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

1. Intensity After First Polaroid (\(P_1\)): - When unpolarized light passes through a polaroid, its intensity is reduced by half:

\[ I_1 = \frac{I_0}{2} = \frac{256}{2} = 128 \, \frac{\text{W}}{\text{m}^2}. \]

2. Intensity After Second Polaroid (\(P_2\)): - The intensity after \(P_2\) is given by Malus’s Law:

\[ I_2 = I_1 \cos^2 60^\circ. \]

- Substituting \(\cos 60^\circ = \frac{1}{2}\):

\[ I_2 = 128 \cdot \left(\frac{1}{2}\right)^2 = 128 \cdot \frac{1}{4} = 32 \, \frac{\text{W}}{\text{m}^2}. \]

3. Intensity After Third Polaroid (\(P_3\)): - The intensity after \(P_3\) is again reduced according to Malus’s Law:

\[ I_3 = I_2 \cos^2 30^\circ, \]

where the relative angle between \(P_2\) and \(P_3\) is \(30^\circ\) (since \(90^\circ - 60^\circ = 30^\circ\)).

- Substituting \(\cos 30^\circ = \frac{\sqrt{3}}{2}\):

\[ I_3 = 32 \cdot \left(\frac{\sqrt{3}}{2}\right)^2 = 32 \cdot \frac{3}{4} = 24 \, \frac{\text{W}}{\text{m}^2}. \]

Final Answer: \(24 \, \frac{\text{W}}{\text{m}^2}\).