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

We are given that the incident beam is already plane-polarized with a uniform intensity $$I_0 = 3.3\;{\rm W\,m^{-2}}$$ and that it falls normally on a polarizer whose clear cross-sectional area is $$A = 3 \times 10^{-4}\;{\rm m^{2}}.$$

First we find the incident power. Power is related to intensity by the elementary relation $$P = I \,A.$$ Substituting the given numbers we have

$$P_0 = I_0\,A = \bigl(3.3\;{\rm W\,m^{-2}}\bigr)\,\bigl(3 \times 10^{-4}\;{\rm m^{2}}\bigr) = 9.9 \times 10^{-4}\;{\rm W}.$$

Now the transmission through a polarizer obeys Malus’ Law, which states

$$I = I_0 \cos^{2}\theta,$$

where $$\theta$$ is the angle between the incident light’s polarization direction and the transmission axis of the polarizer. Because the polarizer is rotating steadily, this angle changes uniformly from $$0$$ to $$2\pi$$ during one full revolution. Hence the transmitted intensity keeps oscillating, and we want the average value over an entire turn.

The time (or angular) average of $$\cos^{2}\theta$$ over one full cycle is

$$\langle\cos^{2}\theta\rangle = \frac{1}{2}.$$ Therefore the average transmitted intensity is one half of the incident intensity:

$$\langle I\rangle = \frac{I_0}{2} = \frac{3.3}{2}\;{\rm W\,m^{-2}} = 1.65\;{\rm W\,m^{-2}}.$$

Correspondingly, the average transmitted power becomes

$$\langle P\rangle \;=\; \langle I\rangle\,A = \bigl(1.65\;{\rm W\,m^{-2}}\bigr)\,\bigl(3 \times 10^{-4}\;{\rm m^{2}}\bigr) = 4.95 \times 10^{-4}\;{\rm W}.$$

Next we must find how long one revolution of the polarizer takes. The angular speed is given as $$\omega = 31.4\;{\rm rad\,s^{-1}}.$$ One complete turn corresponds to an angle of $$2\pi\;{\rm rad}.$$ Using the basic relation $$T = \dfrac{2\pi}{\omega}$$ we have

$$T = \frac{2\pi}{31.4} = \frac{6.28}{31.4} = 0.20\;{\rm s}.$$

Finally, energy transmitted in one revolution equals the average power multiplied by the time for one revolution:

$$E = \langle P\rangle\,T = \bigl(4.95 \times 10^{-4}\;{\rm W}\bigr)\,\bigl(0.20\;{\rm s}\bigr) = 9.9 \times 10^{-5}\;{\rm J}.$$ Numerically $$9.9 \times 10^{-5}\;{\rm J} \approx 1.0 \times 10^{-4}\;{\rm J}.$$

Hence, the correct answer is Option B.