A beam of plane polarised light of large
cross-sectional area and uniform intensity
of 3.3 Wm-2 falls normally on a
polariser (cross sectional area 3 $\times$ 10-4 m2) which
rotates about its axis with an angular speed
of 31.4 rad/s. The energy of light passing through
the polariser per revolution, is close to :
Solution
Intensity, I = 3.3 Wm<sup>–3</sup>
<br><br>Area, A = 3 × 10<sup>–4</sup>
m<sup>2</sup>
<br><br>Angular speed, $\omega$ = 31.4 rad/s
<br><br>$I = {I_0}{\cos ^2}(\omega t)$<br><br>$\Rightarrow {I_{av}} = {{{I_0}} \over 2}$
<br><br>$\because$ $\left\langle {{{\cos }^2}\theta } \right\rangle$ = ${1 \over 2}$, in one time period
<br><br>$\therefore$ $E = {{{I_0}} \over 2} \times A \times (\Delta t)$<br><br>and $$\Delta t = {{2\pi } \over \omega } = {{2 \times 3.14} \over {31.4}} = {1 \over 5}s$$<br><br>$\therefore$ $$E = {{3.3} \over 2} \times 3 \times {10^{ - 4}} \times {1 \over 5} = 1 \times {10^{ - 4}}\,J$$
About this question
Subject: Physics · Chapter: Optics · Topic: Reflection and Mirrors
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