Three polaroid sheets P_{1}, P_{2} and P_{3} | vedclass

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

Question

(C) When unpolarised light of intensity $I_{0}$ passes through the first polaroid $P_{1}$,the intensity of the transmitted light becomes $I_{1} = \fra

Vedclass · Accepted Answer

$16 \ Wm^{-2}$

Vedclass · Answer

(C) When unpolarised light of intensity $I_{0}$ passes through the first polaroid $P_{1}$,the intensity of the transmitted light becomes $I_{1} = \frac{I_{0}}{2}$.Given $I_{0} = 128 \ Wm^{-2}$,so $I_{1} = \frac{128}{2} = 64 \ Wm^{-2}$.According to Malus's law,when light of intensity $I_{1}$ passes through a polaroid whose pass axis makes an angle $	heta$ with the incident light's polarization direction,the transmitted intensity is $I = I_{1} \cos^{2} 	heta$.For $P_{2}$,the angle between the pass axis of $P_{1}$ and $P_{2}$ is $	heta_{1} = 45^{\circ}$.Intensity after $P_{2}$ is $I_{2} = I_{1} \cos^{2} 45^{\circ} = 64 	imes (\frac{1}{\sqrt{2}})^{2} = 64 	imes \frac{1}{2} = 32 \ Wm^{-2}$.For $P_{3}$,the angle between the pass axis of $P_{2}$ and $P_{3}$ is $	heta_{2} = 45^{\circ}$.Intensity after $P_{3}$ is $I_{3} = I_{2} \cos^{2} 45^{\circ} = 32 	imes (\frac{1}{\sqrt{2}})^{2} = 32 	imes \frac{1}{2} = 16 \ Wm^{-2}$.

Similar Questions

$A$ polariser is used to

Which property can be used to distinguish between light waves and sound waves?

$A$ beam of natural light falls on a system of $6$ polaroids,which are arranged in succession such that each polaroid is turned through $30^\circ$ with respect to the preceding one. The percentage of incident intensity that passes through the system will be....$\%$

$A$ beam of unpolarised light of intensity $I_0$ is passed through a polaroid $A$ and then through another polaroid $B$ which is oriented so that its principal plane makes an angle of $45^{\circ}$ relative to that of $A$. The intensity of the emergent light is

Vedclass Test Series

Exam Paper Generator

Online Exam Module