JEE Physics 2023 Question with Solution

Step-by-step solution

Standard Method

Given: thickness of glass plate $t = \sqrt{5} \, cm$, refractive index $\mu = \sqrt{2}$, and angle of incidence is equal to the critical angle.

Find: lateral displacement of the ray through the plate.

For the glass-air interface, the critical angle $\theta_c$ is given by

$$\sin \theta_c = \frac{1}{\mu}$$

So,

$$\sin \theta_c = \frac{1}{\sqrt{2}} \implies \theta_c = 45^\circ$$

Hence the angle of incidence is $i = 45^\circ$.

Using Snell's law at the first surface,

$$\sin r = \frac{\sin 45^\circ}{\sqrt{2}} = \frac{1/\sqrt{2}}{\sqrt{2}} = \frac{1}{2}$$

Therefore,

$$r = 30^\circ$$

The lateral displacement through a parallel-sided slab is

$$x = t \sin(i-r) \sec r$$

Substituting the values,

$$x = \sqrt{5} \times \sin(45^\circ - 30^\circ) \times \sec 30^\circ$$ $$x = \sqrt{5} \times \sin 15^\circ \times \sec 30^\circ$$

Using $\sin 15^\circ = 0.26$ and $\sec 30^\circ = \frac{2}{\sqrt{3}}$,

$$x = \sqrt{5} \times 0.26 \times \frac{2}{\sqrt{3}}$$ $$x = \frac{\sqrt{5} \times 0.52}{\sqrt{3}} = 0.52 \, cm$$

Therefore, the lateral displacement is $0.52 \, cm$. The correct option is A.

Common mistakes

• Using the critical angle formula incorrectly as $\sin \theta_c = \mu$. This is wrong because for glass to air, $\sin \theta_c = \frac{1}{\mu}$. First find $\theta_c$ from the reciprocal of refractive index.

• Taking the refracted angle inside the slab as $45^\circ$. This is wrong because Snell's law must still be applied at the first surface. Use $\sin r = \frac{\sin i}{\mu}$ to get $r = 30^\circ$.

• Using an incorrect slab formula for lateral displacement. The displacement is not simply $t \tan(i-r)$ here. Use the standard relation $x = t \sin(i-r) \sec r$ for a parallel glass plate.