Physics - Wave Optics Question with Solution | TestHub

Given,

$\theta ={45}^o$

If$I_0$be the intensity of light before passing through the analyser, then intensity of light after passing through analyser,

$I=I_{0 }{\cos }^2\theta$

$=I_0{\cos }^2{45}^o$

$=I_0\cdot {\left(\frac{1}{\sqrt{2}}\right)}^2$

$I=\frac{I_0}{2}$

Hence, the percentage of polarized light passing through analyser

$=I_0-\frac{\frac{I_0}{2}}{I_0}\times 100$

$=\frac{I_0/2}{I_0}\times 100$

$=\frac{1}{2}\times 100$

$=50\%$