JEE Physics 2023 Question with Solution

Given: The refractive index is $\mu = \sqrt{2}$ and the prism is equilateral, so the prism angle is $A = 60^\circ$.

Find: The angle of minimum deviation $\delta_m$.

For a prism,

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

Substituting the given values,

$$\sqrt{2} = \frac{\sin\left(\frac{60^\circ+\delta_m}{2}\right)}{\sin 30^\circ}$$

Since

$$\sin 30^\circ = \frac{1}{2}$$

we get

$$\sqrt{2} = 2\sin\left(\frac{60^\circ+\delta_m}{2}\right)$$

So,

$$\sin\left(\frac{60^\circ+\delta_m}{2}\right) = \frac{1}{\sqrt{2}}$$

Hence,

$$\frac{60^\circ+\delta_m}{2} = 45^\circ$$

Therefore,

$$60^\circ + \delta_m = 90^\circ$$

and

$$\delta_m = 30^\circ$$

Therefore, the angle of minimum deviation is $30^\circ$.

Given: $\mu = \sqrt{2}$ and for an equilateral prism $A = 60^\circ$.

Find: $\delta_m$.

Using

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

for $A = 60^\circ$, the denominator becomes $\sin 30^\circ = \frac{1}{2}$. So if $\mu = \sqrt{2}$, then the numerator must be $\frac{1}{\sqrt{2}}$, which corresponds to $45^\circ$. Thus,

$$\frac{60^\circ+\delta_m}{2} = 45^\circ$$

which gives

$$\delta_m = 30^\circ$$

The shortcut works because the equilateral prism fixes $A = 60^\circ$, making the denominator a standard trigonometric value.