Wave Optics Question 4
Question 4 - 2024 (27 Jan Shift 2)
When a polaroid sheet is rotated between two crossed polaroids then the transmitted light intensity will be maximum for a rotation of :
(1) $60^{\circ}$
(2) $30^{\circ}$
(3) $90^{\circ}$
(4) $45^{\circ}$
Show Answer
Answer: (4)
Solution:
Let $\mathbf{I}_{0}$ be intensity of unpolarised light incident on first polaroid.
$I_{1}=$ Intensity of light transmitted from $1^{\text {st }}$ polaroid
$=\frac{\mathrm{I}_{0}}{2}$
$\theta$ 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
$\theta+\phi=90^{\circ}\left(\right.$ 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
$\mathrm{I}{3}=\mathrm{I}{2} \cos ^{2} \phi$
$\mathrm{I}{3}=\mathrm{I}{1} \cos ^{2} \theta \cos ^{2} \phi$
$\mathrm{I}{3}=\frac{\mathrm{I}{0}}{2} \cos ^{2} \theta \cos ^{2} \phi$
$\phi=90-\theta$
$\mathrm{I}{3}=\frac{\mathrm{I}{0}}{2} \cos ^{2} \theta \sin ^{2} \theta$
$\mathrm{I}{3}=\frac{\mathrm{I}{0}}{2}\left[\frac{2 \sin \theta \cos \theta}{2}\right]^{2}$
$\mathrm{I}{3}=\frac{\mathrm{I}{0}}{8} \sin ^{2} 2 \theta$
$\mathrm{I}_{3}$ will be maximum when $\sin 2 \theta=1$
$2 \theta=90^{\circ}$
$\theta=45^{\circ}$
