The percentage decrease in the intensity of polarized light when it is passed through an analyser at an angle of \(60^\circ\) is
Show Hint
For polarized light passing through an analyser,
\[
I=I_0\cos^2\theta.
\]
At
\[
\theta=60^\circ,
\]
\[
I=\frac{I_0}{4}.
\]
So \(25\%\) is transmitted and \(75\%\) is lost.