Two polaroids are placed in the path of an un | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Since no light is emitted from the second polaroid $(P_2)$,the first polaroid $(P_1)$ and the second polaroid $(P_2)$ must be crossed,meaning thei

Vedclass · Accepted Answer

$\left( \frac{I_0}{8} ight) \sin^2 2	heta$

Vedclass · Answer

(A) Since no light is emitted from the second polaroid $(P_2)$,the first polaroid $(P_1)$ and the second polaroid $(P_2)$ must be crossed,meaning their polarization axes are perpendicular to each other.Let the initial intensity of the unpolarized light be $I_0$.The intensity of light after passing through the first polaroid $(P_1)$ is $I_1 = \frac{I_0}{2}$.The third polaroid $(P_3)$ is placed at an angle $	heta$ with respect to $P_1$. According to Malus' Law,the intensity of light emerging from $P_3$ is $I_2 = I_1 \cos^2 	heta = \frac{I_0}{2} \cos^2 	heta$.The angle between the polarization axis of $P_3$ and $P_2$ is $(90^\circ - 	heta)$.Applying Malus' Law again for the light passing through $P_2$,the final intensity $I_f$ is:$I_f = I_2 \cos^2(90^\circ - 	heta) = I_2 \sin^2 	heta$$I_f = \left( \frac{I_0}{2} \cos^2 	heta ight) \sin^2 	heta = \frac{I_0}{2} (\sin 	heta \cos 	heta)^2$Using the identity $\sin 2	heta = 2 \sin 	heta \cos 	heta$,we have $\sin 	heta \cos 	heta = \frac{\sin 2	heta}{2}$.$I_f = \frac{I_0}{2} \left( \frac{\sin 2	heta}{2} ight)^2 = \frac{I_0}{2} \cdot \frac{\sin^2 2	heta}{4} = \frac{I_0}{8} \sin^2 2	heta$.

Similar Questions

Which of the following statements regarding cyanobacteria is incorrect?

If a solid sphere of mass $1 \ kg$ and radius $0.1 \ m$ rolls without slipping at a uniform velocity of $1 \ m/s$ along a straight line on a horizontal floor,the kinetic energy is

Find the binomial probability distribution whose mean is $3$ and variance is $2$.

Which of the following substances would quickly absorb oxygen?

$A$ thin circular ring of mass $M$ and radius $R$ is rotating about its axis with a constant angular velocity $\omega$. Two objects,each of mass $m$,are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity:

Vedclass Test Series

Exam Paper Generator

Online Exam Module