Results 1 to 9 of 9

Math Help - Finding coordinates when light hit the surface

  1. #1
    Member
    Joined
    Oct 2008
    Posts
    206

    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 . 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 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 . 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
    Attached Thumbnails Attached Thumbnails Finding coordinates when light hit the surface-ray-tracing.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,812
    Thanks
    116
    Quote Originally Posted by noob mathematician View Post
    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 . 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 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 . 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 ...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Oct 2008
    Posts
    206
    Quote Originally Posted by undefined View Post
    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.
    Hi thanks for the reply.

    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?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Quote Originally Posted by noob mathematician View Post
    Hi thanks for the reply.

    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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Oct 2008
    Posts
    206
    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?
    Attached Thumbnails Attached Thumbnails Finding coordinates when light hit the surface-ray-tracing-2.jpg  
    Last edited by noob mathematician; May 26th 2010 at 11:54 PM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    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}.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Oct 2008
    Posts
    206
    Quote Originally Posted by undefined View Post
    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.
    Attached Thumbnails Attached Thumbnails Finding coordinates when light hit the surface-ray-tracing3.jpg  
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Quote Originally Posted by noob mathematician View Post
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding the coordinates when light hit the ellipse
    Posted in the Math Software Forum
    Replies: 0
    Last Post: May 24th 2010, 10:55 PM
  2. Replies: 6
    Last Post: May 20th 2009, 05:28 AM
  3. surface area of a light bulb
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 3rd 2008, 01:58 AM
  4. Replies: 5
    Last Post: October 7th 2007, 07:10 PM
  5. Surface Area in Polar Coordinates
    Posted in the Calculus Forum
    Replies: 4
    Last Post: August 2nd 2006, 10:14 AM

Search Tags


/mathhelpforum @mathhelpforum