JEE Physics 2026 Question with Solution

Given: For the prism, the angle of minimum deviation equals the refracting angle, so $\delta = A$.

Find: The range of the refractive index $n$.

Use the prism formula:

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

Substitute $\delta=A$:

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

Using

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

we get

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

For a prism,

$$0$$

Given: $\delta=A$.

Find: Which option matches the refractive index range.

At minimum deviation,

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

Putting $\delta=A$,

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

Now use the identity

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

So,

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

Since $0<\frac{A}{2}<\frac{\pi}{2}$, the cosine is positive and less than $1$. Thus $n<2$.

The solution concludes with the final result

$$\sqrt{2}$$