A point source S emits unpolarized light uniformly in all directions. At two points A and B, the ratio $r = I_A/I_B$ of the intensities of light is 2. If a set of two polaroids having $45^\circ$ angle between their pass-axes is placed just before point B, then the new value of $r$ will be _____....

The correct numerical answer is 8. Click to view the complete step-by-step calculation.

JEE Advanced 2024 Paper 1 Physics Question 10 Solution

Step 1: Initial state. $r = \frac{I_A}{I_B} = 2 \implies I_A = 2 I_B$ .

Step 2: Effect of polaroids at B. Light from source is unpolarized with intensity $I_B$ . After first polaroid: $I_1 = \frac{I_B}{2}$ . After second polaroid (at angle $\theta = 45^\circ$ relative to first): $I'_B = I_1 \cos^2(45^\circ)$ . $I'_B = \frac{I_B}{2} \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{I_B}{2} \times \frac{1}{2} = \frac{I_B}{4}$

Step 3: New ratio $r'$ . $r' = \frac{I_A}{I'_B} = \frac{2 I_B}{I_B/4} = 2 \times 4 = 8$ . Final answer is 8.

Question 10

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