Question

A beam of plane polarised light of large cross-sectional area and uniform intensity of $3.3{\text{Wm}}^{-2}$ falls normally on a polariser (cross-sectional area $=3\times {10}^{-4}{\text{m}}^2$) which rotates about its axis with an angular speed of $31.4{\text{rad s}}^{-1}.$ The energy of light passing through the polariser per revolution, is close to-

• A. $1.0\times {10}^{-5}\text{J}$
• B. $1.0\times {10}^{-4}\text{J}$
• C. $1.5\times {10}^{-4}\text{J}$
• D. $5.0\times {10}^{-4}\text{J}$

Answer 1

The correct option is B $1.0\times {10}^{-4}\text{J}$

Given,
$I_0=3.3{\text{Wm}}^{-2}$
$A=3\times {10}^{-4}{\text{m}}^2$
$\omega =31.4{\text{rad s}}^{-1}$

From Malu's law, $I=I_0{\cos }^2\theta$

$\text{Average energy}<E>=IAt=I_{0}At<{\cos }^2\theta >$
$\because <{\cos }^2\theta >=\frac{1}{2}$ per revolution

$\therefore \text{Average energy}=\frac{\left(3.3\right)\times \left[\frac{2\pi }{31.4}\right]\times (3\times {10}^{-4})}{2}\simeq 1\times {10}^{-4}\text{J}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.