JEE Physics 2025 Question with Solution

Given: Refractive index of the prism material is $n = 3$ and the angle of minimum deviation is equal to the angle of the prism, so $\delta_m = A$.

Find: The angle of the prism $A$.

For a prism at minimum deviation,

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

Substitute $\delta_m = A$:

$$n = \frac{\sin\left(\frac{A + A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin A}{\sin\left(\frac{A}{2}\right)}$$

Now use $n = 3$:

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

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

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

the solution concludes that the prism angle is $A = 60^\circ$, which matches option B.

Therefore, the correct option is B.

Given: $n = 3$ and $D_{\text{min}} = A$.

Find: Prism angle $A$.

The prism formula is

$$n = \frac{\sin\left(\frac{A + D_{\text{min}}}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$

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

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

Substituting $n = 3$,

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

From the provided solution, solving this gives

$$A = 60^\circ$$

So the angle of the prism is $60^\circ$ and the correct option is B.