Question

In a Young's double slit experiment, three polarizers are kept as shown in the figure. The transmission axes of \( P_1 \) and \( P_2 \) are orthogonal to each other. The polarizer \( P_3 \) covers both the slits with its transmission axis at \( 45^\circ \) to those of \( P_1 \) and \( P_2 \). An unpolarized light of wavelength \( \lambda \) and intensity \( I_0 \) is incident on \( P_1 \) and \( P_2 \). The intensity at a point after \( P_3 \), where the path difference between the light waves from \( S_1 \) and \( S_2 \) is \( \frac{\lambda}{3} \), is: 

• A. \( I_0 \)
• B. \( \frac{I_0}{3} \)
• C. \( \frac{I_0}{2} \)
• D. \( \frac{I_0}{4} \)

Hint:

In a double-slit experiment with polarizers, the intensity of the light after each polarizer can be determined using Malus's law. The total intensity depends on the relative angle between the transmission axes of the polarizers.

Answer 1

The Correct Option is C

In this experiment, the unpolarized light first passes through polarizer \( P_1 \), which will polarize the light. Since the light is unpolarized, the intensity after \( P_1 \) is:

\[ I_1 = \frac{I_0}{2}. \]

The light then passes through polarizer \( P_2 \), which is orthogonal to \( P_1 \). Since the polarizers are at right angles to each other, the intensity after \( P_2 \) will be zero, as no light passes through orthogonal polarizers. However, \( P_3 \) rotates the transmission axis to \( 45^\circ \) relative to \( P_1 \) and \( P_2 \), allowing light to pass through.

The path difference is \( \frac{\lambda}{3} \), which corresponds to a phase difference between the two light waves. Since the polarizer \( P_3 \) is at \( 45^\circ \), it ensures that the intensity is maximized for this configuration. Therefore, the resulting intensity after \( P_3 \) is:

\[ I = \frac{I_0}{2}. \]

Final Answer: \( \frac{I_0}{2} \).