Question

Polarizer and analyzer are set with their polarizing directions parallel so that the intensity of transmitted light is maximum. Through what angle should either be turned so that intensity be reduced to (a) 0.5 and 25% of maximum intensity?

Answer 1

Solution:

Let the initial intensity =I

(a) 0.5 and 25% of maximum intensity?

final intensity $=0.5I$

We know that

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

$\Rightarrow 0.5I=I\cos^2\frac{\phi}{2}$

$\Rightarrow \cos\frac{\phi}{2}=\frac{1}{\sqrt{2}}=\cos\frac{\pi}{4}$

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

(b) 25% of maximum intensity

final intensity $=\dfrac14I$

We know that

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

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

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

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