Two beams,A and B,of plane polarized light wi | vedclass

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

Question

(C) According to Malus's law,the intensity of the emerging beam is given by $I = I_0 \cos^2 	heta$.Let the initial intensity of beam $A$ be $I_A$ and

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(C) According to Malus's law,the intensity of the emerging beam is given by $I = I_0 \cos^2 	heta$.Let the initial intensity of beam $A$ be $I_A$ and beam $B$ be $I_B$. When the polaroid is at the position where beam $A$ has maximum intensity,the angle between the transmission axis of the polaroid and the plane of polarization of beam $A$ is $0^{\circ}$,and for beam $B$ it is $90^{\circ}$.After rotating the polaroid by $30^{\circ}$,the new angle for beam $A$ is $	heta_A = 30^{\circ}$ and for beam $B$ is $	heta_B = 90^{\circ} - 30^{\circ} = 60^{\circ}$.The intensities of the beams after passing through the polaroid are:$I_A' = I_A \cos^2(30^{\circ}) = I_A \left(\frac{\sqrt{3}}{2}ight)^2 = I_A \cdot \frac{3}{4}$$I_B' = I_B \cos^2(60^{\circ}) = I_B \left(\frac{1}{2}ight)^2 = I_B \cdot \frac{1}{4}$Given that the beams appear equally bright,$I_A' = I_B'$:$I_A \cdot \frac{3}{4} = I_B \cdot \frac{1}{4}$$\frac{I_A}{I_B} = \frac{1}{3}$

Similar Questions

When unpolarized light of intensity $I_0$ is incident on a polarizing sheet,the intensity of light that does not pass through is:

In an experiment,two polaroids are arranged such that the intensity of the polarised light emerged from the second polaroid is $37.5 \%$ of the intensity of the unpolarised light incident on the first polaroid. Then the angle between the axes of the two polaroids is (in $^{\circ}$)

Show that when unpolarized light passes through a polarizer,the intensity of the emerging light is half that of the intensity of the incident light.

Vedclass Test Series

Exam Paper Generator

Online Exam Module