First slide

Diffraction of light-Wave Optics

Question

A beam of unpolarized light having flux 10^-3 W falls normally on a polarizer of cross-sectional area 3 $\times$ 10^-4 m². The polarizer rotates with an angular frequency of 31.4 rads^-1. The energy of light passing through the polarizer per revolution will be

Moderate

A

$10^{- 4} J$
B

$10^{- 3} J$
C

$10^{- 2} J$
D

$10^{- 1} J$

Solution

Here, $ω = 31.4 {rads}^{- 1}$
$∴$ Time period of revolution,
$T = \frac{2 π}{ω} = \frac{2 \times 3 .14}{31 .4} = 0 .2 s$
Energy transmitted/revolution,
$\begin{array}{l} = (IA) T = (\frac{I_{0}}{2} A) T \\ = \frac{ϕ_{0} T}{2} = \frac{10^{- 3} \times 0 .2}{2} = 10^{- 4} J \end{array}$

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