Three polarising sheets are placed co-axially | vedclass

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

Question

(B) When unpolarised light of intensity $I_0$ passes through the first polaroid,the intensity of the emergent light is $I_1 = \frac{I_0}{2}$.According

Vedclass · Accepted Answer

$\frac{3 I_0}{32}$

Vedclass · Answer

(B) When unpolarised light of intensity $I_0$ passes through the first polaroid,the intensity of the emergent light is $I_1 = \frac{I_0}{2}$.According to Malus' Law,the intensity of light emerging from a polaroid is $I = I_{incident} \cos^2 	heta$,where $	heta$ is the angle between the pass axis of the polaroid and the plane of polarisation of the incident light.For the second polaroid,the angle between its pass axis and the pass axis of the first polaroid is $	heta_1 = 30^{\circ}$. Thus,the intensity of light emerging from the second polaroid is $I_2 = I_1 \cos^2 30^{\circ} = \left(\frac{I_0}{2}ight) \left(\frac{\sqrt{3}}{2}ight)^2 = \left(\frac{I_0}{2}ight) \left(\frac{3}{4}ight) = \frac{3 I_0}{8}$.For the third polaroid,the angle between its pass axis and the pass axis of the second polaroid is $	heta_2 = 90^{\circ} - 30^{\circ} = 60^{\circ}$. Thus,the intensity of light emerging from the third polaroid is $I_3 = I_2 \cos^2 60^{\circ} = \left(\frac{3 I_0}{8}ight) \left(\frac{1}{2}ight)^2 = \left(\frac{3 I_0}{8}ight) \left(\frac{1}{4}ight) = \frac{3 I_0}{32}$.

Similar Questions

Polarized glass is used for sunglasses because...

Through which characteristic can we distinguish light waves from sound waves?

An unpolarised beam of intensity $I_{0}$ falls on a polaroid. The intensity of the emergent light is

Vedclass Test Series

Exam Paper Generator

Online Exam Module