Question

Unpolarized light of intensity ${\mathrm{I}}_0$is incident on three polarizing filters. The axis of the first is vertical, that of the second is $45°$from vertical, and that of the third is horizontal. What light intensity emerges from the third filter$?$

Short answer

Light intensity emerges from the third filter is$0.1175\left(\frac{{\mathrm{I}}_{\mathrm{o}}}{2}\right)$.

Step-by-step solution

Formula and Calculation for angle

Given info:

First Angle is ${\mathrm{\theta }}_1={45}^{°}$.

Second angle is ${\mathrm{\theta }}_2={90}^{°}$.

Intensity is,

${\mathrm{I}}_1=\left(\frac{{\mathrm{I}}_{\mathrm{o}}}{2}\right){\cos }^2\mathrm{\theta }$

Calculation for intensity

Intensity for third filter is,

${\mathrm{I}}_3=\left(\frac{{\mathrm{I}}_{\mathrm{o}}}{2}\right)\mathrm{cos\theta }$

${\mathrm{I}}_3=0.235\left(\frac{{\mathrm{I}}_{\mathrm{o}}}{2}\right)0.5$

${\mathrm{I}}_3=0.1175\left(\frac{{\mathrm{I}}_{\mathrm{o}}}{2}\right)$