Question

Three identical polaroids \(P_1\), \(P_2\), and \(P_3\) are placed one after another. The pass axis of \(P_2\) and \(P_3\) are inclined at angles of \(60^\circ\) and \(90^\circ\) with respect to the axis of \(P_1\). The source \(S\) has an intensity of \(\frac{256 \, \text{W}}{\text{m}^2}\). The intensity of light at point \(O\) is _____ \(\frac{\text{W}}{\text{m}^2}\).

Hint:

For intensity through multiple polaroids:

• After the first polaroid, intensity is reduced to half.
• Apply Malus’s Law \(I = I_0 \cos^2 \theta\) successively for each additional polaroid.
• The angle used in Malus’s Law is the relative angle between the polaroids’ pass axes.

Answer 1

Correct Answer: 24

To determine the intensity of light at point \(O\) after passing through the polaroids, we utilize Malus's law. This law states that the intensity \(I\) of polarized light after passing through a polarizer is given by:

\(I = I_0 \cos^2 \theta\)

where \(I_0\) is the initial intensity, and \(\theta\) is the angle between the light's polarization direction and the polarizer's axis.

1. Initial Intensity: \(I_0 = \frac{256 \, \text{W}}{\text{m}^2}\).
2. Passing through \(P_1\): Since the light is initially unpolarized, intensity after \(P_1\) becomes half:
   \(I_1 = \frac{I_0}{2} = \frac{256}{2} = 128 \, \text{W/m}^2\).
3. Passing through \(P_2\): The angle with \(P_1\) is \(60^\circ\):
   \(I_2 = I_1 \cos^2 60^\circ = 128 \times \left(\frac{1}{2}\right)^2 = 32 \, \text{W/m}^2\).
4. Passing through \(P_3\): The angle with \(P_2\) is \(90^\circ - 60^\circ = 30^\circ\):
   \(I_3 = I_2 \cos^2 30^\circ = 32 \times \left(\frac{\sqrt{3}}{2}\right)^2 = 24 \, \text{W/m}^2\).

The intensity of light at point \(O\) is 24 \(\frac{\text{W}}{\text{m}^2}\), which falls within the range (24, 24).