year by year.

Browse all Physics topics

Solution

Let the refractive index of the prism be $n_1 = n$ and the refractive index of the coating on the exit surface be $n_2 = \frac{n}{2}$ .

The problem states that at the exit surface, the light meets the condition of critical angle. Therefore, $r_2 = \theta_c$ .

The critical angle $\theta_c$ for the interface between the prism and the coating is given by $\sin \theta_c = \frac{n_2}{n_1}$ .

Substituting the values, $\sin \theta_c = \frac{n/2}{n} = \frac{1}{2}$ .

Thus, $\theta_c = 30^\circ$ .

Since $r_2 = \theta_c$ , we have $r_2 = 30^\circ$ .

Using the relation $r_1 = r_2 = \frac{A}{2}$ for minimum deviation, we get $\frac{A}{2} = 30^\circ$ .

Therefore, $A = 60^\circ$ .

JEE Main 2026 — Physics Optics

Options

Did you get this right?

Solution

Your note

JEE Main Optics in other years

More JEE Main Optics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Modern Physics

Thermodynamics

Waves & Oscillations

Keep practicing Optics

JEE Main 2026 — Physics Optics

Options

Did you get this right?

Solution

Your note

JEE Main Optics in other years

More JEE Main Optics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Modern Physics

Thermodynamics

Waves & Oscillations

Keep practicing Optics