Laser in front of a wall - mathemathical physics

  • Thread starter skrat
  • Start date
  • #1
746
8

Homework Statement


In front of a long, straight, white wall a laser is hanged. Laser is parallel to the wall when we decide to rotate it with constant angular velocity for ##\pi ## so that the laser spot travels over the entire wall. The duration of the turn lasts just as long as it takes for the laser light to travel to the wall and back to the laser. The wall radiates the light equally in all directions.

An observer standing next to the laser watches the light on the wall using a high speed camera. When will he first notice a bright spot and from which direction? How fast does the spot on the wall move at ##\pi / 4## (angle next to the perpendicular line)?

Homework Equations





The Attempt at a Solution



Let's start by saying that the laser is distance ##a## from the wall. The rotation therefore lasts for ##t=\frac{2a}{c_0}##, where ##c_0## is of course the speed of light.

##\pi =\omega t=\omega \frac{2a}{c_0}## and from here ##\omega =\frac{\pi c_0}{2a}##.

So finally ##\varphi (t)=\frac{\pi c_0}{2a}t## for ##t\in [0, \frac{2a}{c_0}]##.

I assume, that every spot on the wall that laser hits while rotating, becomes a light source that radiates light in all directions. Now I THINK I am somehow supposed to use Lagrangian multiplier to find the extreme value of time that light needs to come back to the laser. But I am having some really big troubles to find out what ##u## and ##v## are in ##min(u+\lambda v)##.

Is this even the right idea? Could somebody please help?
 

Answers and Replies

  • #2
6,054
391
With respect to what is the angle measured? And, given the angle, what is the duration of the round trip? And what is the time when that round trip ends?
 
  • #3
746
8
Am... I guess i wasn't clear enough.
At ##t=0## the laser is parallel to the wall, therefore the angle of rotation is ##\varphi =0## and there is no bright spot on the wall. The whole rotation of the laser for ##\pi ## will take ##t= \frac{2a}{c_0}## and is calculated above.

For example at ##t/2## the laser will be perpendicular to the wall.

I hope that answers your question or maybe I completely missunderstood you. If that is the case, I will prepare a sketch of the problem as soon as i get to the computer.
 
  • #4
6,054
391
That only answers the first question. What about the other two?
 
  • #5
746
8
With respect to what is the angle measured? And, given the angle, what is the duration of the round trip? And what is the time when that round trip ends?



The duration of the round trip is equal to the duration that light needs if it travels from the laser, to the closest point on the wall and back. That is the same as if ##\varphi = \pi /2##. Let's mark that distance with ##a##.

The time is than ##t=\frac{2a}{c_0}##. But that is exactly what i have written in the previous two posts so i guess you want to hear something else.
 
  • #6
6,054
391
$$ t = \frac {2a} {c_0} $$ only when the laser is illuminating the wall at the right angle with the wall. At any other time, the round trip distance is not ##2a## and so the round trip duration is different. What is it at an arbitrary angle?
 
  • #7
746
8
Aaaaa, you mean round trip od the light and not the laser. :D hehe, sorry, english is not my mother language but i try.

Of course ##t_L=\frac{2a}{c_0sin(\varphi (t))}##.
 
  • #8
6,054
391
So when does the light emitted by the laser at time ##t## comes back to its origin? And what is minimal value of that?
 
  • #9
746
8
So when does the light emitted by the laser at time ##t## comes back to its origin? And what is minimal value of that?

##t_L=\frac{2a}{c_0sin(\varphi (t))}## later.

Which is minimum for ##sin\varphi =1##.
 
  • #10
6,054
391
##t_L## is the time later. What is the total time of that?
 
  • #11
746
8
##t_0=t_L+t## where ##t## is time measured from the beginning of rotation and ##t\in [0,\frac{2a}{c_0}]##.
 
  • #12
6,054
391
And the minimum is?
 
  • #13
746
8
Hmmm... I really hope I am not writing stupidity right now:

##t_0=t_L+t##

##t_0=\frac{2a}{c_0sin(\varphi (t))}+t##

##\frac{\partial t_0}{\partial t}=\frac{2acos(\varphi (t))dt}{c_0sin^2(\varphi (t))}+1=0##

##\frac{2acos(\frac{\pi c_0}{2a}t)\frac{\pi c_0}{2a}}{c_0sin^2(\frac{\pi c_0}{2a}t)}+1=0##

##cos(\frac{\pi c_0}{2a}t)=\frac{1}{\pi }sin^2(\frac{\pi c_0}{2a}t)##

I can probably assume that ##t## is very close to zero which hopefully gives me a green light to use Taylor series...

##1-\frac{1}{2}(\frac{\pi c_0}{2a})^2t^2=\frac{1}{\pi }(\frac{\pi c_0}{2a})^2t^2##

##t=\sqrt{\frac{1}{(\frac{\pi c_0}{2a})^2(\frac{1}{2}+\frac{1}{\pi })}}##

:D
Right now I just hope I didn't do anything embarrassing. :D
 
  • #14
6,054
391
Hmmm... I really hope I am not writing stupidity right now:

##t_0=t_L+t##

##t_0=\frac{2a}{c_0sin(\varphi (t))}+t##

##\frac{\partial t_0}{\partial t}=\frac{2acos(\varphi (t))dt}{c_0sin^2(\varphi (t))}+1=0##

Sign error here, and the derivative of ##\phi## is missing. Recall the chain rule.
 
