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Thread: Laser pointed at mirror - how to find the point it reflects back to?

  1. #1
    Sep 2013
    Longmont, CO

    Laser pointed at mirror - how to find the point it reflects back to?

    Hey everyone, new to the forum, taking Calc 3, just looking for some help on some problems, will help if I can. Not sure how to write equations on this thing but I'll do the best that I can.

    Consider a room ten units long in the x,y, and z directions. Specifically, the walls of the room are the four planes,
    x= 0, x= 10, y= 0, and y= 10, and the floor and ceiling are z= 0 and z= 10, respectively. A at triangular mirror is
    mounted in one of the corners of the ceiling. The corners of the mirror are at locations (10,9,10), (10,10,9), and (9,10,10).
    You are sitting at location (5,0,0) playing with your new green laser pointer.

    (a) If you aim your laser pointer directly at the corner of the room with coordinates (10,10,10),

    determine the coordinates where the re-directed beam will hit the walls, or floor, of the room.

    (Hint: an incoming ray of light, and the surface normal where the ray hits the surface, form a
    plane. The re-directed ray is in the same plane. Also, the angle between the incoming ray and the

    normal is the same as the angle between the normal and the re-directed ray.

    My attempt at a solution:
    So I took the starting point and where I am aiming the laser and made it into a vector to be the laser beam. <5,10,10>, then reduced it by a scale of 5 to have a more workable but still parallel vector <1,2,2>. I then parameterized it to start at (5,0,0).
    x(t)=5+t, y(t)=2t, z(t)=2t

    Next I found the normal vector to the plane of the mirror <1,1,1>, and used 1 point (9, 10, 10) to find the equation for the plane: x+y+x=29

    Plugged my parameter equations in for x,y, and z to find where the laser hits the mirror: (9.8, 9.6, 9.6)

    Since I'm told the the normal vector to the mirror and the laser vector form a plane that the redirected beam will stay on, I took the cross product of the two vectors to find the normal of that plane. So <1,1,1> X <1,2,2>, which I found to be equal to: <0,-1,1>.

    I know that this plane must exist at the point (9.8, 9.6, 9.6) so I plug all of this in to find the standard equation for a plane. I end up with y=z.

    I am also told that the angle between the incoming ray and the normal of the mirror is the same as the outgoing ray and the normal of the mirror.
    So cos(theta) = (normal dot laser)/(|n||L|), I work this out to find that theta is: 15.8 degrees.

    This is where I'm stuck. I think that I can use that same equation but use the reflected beam: cos(15.8) = (normal dot reflected beam)/(|n||R|) to find it, so I put in <x,y,z> for the reflected beam but this is where I'm stuck. I have two equations: y=z and cos(15.8)=vector stuff but I have 3 unknowns. Can I assume that it will hit at x=0? If I plug that in and y=z in, then I get cos(15.8)=2/sqrt(6), because the y's or z's end up cancelling.

    Not sure where I went wrong here, if anyone can shed some light it would be awesome. Thanks!
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  2. #2
    MHF Contributor
    Sep 2012

    Re: Laser pointed at mirror - how to find the point it reflects back to?

    Hey caseym33.

    Once you translate the point of intersection to the origin you then have a normal vector and a direction vector of the incoming light ray. If you reverse the direction vector, normalize it and calculate <n,d> you will get cos(theta).

    Now you need to calculate a new direction vector that is going out from the surface. To do this you must reflect the vector about the normal vector.

    This is because the reversed direction vector (from the incoming vector of light) and the final direction vector are symmetrical with respect to the normal.

    The rotated vector needs to be rotated 2*theta along the plane formed by the normal vector and the reversed direction vector. The plane has normal r x n (r is reversed incoming vector, n is normal) which is equal to n x i (incoming vector) where all vectors are unit length.

    Then you rotate the direction vector about this axis of 2*theta and you're done. Also you don't need to calculate theta directly: you can use cos(theta) calculated from your inner product (and you can calculate sin(theta) from cross product) and substitute this into the rotation matrices directly.

    So in summary:

    1) Translate point of contact to the origin (subtract the value of the contact point)
    2) Form vectors for normal of surface and reversed direction of incoming light
    3) Calculate cos(theta) and sin(theta) using dot and cross products
    4) Form rotation matrix that rotates reverse direction vector about normal formed by normal of surface and reversed direction vector
    5) Obtain final direction vector
    6) Translate this vector back to original co-ordinate system (add the contact point obtained from 1) and repeat process for further rays.
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