Two polaroids $P_1$ and $P_2$ are placed with their axis perpendicular to each other. Unpolarised light $I_o$

Question

Two polaroids $P_1$ and $P_2$ are placed with their axis perpendicular to each other. Unpolarised light $I_o$ 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

Accepted Answer

Let the intensity of unpolarised light be $I_o$.
When unpolarised light passes through the first polaroid $P_1$, its intensity becomes half:
$I_1 = \frac{I_o}{2}$
The transmitted light is now plane-polarised.
The third polaroid $P_3$ is placed between $P_1$ and $P_2$ with its pass axis at an angle of $45^\circ$ with respect to $P_1$. According to Malus's Law, the intensity of light transmitted through $P_3$ is:
$I_3 = I_1 \cos^2(45^\circ) = \frac{I_o}{2} 	imes \left(\frac{1}{\sqrt{2}}ight)^2 = \frac{I_o}{2} 	imes \frac{1}{2} = \frac{I_o}{4}$
The polaroids $P_1$ and $P_2$ are crossed (their axes are perpendicular to each other). Since $P_3$ makes an angle of $45^\circ$ with $P_1$, the angle between the transmission axes of $P_3$ and $P_2$ is $90^\circ - 45^\circ = 45^\circ$.
Again, applying Malus's Law, the intensity of light transmitted through $P_2$ is:
$I_2 = I_3 \cos^2(45^\circ) = \frac{I_o}{4} 	imes \left(\frac{1}{\sqrt{2}}ight)^2 = \frac{I_o}{4} 	imes \frac{1}{2} = \frac{I_o}{8}$

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