# Finding coordinates when light hit the surface

• May 25th 2010, 06:29 PM
noob mathematician
Finding coordinates when light hit the surface
Hi all,

I am stuck with the following question. This is a question posted to me by my physicist friend

Attached is the figure of a ellipse with a formula of http://www.mathhelpforum.com/math-he...3e3513f0-1.gif. In other word, it has a horizontal radius of 1.5 units and vertical radius of 1 unit.

Now, suppose I launch a light from the coordinates (-1.5,0.0) at an angle alpha http://www.mathhelpforum.com/math-he...1b6d5f09-1.gif in such a way that the light will experience reflection on the surface. How am I going to find all the coordinates when the light hit the surface.

My approach is to find the derivative. Hence I got http://www.mathhelpforum.com/math-he...73bd1a69-1.gif. Then I extend the line until it hit the surface and I check the coordinates. Draw the tangent line and reflect it with protractor manually.

The process is rather tedious since extending the line is not really a rigorous proven way. I hope to hear whether is there any alternative better way of finding the coordinates

Thanks and hope for your help :)
• May 26th 2010, 10:10 AM
earboth
Quote:

Originally Posted by noob mathematician
Hi all,

I am stuck with the following question. This is a question posted to me by my physicist friend

Attached is the figure of a ellipse with a formula of http://www.mathhelpforum.com/math-he...3e3513f0-1.gif. In other word, it has a horizontal radius of 1.5 units and vertical radius of 1 unit.

Now, suppose I launch a light from the coordinates (-1.5,0.0) at an angle alpha http://www.mathhelpforum.com/math-he...1b6d5f09-1.gif in such a way that the light will experience reflection on the surface. How am I going to find all the coordinates when the light hit the surface.

My approach is to find the derivative. Hence I got http://www.mathhelpforum.com/math-he...73bd1a69-1.gif. Then I extend the line until it hit the surface and I check the coordinates. Draw the tangent line and reflect it with protractor manually.

The process is rather tedious since extending the line is not really a rigorous proven way. I hope to hear whether is there any alternative better way of finding the coordinates

Thanks and hope for your help :)

There is something weird with this question: The point from which the light is emitted is placed exactly at the left vertex of the ellipse, that means it is a point of the ellipse from which the light comes.

So in my opinion the question can't be answered ... (Thinking)
• May 26th 2010, 11:19 AM
undefined
If I'm reading correctly, then this problem is usually stated as: There is a very small slit at (the given point) such that a beam of light can enter, etc.

It seems almost exactly like Problem 144 on Project Euler. I don't want to give away answers, but I see nothing wrong with your approach. But rather than using a protractor, why not just calculate the numbers? My solution to problem 144 was a bit messy but only took about 30 lines of Java code. If you can solve problem 144, you will get access to the solution forum with some nice explanations.

And in case you're stuck on one of the more basic issues (it was hard to tell from your post) -- you have an equation of a line, and an equation of an ellipse. To find points of intersection, solve the system of equations.
• May 26th 2010, 06:02 PM
noob mathematician
Quote:

Originally Posted by undefined
If I'm reading correctly, then this problem is usually stated as: There is a very small slit at (the given point) such that a beam of light can enter, etc.

It seems almost exactly like Problem 144 on Project Euler. I don't want to give away answers, but I see nothing wrong with your approach. But rather than using a protractor, why not just calculate the numbers? My solution to problem 144 was a bit messy but only took about 30 lines of Java code. If you can solve problem 144, you will get access to the solution forum with some nice explanations.

And in case you're stuck on one of the more basic issues (it was hard to tell from your post) -- you have an equation of a line, and an equation of an ellipse. To find points of intersection, solve the system of equations.

I aware that by finding the equations of both figures will allow me to obtain the coordinates as required

Actually I am having difficulty in finding the equations of all my incident rays.
Take the first ray for example, will I be able to obtain the equation of the incident ray given only one coordinate (0,-1.5) of the line and the angle $\alpha$?
• May 26th 2010, 09:43 PM
undefined
Quote:

Originally Posted by noob mathematician

I aware that by finding the equations of both figures will allow me to obtain the coordinates as required

Actually I am having difficulty in finding the equations of all my incident rays.
Take the first ray for example, will I be able to obtain the equation of the incident ray given only one coordinate (0,-1.5) of the line and the angle $\alpha$?

From the angle, you can obtain the slope, then use point-slope equation.
• May 26th 2010, 10:49 PM
noob mathematician
Ok this is what I manage to obtain for my 1st ray traveling at $\alpha=60^o$

I let $tan60^o$= my slope for the ray

Then $Y=(tan60^o)X+C$,
$0=(tan60^o)(-1.5)+C$,
C $\sim 2.6$

Equating $Y=(tan60^o)X+2.6$ and $y^2+\frac{x^2}{2.25}=1$ I get (-1.12,0.660)?

Nonetheless, the subsequent stages will be hard. Unlike the case for my 1st ray, I can't seem to find the the incident angle to calculate for the slope of my second ray. Please see picture to know what I mean. Is there a guided algorithm to find out these angles?
• May 27th 2010, 07:23 AM
undefined
It's easier to use point-slope equation than slope-intercept. Also I don't see why you didn't replace $\tan 60^\circ$ with $\sqrt{3}$.

The point of intersection I get, to two decimal places, is (-1.11, 0.67). But depending on your purposes you might want more than two decimal places of precision.

To find the angle of the reflected ray, you can use the derivative like you mentioned in the first post. By the way, the derivative you wrote in the first post isn't correct. Differentiate implicitly and you should get $\frac{dy}{dx} = \frac{-2x}{3y}$.
• May 28th 2010, 01:31 AM
noob mathematician
Quote:

Originally Posted by undefined
It's easier to use point-slope equation than slope-intercept. Also I don't see why you didn't replace $\tan 60^\circ$ with $\sqrt{3}$.

The point of intersection I get, to two decimal places, is (-1.11, 0.67). But depending on your purposes you might want more than two decimal places of precision.

To find the angle of the reflected ray, you can use the derivative like you mentioned in the first post. By the way, the derivative you wrote in the first post isn't correct. Differentiate implicitly and you should get $\frac{dy}{dx} = \frac{-2x}{3y}$.

Hi can you check my mistake for the derivative below:

$y^2+\frac{x^2}{1.5^2}=1$
differentiate to obtain
$2y(\frac{dy}{dx})+\frac{2x}{1.5^2}=0$
$\frac{dy}{dx}= -\frac{2x}{1.5^2}*\frac{1}{2y}$
$\frac{dy}{dx}= -\frac{x}{y}*\frac{1}{1.5^2}$
$\frac{dy}{dx}= -\frac{4x}{9y}$

Neverthelss, I still don't know how to get the angle by using the derivative of the ellipse. I hope it's ok for you to show me the next angle (attached diagram) with some explanation. :)
• May 28th 2010, 06:42 AM
undefined
Quote:

Originally Posted by noob mathematician
Hi can you check my mistake for the derivative below:

$y^2+\frac{x^2}{1.5^2}=1$
differentiate to obtain
$2y(\frac{dy}{dx})+\frac{2x}{1.5^2}=0$
$\frac{dy}{dx}= -\frac{2x}{1.5^2}*\frac{1}{2y}$
$\frac{dy}{dx}= -\frac{x}{y}*\frac{1}{1.5^2}$
$\frac{dy}{dx}= -\frac{4x}{9y}$

Neverthelss, I still don't know how to get the angle by using the derivative of the ellipse. I hope it's ok for you to show me the next angle (attached diagram) with some explanation. :)

Ah, you're right, dumb mistake on my part. I neglected to apply the chain rule on $\left(\frac{x}{1.5}\right)^2$.

As for finding the angle you're after, consider the normal to the curve (look up definition of normal if you're unsure), and use trigonometry or linear algebra.

I really don't like giving away answers to Project Euler problems. The philosophy over there is, if you can't solve it, you can't solve it. There is a competitive or pseudo-competitive aspect to the site, and it just wouldn't be the same if the answers were simply given. To me, not knowing is part of the fun of it, and the challenge of it. So it's mixed luck that your physicist friend posed a question extremely similar to a problem over there. I will not answer any more of your questions. Maybe someone else will.

If you're really motivated to solve this without help, then in my opinion, you will solve it without help. You just may need patience to gain the skills necessary to do so. Work through easier Euler problems and read the solution forums, read/review math books, search Wikipedia and MathWorld and whatnot, solve other problems, etc.

Good luck.