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:
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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.
In this experiment, unpolarized light undergoes polarization upon passing through polarizer \( P_1 \). Consequently, the intensity of the light after \( P_1 \) is given by:
\[ I_1 = \frac{I_0}{2}. \]
The light then encounters polarizer \( P_2 \), oriented orthogonally to \( P_1 \). Ideally, no light would pass through these orthogonal polarizers, resulting in zero intensity. However, polarizer \( P_3 \), rotated to a \( 45^\circ \) angle relative to \( P_1 \) and \( P_2 \), permits light transmission.
A path difference of \( \frac{\lambda}{3} \) establishes a phase difference between the two light waves. The \( 45^\circ \) orientation of polarizer \( P_3 \) optimizes the intensity for this setup. The resultant intensity after \( P_3 \) is therefore:
\[ I = \frac{I_0}{2}. \]
Final Answer: \( \frac{I_0}{2} \).
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