Collisions, using conservation of Energy

1. Jul 31, 2015

Prannoy Mehta

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

A small mirror of area A and mass m is suspended in a vertical plane by a weightless string. A beam of light of intensity I falls normally on the mirror and the string is deflected from the vertical by a very small angle A. Assuming the mirror to be perfectly reflecting, to obtain an expression for A.

Ans: 2IA/(mgc), where c is the speed of light.

2. Relevant equations

Energy should be conserved in the entire system. If the mirror moves up by an angle it has gained potential energy, and the only way which it could have got this is from light. But E=Hf (f is the frequency of light) so the light's frequency decreases upon complete reflection. What am I missing ?

3. The attempt at a solution

None, I do not understand why the mirror moves in the first place. I do not understand what am I missing from Conservation of energy. Or is the question totally incorrect.

Thank you for all the possible help.

2. Jul 31, 2015

DEvens

The idea you are missing is that light carries momentum. The momentum light carries is proportional to the energy it transmits. Since you are working on this problem and posted it in the homework section I will presume you can find the proportion in your text.

So the mirror will feel a force. The light presumably reflects with 100% efficiency, so you get a factor of 2.

So you need to work out the force on the mirror from reflecting light. And you do that by working out how much energy per second is reflected, and doing the correct proportion to get force from change in momentum.

Then you need to work out the angle that just balances this force. Remember your small-angle formulas for trig functions.

3. Jul 31, 2015

Prannoy Mehta

Light carrying momentum, also implies that it has, which is no true. Right? (I am not sure). I understand that in the question we will be applying the change of momentum to the mirror, conservation of momentum will again says it will loose momentum, which does not happen; keeping c constant.

I have assumed the concept to be right, I have done the following but I still seem to be getting the wrong answer:

P=IA --- (1)
P= Fv = 2mv^2/t -- (2)

Substituting (2) in (1) we will obtain in expression: t = 2mv^2/(IA)

tan X = F/mg (Assuming light imparts force F on the mirror, let the angle be X to avoid confusion.)

Since X is very small tan X is approximately X

X = F/mg = 2v^2/gt

Substituting for t, we obtain,

X= IA/ (mgv) = IA/ (mgv), not the correct answer according the text. (I have given the answer in the text in the question)

4. Jul 31, 2015

DEvens

Ok, I presumed wrong when I presumed you could look up the momentum of light as a function of energy. Well, I won't just give it to you. So you must read about photons.

https://en.wikipedia.org/wiki/Photon

Once you know what the momentum of a given amount of energy is in the form of photons, then ask yourself, what happens to the momentum of a photon when it reflects off a mirror? So for a given amount of energy hitting the mirror, how much momentum is imparted to the mirror?

5. Jul 31, 2015

Prannoy Mehta

I am sorry, for not looking it up. I will do it now, and get back to you as soon as possible.

6. Jul 31, 2015

haruspex

You meant, has no mass, right?
This is a key concept, photons have momentum even though they have no mass. When you find the formula, you will discover that the loss of momentum can be accounted for even though c is constant. Also, whether the photons lose momentum depends on your reference frame. In the frame of the mirror, they don't.