# Speed Change Due to Laser

1. Aug 12, 2011

### Gavandeshaq

1. The problem statement, all variables and given/known data
A laser is used to focus 25 MW of light onto a missile for 15 seconds. If the mass of the missile is 200 kg, determine the speed change imparted to the missile in the direction of the laser beam.

2. Relevant equations
E=hf
p=h/$\lambda$
F=dp/dt
$\lambda$f=c
F=ma

3. The attempt at a solution
I figured I'd try and get the power into joules, so that I could find the frequency, and henceforth the wavelength etc.
I found that the laser transfers 1.6x10^7 J across the 15s, when putting this into E=hf I came out with a frequency of 2.4x10^40 Hz (Seems fairly large, but since I know very little about lasers, I'm going with it).
Now I must be missing something here, cos I can't see what other variables I'm able to figure out from here, or even if I've done the steps so far correctly.
This isn't a homework question as such, it's from a past university paper and I'm revising for my resits. I could really use help on this ASAP, cos I'm worrying that I might not pass my resits...

2. Aug 12, 2011

### vela

Staff Emeritus
The equation E=hf applies to a single photon. You can't apply it to the laser beam as a whole. In any case, you don't need to resort to the quantum mechanical nature of light for this problem anyway.

The concept you need is that light has momentum. What you need to find is the relationship between the energy of light and its momentum.

3. Aug 15, 2011

### Gavandeshaq

Ok, so now I'm using the equation p=E/c=hv/c=h/λ.
I found the energy transmitted throughout the 15s of the laser shining on the missile.
25MW*15s= 375MJ, sounds like quite a lot to me...
Anyway, from here I use the E=hv part of the equation, which gives me a value of 5.7x10^41 for the velocity. Which is, of course, impossible.
Could you shed some light on the mistake I have very obviously made?

4. Aug 15, 2011

### vela

Staff Emeritus
At a minimum, you need to understand what the quantities in an equation represent so you can know if it applies or not. Verify your understanding of the various equations and what they mean.

5. Aug 15, 2011

### Gavandeshaq

I understand that the v in this equation represents frequency. I'm used to the equation E=hf and that just threw me for a second. My only qualm is that I have values for f and λ now that don't seem realistic.
Obviously here we're dealing with relativistic momentum because photons don't have mass. The value I've come up with for p=h/λ is 1.25 kgm/s. I'm assuming this represents the laser beam as a whole, as I used the total energy transmitted to find these variables.
So from here, will the differential equation for F=dp/dt work?

6. Aug 15, 2011

### vela

Staff Emeritus
In the equation E=hf, what does E represent? I suggest you check your textbook for the exact definition.

7. Aug 15, 2011

### Gavandeshaq

It's the quantum energy of a photon measure in eV, right?

8. Aug 15, 2011

### vela

Staff Emeritus
If it's the energy of a single quantum, why are you plugging in the energy of the beam, which consists of many quanta?

9. Aug 15, 2011

### Gavandeshaq

Because at this moment in time, I'm getting many conflicting sources of data.
I'm desperately trying to understand this concept, as my exam looms nearer.
One source tells me the E is measured in Joules, which leads me to quantify the laser as a whole.
Another now says that it's measured in eV, so I've converted the energy into eV and attempted the working from there. Yet again, I've failed, coming out with a photon λ of 1.17x10^52

10. Aug 15, 2011

### vela

Staff Emeritus
Joules and electron-volts are simply different units of energy. The unit doesn't tell you anything about what E represents. If you're told the energy of a photon is 1 eV or 1.6x10-19 J, you've been given the same information either way.

11. Aug 15, 2011

### vela

Staff Emeritus
Anyway, to get back to what I implied back in post #2, E=hf applies to a single quantum of light. You're not working with a single photon, so the equation isn't relevant to this problem at all. In other words, your approach right now is a dead end. Find another way.

12. Aug 15, 2011

### Gavandeshaq

Ah yeah of course, I've forgotten some basic principles here, gonna have to work on that.
Just needed to know how to get to the correct variables for the equations, think I can figure that out now.
Now that my self esteem and confidence have been significantly lowered I feel I can tackle this exam head on though.
Thanks for your help vela :)

13. Aug 15, 2011

### Gavandeshaq

The laser has a power of 25MW, over 15 seconds this is an energy transfer of 375x10^6J
F= 1/c*dE/dt
F= 3x10^-8 * 375x10^)/15
This gives a force of 1/12N
The missile has a mass of 200kg
F=ma: (1/12)/200=4.17x10^-4 m/s^2
That's the deceleration of the missile due to the laser, correct?

14. Aug 15, 2011

### vela

Staff Emeritus
Where did you get this equation from?
That's not 1/c.
Maybe. Are you supposed to assume the light is absorbed by the missile or that it reflects off the missile?

15. Aug 15, 2011

### Gavandeshaq

I got the equation from another thread on this forum.
Of course it's not, but I only put it like that on this thread to save time, when I calculated it I put 1/c in.
And the missile perfectly absorbs the light. (How does the missile absorbing the light give a different outcome to it reflecting it?)

16. Aug 15, 2011

### vela

Staff Emeritus
Well, it'll give you the right answer, but I can't help suspect you're just looking for an equation that has the right variables rather than understanding if it actually applies to the situation or not.
It doesn't really save time when you write something that's obviously wrong and is a common algebra mistake. You might know what you did, but we can only go off of what you write here. We can't read your mind.
This is essentially a collision problem. If the missile absorbs the light completely, it's a perfectly inelastic collision. If any of the light is reflected, it's not perfectly inelastic since the reflected light will carry away some momentum.

17. Aug 15, 2011

### Gavandeshaq

I came across the equation when looking for an equation that gave me the momentum of a laser beam when energy was a given variable. It so happened that the equation had the correct variables for this situation.
Yeah I'm sorry, I think the "of course" was a misunderstanding. I meant a "whoops, of course it is, how silly of me" kind of thing.
Ok so if the light is completed absorbed, the change in speed decelerates the missile? Does the same thing happen if it is perfectly reflected, since there must be an equal and opposite force?

18. Aug 15, 2011

### vela

Staff Emeritus
It depends on the relative direction of the light and the movement of the missile. If the light hits the missile from directly behind, it'll speed the missile up. If it hits the missile from directly in front, it'll slow it down. Anywhere in between, you get a mix of a change in speed and direction of the missile.

Reflection simply changes the amount of momentum delivered to the missile.

19. Aug 15, 2011

### Gavandeshaq

Ok so say for example, the missile is travelling towards the beam, and it perfectly absorbs it. This would cause a change in speed that would decelerate the missile. But because the missile has absorbed the light, it has gain momentum in one aspect. On the other hand, it's slowed down, so it's lost momentum.
If it perfectly reflects the light, it slows down, and does not gain any momentum like in the previous situation?
Is that accurate? Forgive that it's poorly worded, but I'm sure you can understand what I'm trying to put across.

20. Aug 15, 2011

### vela

Staff Emeritus
Remember momentum is a vector, so if the light hits the missile head on, the momenta point in opposite directions. When the missile absorbs the light, the momentum of the light will cancel part of the momentum of the missile which results in the missile slowing down.