Question

Polarizer and analyzer are set with their polarizing directions parallel, so the intensity
of transmitted light is maximum. Through what angle should either be turned, so that
intensity be reduced to 75% of the maximum intensity?

Answer 1

Let the initial intensity =I

and final intensity $=\frac{3I}{4}$

We know that

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

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

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

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