91Ó°ÊÓ

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

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

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

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

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

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

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

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,

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

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

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

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

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