The refractive index of prism is \mu=\sqrt3 and the ratio of the angle of minimum deviation to the angle of prism is one. The value of angle of prism is \underline\hspace2cm ^\circ .

Geometrical Optics PYQ — JEE Main 2024 · 6 Apr · Shift 1 · Q87 with Solution

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Correct answer: 60

Given data
- Refractive index of prism: $\mu = \sqrt{3}$
- Ratio of angle of minimum deviation to angle of prism is one: $\frac{\delta_m}{A}=1 \implies \delta_m=A$
Formula for refractive index at minimum deviation

For a prism in air, $\mu = \frac{\sin\left(\frac{A+\delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$
Use the condition $\delta_m = A$

Substitute into the formula:
$= \frac{\sin A}{\sin\left(\frac{A}{2}\right)}$$ Since $\sin A = 2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)$, $$\mu = \frac{2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = 2\cos\left(\frac{A}{2}\right)$$$
Substitute $\mu = \sqrt{3}$

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

$\frac{A}{2} = 30^\circ$ $A = 60^\circ$
Final answer

The angle of prism is: $\boxed{60^\circ}$
Comparison with stored correct answer

Stored correct answer = $60$

Our derived answer matches the stored answer.