Two square reflectors, each 1.80 cm on a side and of mass 3.65 g, are located at

Question

Two square reflectors, each 1.80 cm on a side and of mass 3.65 g, are located at opposite ends of a thin, extremely light, 1.00-m rod that can rotate without friction and in a vacuum about an axle perpendicular to it through its center (see figure below). 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 1.00 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?
   _____________rad/s2

can anyone help me out with this question I can't seem to get the right answer

Explanation / Answer

We know that Intensity = force /Area
Intensity will be given as = (1/2)eoEo2c =(1/2)(8.85*10-12)(1)2(3*108)
Intensity = 13.275*10-4 W/m2
And we know that
Pressure for absorbing = intensity/C = 4.425*10-12 N/m2
Pressure for reflecting = 2(intensity/c) = 8.85*10-12 N/m2
Now We know that Pressure = force/area
Force = pressure*area
so on absorbing Force = 4.425*10-12* (1.8)*10-4 N
On Reflecting = 8.85*10-12*(1.8)*10-4 N
Torque = Force *distance
Torque =( 8.85*10-12*(1.8)*10-4 - 4.425*10-12* (1.8)*10-4)*0.5 = 3.9825*10-16 Nm
We know that Torque = Moment of inertia*angular acceleration
angular acceleration = torque / moment of inertia
moment of inertia = m1r12 + m2r22 here mass of rod is not mentioned therefore considering it as massless
Moment of inertia = 1.825*10-3 kg - m2
Angular acceleration = 2.182 *10-13 rad/s2

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