Question

Unpolarised light of intensity $32W{m}^{-2}$ passes through three polarisers such that the transmission axis of the last polarizer is perpendicular to that of the first. At what angle will the transmitted intensity be maximum?

• A. ${15}^{\circ }$
• B. ${30}^{\circ }$
• C. ${45}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is C ${45}^{\circ }$

$I_0=32W{m}^{-2}$
$I_1=\frac{I_0}{2}$
$I_2=I_{1}co{s}^2\theta =\frac{I_0}{2}co{s}^2\theta$ . . . .(1)
$I_3=I_{2}co{s}^2(90-\theta )=\frac{I_0}{2}si{n}^2\theta co{s}^2\theta$
$=\frac{32}{2\times 4}\left(2^{2}si{n}^2\theta co{s}^2\theta \right)$
$=\frac{32}{8}(sin2\theta {)}^2$
$I_3=4[sin2\theta {]}^2$
For maximum $sin2\theta =1$
$2\theta ={90}^{\circ }$
$\theta ={45}^{\circ }$