Question

A source of light is placed in front of a screen. Intensity of light on the screen is I. Two Polaroids P\(_1\) and P\(_2\) are so placed in between the source of light and screen that the intensity of light on screen is I/2. P\(_2\) should be rotated by an angle of __________ (degrees) so that the intensity of light on the screen becomes \(\frac{3I}{8}\).

Hint:

The key to solving problems with multiple polaroids is to apply the rules sequentially. First polaroid halves the intensity of unpolarized light. Subsequent polaroids follow Malus's Law, \(I_{out} = I_{in} \cos^2\theta\), where \(\theta\) is the angle between the polarization of the incoming light and the axis of the polaroid.

Answer 1

Correct Answer: 30

Step 1: Understanding the Question:
This problem involves the change in intensity of light after passing through two polaroids. We need to find the angle of rotation of the second polaroid to achieve a specific final intensity. Let's assume the initial light from the source is unpolarized.
Step 2: Key Formula or Approach:
1. When unpolarized light of intensity I\(_0\) passes through a polaroid, the intensity of the transmitted light is I\(_1\) = I\(_0\)/2.
2. When polarized light of intensity I\(_1\) passes through a second polaroid (analyzer) whose pass axis is at an angle \(\alpha\) with the first, the final intensity is given by Malus's Law: I\(_2\) = I\(_1\) cos\(^2\alpha\).
Step 3: Detailed Explanation:
Let the intensity of the unpolarized light from the source be `I`.
After passing through the first polaroid, P\(_1\), the intensity becomes:
\[ I_1 = \frac{I}{2} \] This light is now plane-polarized. It then passes through the second polaroid, P\(_2\). Let the angle between the pass axes of P\(_1\) and P\(_2\) be \(\alpha\). The final intensity on the screen is:
\[ I_{final} = I_1 \cos^2\alpha = \left(\frac{I}{2}\right) \cos^2\alpha \] Initial Condition:
We are given that initially, the final intensity is I/2.
\[ \frac{I}{2} = \left(\frac{I}{2}\right) \cos^2\alpha \] This implies \( \cos^2\alpha = 1 \), which means \( \cos\alpha = \pm 1 \). So, the initial angle between the polaroids is \(\alpha = 0^\circ\) (or 180\(^\circ\)). They are parallel.
After Rotation:
Now, the polaroid P\(_2\) is rotated by an angle \(\theta\). The new angle between the pass axes of P\(_1\) and P\(_2\) is \( \alpha' = \alpha + \theta = 0 + \theta = \theta \).
The new final intensity, I'\(_{final}\), is given as \(\frac{3I}{8}\).
Using Malus's Law again:
\[ I'_{final} = I_1 \cos^2\theta = \left(\frac{I}{2}\right) \cos^2\theta \] \[ \frac{3I}{8} = \left(\frac{I}{2}\right) \cos^2\theta \] Cancel `I` from both sides:
\[ \frac{3}{8} = \frac{1}{2} \cos^2\theta \] \[ \cos^2\theta = 2 \times \frac{3}{8} = \frac{6}{8} = \frac{3}{4} \] \[ \cos\theta = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \] The angle \(\theta\) for which \( \cos\theta = \frac{\sqrt{3}}{2} \) is \( \theta = 30^\circ \).
Step 4: Final Answer:
P\(_2\) should be rotated by an angle of 30 degrees.