Two square reflectors, each 1.50 cm on a side and of mass 4.00 g, are located at opposite ends of a thin, extremely light, 1.00-m rod that can rotate without friction and in vacuum about an axle perpendicular to it through its centre (Fig. P32.39). These reflectors are small enough to be treated as point masses in moment-of-inertia calculations. Both reflectors are illuminated on one face by a sinusoidal light wave having an electric field of amplitude 1.25 N/C that falls uniformly on both surfaces and always strikes them perpendicular to the plane of their surfaces. One reflector is covered with a perfectly absorbing coating, and the other is covered with a perfectly reflecting coating. What is the angular acceleration of this device?

The average intensity of a plane wave is${\mathbf{l}}_{\mathbf{avg}}\mathbf{=}\frac{{\mathbf{E}}_{\mathbf{max}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{\mu }}_{\mathbf{0}}\mathbf{c}}{\mathbf{E}}_{\mathbf{max}}$is electric field amplitude,${\mathbf{\mu }}_{\mathbf{0}}$is relative permeability in free space, c is speed of light.

The average intensity of a plane wave is${\mathrm{l}}_{\mathrm{avg}}=\frac{{\mathrm{E}}_{\max }^2}{2{\mathrm{\mu }}_0\mathrm{c}}{\mathrm{E}}_{\max }$.Substitute the values we have

$$\begin{gathered} {\mathrm{l}}_{\mathbf{avg}}=\frac{1.25}{2\left(4\mathrm{\pi }\times {10}^{-7}\right)\left(3\times {10}^8\right)} \\ =2.702\times {10}^{-3}\mathrm{W}/\mathrm{m} \end{gathered}$$

To calculate the force,$\mathrm{F}=\frac{{\mathrm{l}}_{\mathrm{avg}}\mathrm{A}}{\mathrm{c}}$.Substitute the values we get,

$$\begin{gathered} \mathrm{F}=\frac{2.072\times {10}^{-3}\mathrm{W}/{\mathrm{m}}^2\left(2.25\times {10}^{-4}{\mathrm{m}}^2\right)}{3\times {10}^8} \\ =1.554\times {10}^{-15}\mathrm{N} \end{gathered}$$

To calculate the area of a square reflector is$\mathrm{A}={\mathrm{a}}^2$

$$\begin{gathered} \mathrm{A}={\left(1.5\times {10}^{-2}\mathrm{m}\right)}^2 \\ =2.25\times {10}^{-4}{\mathrm{m}}^2 \end{gathered}$$

The inertia for the square wave is, I = Inertia of absorbing side + Inertia of reflecting side

$$\begin{gathered} l=m{\left(\frac{L}{2}\right)}^2+m{\left(\frac{L}{2}\right)}^2 \\ =\frac{m{L}^2}{2} \end{gathered}$$

Substitute the values we get,

$$\begin{gathered} \mathrm{l}=\frac{4\times {10}^{-4}\left(1\right)}{2} \\ =2\times {10}^{-3}\mathrm{kg}.{\mathrm{m}}^2 \end{gathered}$$

To calculate the net torque using Newton's second law, ${\mathrm{\tau }}_{\mathrm{net}}=\frac{\mathrm{FL}}{2}$The net torque is also represented as ${\mathrm{\tau }}_{\mathrm{net}}=\mathrm{\alpha l}$where, $\alpha$is angular acceleration, l is the inertia. Compare the equations we have,$\frac{\mathrm{FL}}{2}=\mathrm{\alpha l}$

$\mathrm{\alpha l}=\frac{\mathrm{FL}}{2l}$Substitute the values we get,

$$\begin{gathered} \mathrm{\alpha }=\frac{\left(1.554\times {10}^{-15}\mathrm{N}\right)\left(1\mathrm{m}\right)}{2\left(2\times {10}^{-3}{\mathrm{kgm}}^2\right)} \\ =3.89\times {10}^{-13}\mathrm{rad}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the angular acceleration of the two square reflectors located at opposite end of a light is$3.89\times {10}^{-13}\mathrm{rad}/{\mathrm{s}}^2$