When a polaroid sheet is rotated between two crossed polaroids then the transmitted light intensity will be maximum for a rotation of :

\(I_1=\) Intensity of light transmitted from \(1^{\text {st }}\) polaroid

\(=\frac{I_0}{2}\)

\(\theta\) be the angle between \(1^{\text {st }}\) and \(2^{\text {nd }}\) polaroid

\(\phi\) be the angle between \(2^{\text {nd }}\) and \(3^{\text {rd }}\) polaroid

\(\theta+\phi=90^{\circ}\) (as \(1^{\text {st }}\) and \(3^{\text {rd }}\) polaroid are crossed)

\(\phi=90^{\circ}-\theta\)

\(\mathrm{I}_2=\) Intensity from \(2^{\text {nd }}\) polaroid

\(I_2=I_1 \cos ^2 \theta=\frac{I_0}{2} \cos ^2 \theta\)

\(\mathrm{I}_3=\) Intensity from \(3^{\text {rd }}\) polaroid

\(\mathrm{I}_3=\mathrm{I}_2 \cos ^2 \phi\)

\(\mathrm{I}_3=\mathrm{I}_1 \cos ^2 \theta \cos ^2 \phi\)

\(\mathrm{I}_3=\frac{\mathrm{I}_0}{2} \cos ^2 \theta \cos ^2 \phi\)

\(\phi=90-\theta\)

\(I_3=\frac{I_0}{2} \cos ^2 \theta \sin ^2 \theta\)

\(I_3=\frac{I_0}{2}\left[\frac{2 \sin \theta \cos \theta}{2}\right]^2\)

\(I_3=\frac{I_0}{8} \sin ^2 2 \theta\)

\(\mathrm{I}_3\) will be maximum when \(\sin 2 \theta=1\)

\( 2 \theta=90^{\circ} \)

\( \theta=45^{\circ}\)