KCET · Physics · Wave Optics
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
- A \( 45^{\circ} \)
- B \( 60^{\circ} \)
- C \( 30^{\circ} \)
- D \( 0^{\circ} \)
Answer & Solution
Correct Answer
(A) \( 45^{\circ} \)
Step-by-step Solution
Detailed explanation
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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