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Force exerted by a laser beam

  • Thread starter Matthaeus_
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  • #1
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Homework Statement



A laser beam ([tex]\mathrm{Power} = 1\ \mathrm{W})[/tex] is completely reflected by a mirror perpendicular to the beam. Light is made of photons, and each photon carries an energy [tex]E = h\nu[/tex] and a momentum [tex]P = h/\lambda[/tex], where [tex]\nu[/tex] is the frequency, [tex]\lambda[/tex] is the wavelength and [tex]h[/tex] is Planck's constant. Find the force with which light pushes the mirror.

Homework Equations


Apart from those already present in the problem statement, I have:
[tex]\lambda \nu = c[/tex]

[tex]F = \frac{dp}{dt}[/tex]

The Attempt at a Solution


Each second, the light source emits [tex]n[/tex] photons, each one carries an energy [tex]E = h\nu = hc/\lambda[/tex], for a total power of [tex]1\ \mathrm{W}[/tex]. This gives:

[tex]\displaystyle n = \frac P E = \frac{\lambda}{hc}[/tex]

In one second then, [tex]n[/tex] photons hit the mirror and bounce back, which gives:

[tex]\displaystyle F = \frac{dp}{dt} = n \cdot 2p = 2 \frac{\lambda}{hc}\cdot \frac{h}{\lambda} = \frac 2 c \approx 6.67\cdot 10^{-9}\ \mathrm{N}[/tex]

The result is somewhat intuitively pleasing, can you check it is correct, please?
 

Answers and Replies

  • #2
Doc Al
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Looks good to me.
 
  • #3
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Whew... Thank you Doc for checking :)
 
  • #4
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Sorry to dig up such an old question, but after looking at this question, I don't understand were the 2 comes from in the F=n2p. Could someone explain this to me?
Thanks
 
  • #5
mgb_phys
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The momentum is twice the incoming photon's because it bounces back
Think of a ball, if you throw it to hit a wall and stop then you need twice as much force for it to hit the wall and come back at the same speed.
 
  • #6
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That is to say, if it was not a mirror, and the light did not reflect off the surface, the 2 would be a 1 instead?
 
  • #7
Doc Al
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Sorry to dig up such an old question, but after looking at this question, I don't understand were the 2 comes from in the F=n2p.
It comes from the fact that the change in momentum is twice the original momentum.
 
  • #8
Redbelly98
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That is to say, if it was not a mirror, and the light did not reflect off the surface, the 2 would be a 1 instead?
Yes. If the light is absorbed then it's a 1. As Doc Al said, it's all about change in momentum.
 

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