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.
Initially, unpolarized light enters polarizer \( P_1 \), polarizing it. The intensity after \( P_1 \) is \( I_1 = \frac{I_0}{2} \). Subsequently, the light encounters polarizer \( P_2 \), oriented orthogonally to \( P_1 \). Due to their perpendicular alignment, no light passes through \( P_2 \), resulting in zero intensity. However, polarizer \( P_3 \) rotates the transmission axis to \( 45^\circ \) relative to \( P_1 \) and \( P_2 \), enabling light transmission. A path difference of \( \frac{\lambda}{3} \) introduces a corresponding phase difference. The \( 45^\circ \) orientation of \( P_3 \) maximizes the intensity for this setup. Consequently, the final intensity after \( P_3 \) is \( I = \frac{I_0}{2} \). Final Answer: \( \frac{I_0}{2} \).
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