  • #15
746
8
Sorry, I have the right sign in the notes. There should also stand ##\frac{d\varphi }{dt}## and not ##dt##.

##\frac{\partial t_0}{\partial t}=-\frac{2acos(\varphi (t))\frac{d\varphi }{dt} }{c_0sin^2(\varphi (t))}+1=0##

Where ##\varphi (t)=\frac{\pi c_0}{2a}t## and ##\frac{d\varphi }{dt}= \frac{\pi c_0}{2a}##

However, the the rest should be fine.
 
  • #16
6,054
391
You do not have to assume that ##t## is small. And, even if it is small, the angle need not be small.

You can solve the equation exactly.
 
  • #17
746
8
You do not have to assume that ##t## is small. And, even if it is small, the angle need not be small.

Sorry to ask, but why would the angle have to be small? How did you guess that?
 
  • #18
6,054
391
I am not guessing. I am saying that you do not know a priory that it will be small. And there is a way to solve that exactly, so there is no need for that guesswork.

P.S. Sometimes you cannot solve an equation exactly. Then you would have to think how it can be approximated in a physically valid way. But that is not guessing, and not required for this problem anyway.
 
  • #19
746
8
Understood.

Well, the reason I thought about approximation is because I have no idea at all on how to exactly solve this... Could you give me a hint or two?

##\pi cos(\frac{\pi c_0}{2a}t)=sin^2(\frac{\pi c_0}{2a}t)##
 
  • #20
746
8
Or... I can:

##\pi cos(\gamma t)=sin^2(\gamma t)## for ##\gamma =\frac{\pi c_0}{2a}##.

##\pi cos(\gamma t)= 1-cos^2(\gamma t)## which is quadratic equation for ##cos(\gamma t)##.

For ##A=cos(\gamma t)## the equation is ##A^2+\pi A-1=0##.

##A_{1,2}=\frac{-\pi \pm \sqrt{\pi^2+4}}{2}##

##\gamma t=arccos[\frac{-\pi \pm \sqrt{\pi^2+4}}{2}]##

##t=\frac{1}{\gamma}arccos[\frac{-\pi \pm \sqrt{\pi^2+4}}{2}]##
 
  • #21
6,054
391
Good! Note that the value of the arccosine gives you the angle. Now you can check whether your idea that you could use the Taylor expansion was good or not.
 
  • #22
746
8
Already checked that. =)

If ##a=2 m## the laser will rotate for ##1.33\cdot 10^{-8}s##.

Observer next to the laser should see the bright spot after ##5.41\cdot 10^{-9}s## cacluated exactly and ##4.69\cdot 10^{-9}s## calculated using Taylor expansion which is for about 13% wrong. But, still better than I expected.

Anyway, thank you for your awesome guidance, voko!

One questioin still remains: How fast does the spot on the wall move at ##\varphi = \pi /4##?

Is it ok if I say that ##v=\omega R## (as if it was circling) I am afraid that by saying that I assume ##a\gg1##.

But it kinda does make sense, because ##v=\omega \frac{a}{sin(\varphi(t))}=\frac{\pi c_0}{2}\frac{1}{sin(\varphi(t))}##. Velocity is therfore the smallest for ##\varphi=\pi /2## and goes to infinity if ##\varphi=0## or ##\varphi =\pi /2##.
 
  • #23
6,054
391
Observe that the angle from which the first reflection arrives is independent of ##\gamma## and is 1.2752 (73°); clearly such an angle is not small and just one or two terms of the Taylor expansion give a very coarse approximation.

Regarding your approach for the second part, I would first like to understand what exactly "how fast" means. How is it measured? Do we stick a ruler next to the spot and observe how fast it moves from the vicinity of the spot, or do we observe the spot and the ruler from the original location? Or do we measure it in some other way?
 
  • #24
746
8
No ruler is mentioned. I assume that we observe the spot from the original location (next to the laser) and measure the speed of the spot on the wall. I guess adding a ruler would be useful.
 
  • #25
6,054
391
The question is, how do you measure the speed of a distant spot? If we have a ruler there, then we read (at origin) at time ##t_1## position ##p_1##, and at time ##t_2## position ##p_2##. Assuming these are close to each other, the speed is given by $$ p_2 - p_1 \over t_2 - t_1 $$ Note, however, that if the times are not taken at the origin (after the round trip), but at the wall (after only the forward leg of the trip), then ##t_2' - t_1'## is roughly half ##t_2 - t_1##, so the speed is twice greater. That is why you need to choose whether the speed is measured locally or remotely.
 

Related Threads on Laser in front of a wall - mathemathical physics

  • Last Post
Replies
11
Views
1K
Replies
5
Views
1K
  • Last Post
Replies
2
Views
980
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
1
Views
6K
  • Last Post
Replies
2
Views
920
Replies
0
Views
1K
  • Last Post
Replies
4
Views
15K
Top