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Calculating force of a laser

 
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Jul18-12, 07:43 AM   #1
 

Calculating force of a laser


Hi

1. The problem statement, all variables and given/known data

I have to calculate the force of a laser. Given are the power P= 100W and the velocity of light.



2. Relevant equations

equations I considered using:

[tex] E=f \cdot h [/tex] and [tex] p=\frac {h}{\lambda} [/tex]



3. The attempt at a solution

Actually that's all I got so far. I also considered P=F*v but I don't think I'm allowed to use this formula (I'm not familiar with the physics of lasers). Does anyone have some hints for me?

Thank you
 
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Jul18-12, 08:20 AM   #2
 
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Quote by Lindsayyyy View Post
Hi

1. The problem statement, all variables and given/known data

I have to calculate the force of a laser. Given are the power P= 100W and the velocity of light.



2. Relevant equations

equations I considered using:

[tex] E=f \cdot h [/tex] and [tex] p=\frac {h}{\lambda} [/tex]



3. The attempt at a solution

Actually that's all I got so far. I also considered P=F*v but I don't think I'm allowed to use this formula (I'm not familiar with the physics of lasers). Does anyone have some hints for me?

Thank you
I guess you're calculating the force exerted by a laser on a body that absorbs all the incident radiation. The collisions of photons against this body can be said to be inelastic.

The energy of a single photon E = pc.

What's the energy of N photons?

Let's say those N photons are emitted uniformly over time T.

Can you think of how to derive the power of the laser from that equation?

What's the definition of force (hint: rate of change of _____)?

Can you now see the relationship between power and force? It's essentially the same as P = Fv, except we derived it without making Newtonian assumptions (which don't really apply here).
 
Jul18-12, 08:32 AM   #3
 
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The laser is creating a stream of photons where each photon carries a definite amount of momentum. Newton's second law of motion can be expressed in terms of momentum as force equals rate of change of momentum. So, in the case of the laser, the force will be the total amount of photon momentum created every second.

[Note added: Sorry, I did not see Curious3141's comment. He posted his while I was still constructing my comment. Also, I was thinking of the reaction force on the laser rather than the force on the object that is hit by the laser beam. The forces are the same if the object completely absorbs the laser light, as Curious3141 says.]
 
Jul18-12, 08:42 AM   #4
 

Calculating force of a laser


yeah sorry, forget to write that aswell, it's getting absorbed.



[tex] P=\frac {dE}{dt}[/tex]

and force is equal to [tex] F=\frac {dp}{dt}[/tex]

so [tex] F=\frac {dp}{dE} \cdot P [/tex] ?

I insert the terms for E and P I wrote down in my first post with f= c/lambda

and get [tex] F=\frac {P}{c}[/tex]

is that right?

Thanks for your help

edit: no need to be sorry, I'm happy for every help I can get. Reading the same thing from different people/source makes it easier to understand sometimes:)
 
Jul18-12, 08:46 AM   #5
 
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That's correct. Good.
 
Jul18-12, 08:51 AM   #6
 
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Quote by Lindsayyyy View Post
yeah sorry, forget to write that aswell, it's getting absorbed.



[tex] P=\frac {dE}{dt}[/tex]

and force is equal to [tex] F=\frac {dp}{dt}[/tex]

so [tex] F=\frac {dp}{dE} \cdot P [/tex] ?

I insert the terms for E and P I wrote down in my first post with f= c/lambda

and get [tex] F=\frac {P}{c}[/tex]

is that right?

Thanks for your help

edit: no need to be sorry, I'm happy for every help I can get. Reading the same thing from different people/source makes it easier to understand sometimes:)
Looks good, although I would be careful about defining your terms properly. E refers to the energy of the photon beam, not a single photon, and so forth. I introduced the N term to deal with this. You didn't, which is fine, but it's important to be clear with your definitions when you write these things out.

As I said, your final expression is analogous to F = P/v, as in classical mechanics.
 
Jul18-12, 09:02 AM   #7
 
Ok, thank you
 
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