MCQ
When a polaroid sheet is rotated between two crossed polaroids then the transmitted light intensity will be maximum for a rotation of :
- A$60^{\circ}$
- B$30^{\circ}$
- C$90^{\circ}$
- D$45^{\circ}$
✓
Answer
Let $I_0$ be intensity of unpolarised light incident on first polaroid.
$I_1=$ 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
$\theta+\phi=90^0($ as $1^{\text {st }}$ and $3^{\text {rd }}$ polaroid are crossed$)$
$\phi=90^0-\theta$
$I_2=$ Intensity from $2^{\text {nd }}$ polaroid
$I_2=I_1 \cos ^2 \theta=\frac{I_0}{2} \cos ^2 \theta$
$I_3=$ Intensity from $3^{\text {rd }}$ polaroid
$I_3=I_2 \cos ^2 \phi$
$I_3=I_1 \cos ^2 \theta \cos ^2 \phi$
$I_3=\frac{I_0}{2} \cos ^2 \theta \cos ^2 \phi$
$\phi=90-\theta$
$I_3=\frac{I_0}{2} \cos ^2 \theta \sin ^2 \theta$
$I_3=\frac{I_0}{2}\left[\frac{2 \sin \theta \cos \theta}{2}\right]^2$
$I_3=\frac{I_0}{8} \sin ^2 2 \theta$
$I _3$ will be maximum when $\sin 2 \theta=1$
$2 \theta=90^{\circ}$
$\theta=45^{\circ}$
$I_1=$ 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
$\theta+\phi=90^0($ as $1^{\text {st }}$ and $3^{\text {rd }}$ polaroid are crossed$)$
$\phi=90^0-\theta$
$I_2=$ Intensity from $2^{\text {nd }}$ polaroid
$I_2=I_1 \cos ^2 \theta=\frac{I_0}{2} \cos ^2 \theta$
$I_3=$ Intensity from $3^{\text {rd }}$ polaroid
$I_3=I_2 \cos ^2 \phi$
$I_3=I_1 \cos ^2 \theta \cos ^2 \phi$
$I_3=\frac{I_0}{2} \cos ^2 \theta \cos ^2 \phi$
$\phi=90-\theta$
$I_3=\frac{I_0}{2} \cos ^2 \theta \sin ^2 \theta$
$I_3=\frac{I_0}{2}\left[\frac{2 \sin \theta \cos \theta}{2}\right]^2$
$I_3=\frac{I_0}{8} \sin ^2 2 \theta$
$I _3$ will be maximum when $\sin 2 \theta=1$
$2 \theta=90^{\circ}$
$\theta=45^{\circ}$
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