Question

A polarizer and an analyzer are oriented so that the maximum amount of lights is transmitted. Fraction of its maximum value is the intensity of the transmitted light reduced when the analyzer is rotated through (intensity of incident light $=I_o$)
a) 30${}^{\circ }$ b) 45${}^{\circ }$ c) 60${}^{\circ }$

• A. $0.375I_0,0.25I_0,0.125I_0$
• B. $0.25I_0,0.375I_0,0.125I_0$
• C. $0.125I_0,0.25I_0,0.0375I_0$
• D. $0.125I_0,0.375I_0,0.25I_0$

Answer 1

The correct option is A $0.375I_0,0.25I_0,0.125I_0$

If $I_0$ is incident, then after passing through
the polarizer then Intensity becomes $I_0/2$
Now when the analyzer is rotated the transmitted
intensity reduces.
$I=\frac{I_0}{2}{\cos }^2\theta$ . If $\theta$ is the angle of rotation
a)$\theta ={30}^0;I=\frac{I_0}{2}\times co{s}^{2}30$
$=0.375I_0$
b)$\theta ={45}^0;I=\frac{I_0}{2}co{s}^{2}45$
$=0.25I_0$
c)$\theta ={60}^0;I=\frac{I_0}{2}co{s}^{2}60$
$=0.125I_0$