Question

A polarizer and analyzer are oriented so that the amount of transmitted light is maximum. Through what angle should either be turned so that the intensity of transmitted light is reduced to (a) 0.75 and (b) 0.25 times the maximum intensity?

Answer 1

Solution:

(a) 0.75 times the maximum intensity?

Let the initial intensity =I

and final intensity $=\frac{3I}{4}$

We know that

$I=I_o\cos^2\frac{\phi}{2}$

$\frac{3I}{4}=I\cos^2\frac{\phi}{2}$

$\cos\frac{\phi}{2}=\frac{\sqrt{3}}{2}=\cos\frac{\pi}{6}$

$\phi = \frac{\pi}{3}$

(b) 0.25 times the maximum intensity?

25% of maximum intensity

final intensity $=\dfrac14I$

We know that

$I=I_o\cos^2\frac{\phi}{2}$

$\Rightarrow \dfrac14I=I\cos^2\frac{\phi}{2}$

$\Rightarrow \cos\frac{\phi}{2}=\frac{1}{{2}}=\cos\frac{\pi}{3}$

$\Rightarrow \phi = \frac{2\pi}{3}$