QJEEM25Phad27: JEE Main 2025 Physics - Class 12 JEEM25Phad27 Solution: The refractive index of the material of a glass prism is . The…

$\textbf{Step 1: Symmetry at Minimum Deviation}$

In a prism, when the deviation is minimum, the path of light is symmetric. This means that the angle of incidence ( $i$ ) is equal to the angle of emergence ( $i'$ ), and the light inside the prism makes equal angles with the prism faces. If the prism angle is $A$ , then the refracted angle at each interface is:

$r = \frac{A}{2}$

$\textbf{Step 2: Relating Deviation to the Angles}$

The formula for the deviation ( $D$ ) in a prism is given by:

$D = i + i' - A$

At minimum deviation, $i = i'$ , so:

$D_{\text{min}} = 2i - A$

The special condition given is that the minimum deviation is equal to the prism angle:

$D_{\text{min}} = A$

Thus:

$A = 2i - A \quad \Longrightarrow \quad 2i = 2A \quad \Longrightarrow \quad i = A$

$\textbf{Step 3: Applying Snell's Law}$

At the first surface, Snell's law gives:

$\sin i = \mu \sin r$

Substitute the values $i = A$ and $r = \frac{A}{2}$ :

$\sin A = \mu \sin\frac{A}{2}$

Given that the refractive index $\mu = \sqrt{3}$ , we have:

$\sin A = \sqrt{3} \sin\frac{A}{2}$

$\textbf{Step 4: Solving the Equation}$

Recall the double-angle formula for sine:

$\sin A = 2 \sin\frac{A}{2} \cos\frac{A}{2}$

Substitute this into the previous equation:

$2 \sin\frac{A}{2} \cos\frac{A}{2} = \sqrt{3} \sin\frac{A}{2}$

Assuming $\sin\frac{A}{2} \neq 0$ , we can divide both sides by $\sin\frac{A}{2}$ :

$2 \cos\frac{A}{2} = \sqrt{3}$

Solve for $\cos\frac{A}{2}$ :

$\cos\frac{A}{2} = \frac{\sqrt{3}}{2}$

Since:

$\cos 30^\circ = \frac{\sqrt{3}}{2}$

It follows:

$\frac{A}{2} = 30^\circ \quad \Longrightarrow \quad A = 60^\circ$

$\textbf{Final Answer:}$

The angle of the prism is $60^\circ$ .