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In a system of two crossed polarisers, it is found that the intensity of light from the second
polariser is half from that of first polariser. The angle between their pass axes is
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polariser is half from that of first polariser. The angle between their pass axes is
Solution:
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Verified Answer
The correct answer is:
\( 45^{\circ} \)
By Malus Law, we have
\[
I=I_{0} \cos \theta
\]
Given, intensity of light from second polariser is half of that from the first polariser, so
\[
\begin{array}{l}
\Rightarrow \frac{I_{0}}{2}=I_{0} \cos ^{2} \theta \\
\Rightarrow \cos ^{2} \theta=\frac{1}{2} \\
\Rightarrow \cos \theta=\frac{1}{\sqrt{2}} \\
\Rightarrow \theta=\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right) \\
\Rightarrow \theta=45^{\circ}
\end{array}
\]
Therefore, angle between the pass axes of polarisers is \( 45^{\circ} \).
\[
I=I_{0} \cos \theta
\]
Given, intensity of light from second polariser is half of that from the first polariser, so
\[
\begin{array}{l}
\Rightarrow \frac{I_{0}}{2}=I_{0} \cos ^{2} \theta \\
\Rightarrow \cos ^{2} \theta=\frac{1}{2} \\
\Rightarrow \cos \theta=\frac{1}{\sqrt{2}} \\
\Rightarrow \theta=\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right) \\
\Rightarrow \theta=45^{\circ}
\end{array}
\]
Therefore, angle between the pass axes of polarisers is \( 45^{\circ} \).
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