Miscellaneous: Practice Problem & Solution
Two Polaroids $P_1$ and $P_2$ are placed with their axis perpendicular to each other. Unpolarised light $I_0$ is incident on $P_1$. A third polaroid $P_3$ is kept in between $P_1$ and $P_2$ such that its axis makes an angle $45^\circ$ with that of $P_1$. The intensity of transmitted light through $P_2$ is: (2017-Delhi)
Solution Explained:
To solve this problem, we apply the core principles of Miscellaneous. Understanding the underlying formula is key to arriving at the correct answer below:
Intensity after $P_1$ is $I_1 = \frac{I_0}{2}$. $P_3$ is at $45^\circ$ to $P_1$, so intensity after $P_3$ is $I_3 = I_1 \cos^2(45^\circ) = (\frac{I_0}{2})(\frac{1}{2}) = \frac{I_0}{4}$. $P_2$ is perpendicular to $P_1$, so it is at $45^\circ$ to $P_3$. Intensity after $P_2$ is $I_2 = I_3 \cos^2(45^\circ) = (\frac{I_0}{4})(\frac{1}{2}) = \frac{I_0}{8}$.
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