JEE Main 2026 — Jan 24, Shift 1 — Physics solution

JEE Main2026Jan 24, Shift 1Physics

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____

A
$6 0^{\circ}$
B
$1 5^{\circ}$
C
$3 0^{\circ}$
D
$4 5^{\circ}$

Solution

At minimum deviation, the two refraction angles inside the prism are equal: $r_{1} = r_{2} = \frac{A}{2}$ .

The exit face is between the prism ( $n$ ) and the coating ( $n /2$ ). The critical angle for this interface satisfies:

$sin θ_{c} = \frac{n /2}{n} = \frac{1}{2} \Rightarrow θ_{c} = 3 0^{\circ}$ .

The problem states the angle of refraction inside the prism at the exit face just equals $θ_{c}$ :

$\frac{A}{2} = 3 0^{\circ} \Rightarrow A = 6 0^{\circ}$ .

← Back to full paper

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →