# Laser in front of a wall - mathemathical physics

1. Mar 8, 2014

### skrat

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

2. Relevant equations

3. 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 $min(u+\lambda v)$.

2. Mar 8, 2014

### voko

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. Mar 8, 2014

### skrat

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. Mar 8, 2014

5. Mar 8, 2014

### skrat

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. Mar 8, 2014

### voko

$$t = \frac {2a} {c_0}$$ 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. Mar 8, 2014

### skrat

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. Mar 8, 2014

### voko

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. Mar 8, 2014

### skrat

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

Which is minimum for $sin\varphi =1$.

10. Mar 8, 2014

### voko

$t_L$ is the time later. What is the total time of that?

11. Mar 8, 2014

### skrat

$t_0=t_L+t$ where $t$ is time measured from the beginning of rotation and $t\in [0,\frac{2a}{c_0}]$.

12. Mar 8, 2014

### voko

And the minimum is?

13. Mar 8, 2014

### skrat

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. Mar 8, 2014

### voko

Sign error here, and the derivative of $\phi$ is missing. Recall the chain rule.

15. Mar 8, 2014

### skrat

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. Mar 8, 2014

### voko

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. Mar 8, 2014

### skrat

Sorry to ask, but why would the angle have to be small? How did you guess that?

18. Mar 8, 2014

### voko

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. Mar 8, 2014

### skrat

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. Mar 8, 2014

### skrat

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 $A^2+\pi A-1=0$.

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

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

$t=\frac{1}{\gamma}arccos[\frac{-\pi \pm \sqrt{\pi^2+4}}{2}]$