A conceptual question about the force of light waves when reflected

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SUMMARY

The discussion centers on the force exerted by electromagnetic (EM) waves on a card based on its interaction with the waves. When an EM wave with an intensity of 200 W/m² strikes a black card measuring 0.2m x 0.3m, the force exerted by the radiation when absorbed is calculated to be 4 x 10^-8 N. However, when the card reflects 100% of the light, the force doubles due to the need for a greater impulse to reverse the momentum of the incoming wave, resulting in a total force of 8 x 10^-8 N.

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Homework Statement

An EM wave with intensity=200 W/m^2 is normal to a black .2m x .3m card that absorbs 100% of the wave. (a) Find the force exerted on the card by the radiation. (b) Find the force exerted by the same wave if the card reflects 100% of the light.



Homework Equations


Pr=I/c Pressure=F/A


The Attempt at a Solution


Part (a) I solved. (I/c)(0.2*0.3)=F=4*10^-8 N
Part (b) I know is double the force.

My question is conceptual. Why is the force doubled when the EM wave is reflected?

Thank you for any and all replies.
 
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Think about it in terms of momentum. Force is what is required to change momentum; a net change in momentum due to a force is usually called an impulse. In the case of absorption, you only need the card to provide an impulse equal in magnitude to the momentum, which is opposite in direction to the light wave. This will exactly stop the wave. On the other hand, if you want to reflect the wave, you need to not only bring its velocity to zero but then to send it back in the opposite direction at the incoming speed. In other words you need to impart an impulse twice as large because the magnitude of the change in momentum is twice as large.
 
That makes perfect sense. Thank you Steely Dan.

PS Love your handle. Get along Kid Charlemagne.
 

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