Question

laser light operating at 4mW falls normally on a perfect reflector.then the force exerted on the surface?

Answer 1

@Trexy

Answer 2

I think that energy doesn't exert any force .laser is a form of energy.If iam not wrong this is true

Answer 3

CORRECT

Answer 4

thank you

Answer 5

Momentum of 1 photon= \(\frac{h}{\lambda}\)
suppose there are n photons striking the reflector per sec, then momentum of n photons=\[\frac{nh}{\lambda}\] since the photons are reflected back, change in momentum of n photons=\[\frac{2nh}{\lambda}\] and foce exerted = rate of change of momentum=\[F=\frac{2nh}{\lambda t}......(1)\]force exerted due to photons=change in momentum of n photonNow, power=\[\frac{Energy}{time}=\frac{nhc}{\lambda t}=4\times10^{-3}\]\[\frac{nh}{\lambda t}=\frac{4 \times 10^{-3}}{c}......(2)\]From (1) and (2), you get, \[F=\frac{2\times 4\times 10^{-3}}{c}\] where C is velocity of light.


P.S. I expect the power to be 4M W instead of 4m W.. If its 4M W, use 4 * 10^6.