The correct option is (4) 45°.
Let \(I_0\) be intensity of unpolarised light incident on first polaroid.
\(I_1\) = Intensity of light transmitted from 1st polaroid
= \(\frac{I_0}{2}\)
\(\theta\) be the angle between 1st and 2nd polaroid
\(\phi\) be the angle between 2nd and 3rd polaroid
\(\theta\) + \(\phi\) = 90° (as 1st and 3rd polaroid are crossed)
\(\phi\) = 90° - \(\theta\)
\(I_2\) = = Intensity from 2nd polaroid
\(I_2 = I_1 cos^2\theta = \frac{I_0}{2} cos^2 \theta\)
\(I_3\) = Intensity from 3rd polaroid
\(I_3 = I_2 cos^2 \phi\)
\(I_3 = I_1 cos^2 \theta cos^2 \phi\)
\(I_3 = \frac{I_0}{2} cos^2 \theta cos^2 \phi\)
\(\phi = 90 - \theta\)
\(I_3 = \frac{I_0}{2} cos^2 \theta sin^2 \theta\)
\(I_3 = \frac{I_0}{2} [\frac{2sin \theta cos \theta}{2}]^2\)
\(I_3 = \frac{I_0}{8} sin^2 2\theta\)
\(I_3\) will be maximum when \(sin 2\theta = 1\)
\(2\theta = 90^\circ\)
\(\theta = 45^\circ\)
