Question

Two polaroids are placed in the path of unpolarised beam of intensity ' $\text{I}_0$ ' such that no light is emitted from the second polaroid. If a third polaroid whose polarisation axis makes an angle ' $\theta$ ' with the polarisation axis of first polaroid is placed between these polaroids then the intensity of light emerging from the last polaroid will be}

• A. $\frac{\text{I}_0}{4}(\sin 2\theta)^2$
• B. $\frac{\text{I}_0}{8}(\sin 2\theta)^2$
• C. $\frac{\text{I}_0}{4}\sin^2 \theta$
• D. $\frac{\text{I}_0}{8}\sin^2 \theta$

Hint:

$2\sin\theta\cos\theta = \sin 2\theta$.

Answer 1

The Correct Option is B

Step 1: Concept
Law of Malus: $I = I_{incident} \cos^2 \phi$.
Step 2: Analysis
1. After 1st polaroid: $I_1 = I_0/2$.
2. After 3rd (middle) polaroid (at $\theta$): $I_2 = I_1 \cos^2 \theta = \frac{I_0}{2} \cos^2 \theta$.
3. 2nd polaroid is at $90^\circ$ to the 1st. Angle between 3rd and 2nd is $(90 - \theta)$.
Step 3: Calculation
$I_3 = I_2 \cos^2(90 - \theta) = \frac{I_0}{2} \cos^2 \theta \sin^2 \theta$
$I_3 = \frac{I_0}{2} (\sin \theta \cos \theta)^2 = \frac{I_0}{2} \left( \frac{\sin 2\theta}{2} \right)^2 = \frac{I_0}{8} \sin^2 2\theta$.
Step 4: Conclusion
Hence, the final intensity is $\frac{I_0}{8}(\sin 2\theta)^2$.
Final Answer: (B)