Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids?
Example-(10.2)
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Let $I_0$ be the intensity of polarised light after passing through the first polariser $P_1$. Then the intensity of light after passing through second polariser $P_2$ will be
$
I=I_0 \cos ^2 \theta \text {, }
$
where $\theta$ is the angle between pass axes of $P_1$ and $P_2$. Since $P_1$ and $P_3$ are crossed the angle between the pass axes of $P_2$ and $P_3$ will be $(\pi / 2-\theta)$. Hence the intensity of light emerging from $P_3$ will be
$
\begin{aligned}
I & =I_0 \cos ^2 \theta \cos ^2\left(\frac{\pi}{2}-\theta\right) \\
& =I_0 \cos ^2 \theta \sin ^2 \theta=\left(I_0 / 4\right) \sin ^2 2 \theta
\end{aligned}
$
Therefore, the transmitted intensity will be maximum when $\theta=\pi / 4$.
$
I=I_0 \cos ^2 \theta \text {, }
$
where $\theta$ is the angle between pass axes of $P_1$ and $P_2$. Since $P_1$ and $P_3$ are crossed the angle between the pass axes of $P_2$ and $P_3$ will be $(\pi / 2-\theta)$. Hence the intensity of light emerging from $P_3$ will be
$
\begin{aligned}
I & =I_0 \cos ^2 \theta \cos ^2\left(\frac{\pi}{2}-\theta\right) \\
& =I_0 \cos ^2 \theta \sin ^2 \theta=\left(I_0 / 4\right) \sin ^2 2 \theta
\end{aligned}
$
Therefore, the transmitted intensity will be maximum when $\theta=\pi / 4$.
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