Question 91: Wave Optics

Question

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

Accepted Answer

Let $$\mathrm{I}_0$$ be intensity of unpolarised light incident on first polaroid.
$$\mathrm{I}_1=$$ Intensity of light transmitted from $$1^{	ext {st }}$$ polaroid $$=\frac{\mathrm{I}_0}{2}$$
$$	heta$$ be the angle between $$1^{\mathrm{st}}$$ and $$2^{	ext {nd }}$$ polaroid
$$\phi$$ be the angle between $$2^{	ext {nd }}$$ and $$3^{	ext {rd }}$$ polaroid
$$	heta+\phi=90^{\circ}$$ (as $$1^{	ext {st }}$$ and $$3^{	ext {rd }}$$ polaroid are crossed)
$$\phi=90^{\circ}-	heta$$
$$\mathrm{I}_2=$$ Intensity from $$2^{	ext {nd }}$$ polaroid
$$\mathrm{I}_2=\mathrm{I}_1 \cos ^2 	heta=\frac{\mathrm{I}_0}{2} \cos ^2 	heta$$
$$\mathrm{I}_3=$$ Intensity from $$3^{	ext {rd }}$$ polaroid
$$\mathrm{I}_3=\mathrm{I}_2 \cos ^2 \phi$$
$$\mathrm{I}_3=\mathrm{I}_1 \cos ^2 	heta \cos ^2 \phi$$
$$\mathrm{I}_3=\frac{\mathrm{I}_0}{2} \cos ^2 	heta \cos ^2 \phi$$
$$\phi=90-	heta$$
$$\begin{aligned}
& I_3=\frac{I_0}{2} \cos ^2 	heta \sin ^2 	heta \
& I_3=\frac{I_0}{2}\left[\frac{2 \sin 	heta \cos 	heta}{2}ight]^2 \
& I_3=\frac{I_0}{8} \sin ^2 2 	heta
\end{aligned}$$
$$\mathrm{I}_3$$ will be maximum when $$\sin 2 	heta=1$$
$$\begin{aligned}
& 2 	heta=90^{\circ} \
& 	heta=45^{\circ}
\end{aligned}$$

Answer

$$90^\circ$$

Answer

$$30^\circ$$

Answer

$$45^\circ$$

Answer

$$60^\circ$$

Wave Optics

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

Choose the correct answer: