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Geometrical Optics question

2025 · 23 Jan · Shift 2 · Q63

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Geometrical Optics
2025 · 23 Jan · Shift 2 · Q63

Geometrical Optics question

2025 · 23 Jan · Shift 2 · Q63

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1

The refractive index of the material of a glass prism is

\sqrt{3}

. The angle of minimum deviation is equal to the angle of the prism. What is the angle of the prism?

A
$60^{\circ}$
B
$50^{\circ}$
C
$58^{\circ}$
D
$48^{\circ}$

View written solutionFree

Correct answer: A

Use the prism formula at minimum deviation

For a prism in air, at minimum deviation $\delta_m$ ,

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

where:

$\mu$ = refractive index of prism material
$A$ = angle of prism
$\delta_m$ = angle of minimum deviation

Given condition

It is given that

\mu = \sqrt{3}, \qquad \delta_m = A

Substitute $\delta_m = A$ into the formula:

\sqrt{3} = \frac{\sin\left(\frac{A+A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin A}{\sin\left(\frac{A}{2}\right)}

Simplify using identity

Using

\sin A = 2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)

we get

\sqrt{3} = \frac{2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)}

\sqrt{3} = 2\cos\left(\frac{A}{2}\right)

So,

\cos\left(\frac{A}{2}\right) = \frac{\sqrt{3}}{2}

Find $A$

\frac{A}{2} = 30^\circ

Therefore,

A = 60^\circ

Check options

A: $60^\circ$ ✅
B: $50^\circ$
C: $58^\circ$
D: $48^\circ$

Hence, the correct option is A.

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