Q. Unpolarized light of intensity $32Wm^{- 2}$ passes through three polarizers such that transmission axes of the first and second polarizer makes an angle $30{^\circ}$ with each other and the transmission axis of the last polarizer is crossed with that of the first. The intensity of final emerging light will be
A) $32\ Wm^{- 2}$
B) $3\ Wm^{- 2}$
C) $8\ Wm^{- 2}$
D) $4\ Wm^{- 2}$
Explanation
(b) Angle between $P_{1}$ and $P_{2} = 30{^\circ}$ [Given]
Angle between $P_{2}$ and $P_{3} = \theta = 90{^\circ} - 30{^\circ} = 60{^\circ}$
The intensity of light transmitted by $P_{1}$ is
$I_{1} = \frac{I_{0}}{2} = \frac{32}{2} = 16\frac{W}{m^{2}}$
According to Malus law the intensity of light transmitted
by $P_{2}$ is $I_{2} = I_{1}\cos^{2}{30{^\circ} = 16\left( \frac{\sqrt{3}}{2} \right)^{2} = 12\frac{W}{m^{2}}}$
Similarly intensity of light transmitted by $P_{3}$ is
$I_{3} = I_{2}\cos^{2}{\theta = 12\cos^{2}{60{^\circ} = 12\left( \frac{1}{2} \right)^{2} = 3\frac{W}{m^{2}}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.