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To experimentally verify the momentum of a photon, consider the illustration below. A photon pendulum.

Now, once the light hits the mirror, the momentum of the light is being transferred to the mirror. Depending on the momentum, the mirror will move with some velocity $v$ one can work out using conservation of energy.

Now my textbook claims using $$mgd=\frac{1}{2}mv^2$$ I may find the height $d$ the pendulum will swing. But the mirror on the pendulum with length l is constrained to stay inside a circle (approximately). Why is this approach valid, as a sufficiently large $v$ would let $h$ exceed the circle constraints?

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Samuel
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1 Answers1

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You are correct, but the force exerted by a light ray is so small that there is no danger of the distance $d$ coming close to the radius $\ell$. It's more likely that $d$ would be so small it's difficult to measure accurately.

The derivation of the force exerted by a light beam is discussed in my answer to If I'm floating in space and I turn on a flashlight, will I accelerate? The result is:

$$ F = \frac{W}{c} $$

where $W$ is the beam power in watts and $c$ is the speed of light. A $1$ W light beam exerts a force of about $3 \times 10^{-9}$ N so unless the mirror is exceedingly light the vertical displacement is going to be very small.

John Rennie
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