JEE Physics 2025 Question with Solution

Step-by-step solution

Standard Method

Given: Prism angle $A = 60^\circ$ and refractive index $\mu = \sqrt{2}$ under minimum deviation.

Find: The angle of incidence $i$.

For minimum deviation, the angle of incidence is equal to the angle of emergence, so $i = e$.

Use the relation

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

Since $A = 60^\circ$,

$$\therefore \delta_m = 30^\circ$$

Also,

$$\delta_m = 2i - A \quad [\text{as} \ i = e]$$

Therefore,

$$30^\circ = 2i - 60^\circ$$ $$2i = 90^\circ$$ $$i = 45^\circ$$

Therefore, the angle of incidence is $45^\circ$.

Using minimum deviation condition

Given: $A = 60^\circ$, $\mu = \sqrt{2}$.

Find: $i$ when deviation is minimum.

At minimum deviation for a prism,

$$r_1 = r_2 = \frac{A}{2} = 30^\circ$$

and

$$i = e$$

Using Snell's law in prism form,

$$\mu = \frac{\sin i}{\sin r}$$

So,

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

Hence,

$$i = 45^\circ$$

Therefore, the required angle of incidence is $45^\circ$.

Common mistakes

• Using the general prism relation without applying the minimum deviation condition. At minimum deviation, $i = e$ and $r_1 = r_2$. Always use the symmetric path condition first.

• Taking $r = 60^\circ$ instead of $r = \frac{A}{2} = 30^\circ$. The prism angle is split equally inside the prism only at minimum deviation.

• Substituting incorrectly in $\mu = \frac{\sin i}{\sin r}$ by forgetting that $\sin 30^\circ = \frac{1}{2}$. This leads to a wrong value of $i$. Evaluate the trigonometric value carefully before solving.