A ray of light passing through an equilateral prism is having velocity 2.12 × 10^8 m/s in the prism material, then the minimum angle of deviation is _________ degrees.

Correct answer: B. 30 Velocity of light in vacuum, c = 3 × 10^8 m/s Velocity of light in the prism, v = 2.12 × 10^8 m/s Refractive index of the prism material, μ = (c)/(v) = (3 × 10^8)/(2.12 × 10^8) 1.414 2 For an equilateral prism, the angle of the prism A = 60^ . Using the prism formula: μ = ( ((A + _m)/(2)))/( ((A)/(2))) 2 = ( ((60^ + _m)/(2)))/( (30^ )) 2 = ( (30^ + ( _m)/(2)))/((1)/(2)) (30^ + ( _m)/(2)) = 2 2 = 1 2 30^ + ( _m)/(2) = 45^ ( _m)/(2) = 15^ _m = 30^ Answer: 30

JEE Main 20265 AprilMorning Shift

Exam: JEE Main
Year: 2026
Date: 5 April
Shift: Morning Shift
Subject: Physics
Chapter: Ray Optics

JEE Main 2026 · 5 April · Morning Shift · Physics · Ray Optics

A ray of light passing through an equilateral prism is having...

A ray of light passing through an equilateral prism is having velocity

2.12 \times 10^8

m/s in the prism material, then the minimum angle of deviation is _________ degrees.

Options

A $45$
B $30$
C $28$
D $58$

Correct answer

B. $30$

Step-by-step solution

Velocity of light in vacuum,

c = 3 \times 10^8

m/s

Velocity of light in the prism,

v = 2.12 \times 10^8

m/s

Refractive index of the prism material,

\mu = \dfrac{c}{v} = \dfrac{3 \times 10^8}{2.12 \times 10^8} \approx 1.414 \approx \sqrt{2}

For an equilateral prism, the angle of the prism

A = 60^\circ

.

Using the prism formula:

\mu = \dfrac{\sin\left(\dfrac{A + \delta_m}{2}\right)}{\sin\left(\dfrac{A}{2}\right)}

\sqrt{2} = \dfrac{\sin\left(\dfrac{60^\circ + \delta_m}{2}\right)}{\sin(30^\circ)}

\sqrt{2} = \dfrac{\sin\left(30^\circ + \dfrac{\delta_m}{2}\right)}{\dfrac{1}{2}}

\sin\left(30^\circ + \dfrac{\delta_m}{2}\right) = \dfrac{\sqrt{2}}{2} = \dfrac{1}{\sqrt{2}}

30^\circ + \dfrac{\delta_m}{2} = 45^\circ

\dfrac{\delta_m}{2} = 15^\circ

\delta_m = 30^\circ

Answer:

30

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