JEE Physics 2023 Question with Solution

Given: Unpolarised light of intensity $I_0 = 32 \, \text{Wm}^{-2}$ passes through three polaroids. The pass axis of the third polaroid is perpendicular to that of the first. The emerging intensity is $I_{\text{net}} = 3 \, \text{Wm}^{-2}$.

Find: The angle $\theta$ between the pass axes of the first two polaroids.

After the first polaroid, the intensity becomes

$$I_1 = \frac{I_0}{2}$$

If the angle between the first and second polaroids is $\theta$, then after the second polaroid,

$$I_2 = \frac{I_0}{2}\cos^2\theta$$

Since the third polaroid is perpendicular to the first, the angle between the second and third polaroids is $90^\circ - \theta$. Therefore, after the third polaroid,

$$I_{\text{net}} = I_2 \cos^2(90^\circ - \theta)$$

Using $\cos(90^\circ - \theta) = \sin\theta$,

$$I_{\text{net}} = \frac{I_0}{2}\cos^2\theta \sin^2\theta$$

Using $\sin^2(2\theta) = 4\cos^2\theta\sin^2\theta$,

$$I_{\text{net}} = \frac{I_0}{8}\sin^2(2\theta)$$

Substitute $I_{\text{net}} = 3$ and $I_0 = 32$:

$$3 = \frac{32}{8}\sin^2(2\theta)$$ $$3 = 4\sin^2(2\theta)$$ $$\sin^2(2\theta) = \frac{3}{4}$$ $$\sin(2\theta) = \frac{\sqrt{3}}{2}$$

Hence,

$$2\theta = 60^\circ \text{ or } 120^\circ$$ $$\theta = 30^\circ \text{ or } 60^\circ$$

Therefore, the angle between the pass axes of the first two polaroids is $30^\circ$ or $60^\circ$.

Given: Initial unpolarised intensity $32 \, \text{Wm}^{-2}$ and final intensity $3 \, \text{Wm}^{-2}$.

Find: Possible values of $\theta$.

1. Unpolarised light passing through the first polaroid gives

$$I_1 = \frac{32}{2} = 16 \, \text{Wm}^{-2}$$

2. After the second polaroid at angle $\theta$,

$$I_2 = 16\cos^2\theta$$

3. The third polaroid is perpendicular to the first, so its angle with the second is $90^\circ - \theta$. Thus,

$$I_{\text{net}} = 16\cos^2\theta \cos^2(90^\circ - \theta)$$ $$I_{\text{net}} = 16\cos^2\theta \sin^2\theta$$

4. Since the final intensity is $3 \, \text{Wm}^{-2}$,

$$3 = 16\cos^2\theta \sin^2\theta$$

5. Rewrite using

$$4\cos^2\theta \sin^2\theta = \sin^2(2\theta)$$

So,

$$3 = 4\sin^2(2\theta)$$ $$\sin^2(2\theta) = \frac{3}{4}$$

6. Therefore,

$$\sin(2\theta) = \frac{\sqrt{3}}{2}$$

which gives

$$2\theta = 60^\circ, 120^\circ$$

Hence,

$$\theta = 30^\circ, 60^\circ$$

So the required numerical values are $30$ and $60$.