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Two Mirrors (Posted on 2006-10-08) Difficulty: 2 of 5
Two mirrors are placed in the first quadrant of the xy-plane (perpendicular to the plane). The first mirror along the line y = b-x (for some b > 0) and facing the point (0,0). The second mirror along the line y = mx (where m > 0) and facing the point (b,0). A light source at point (a,0), 0 < a < b, shoots a beam of light into the first quadrant parallel to the first mirror. Find m such that when the beam is reflected exactly once by each mirror, it passes through the original light source at point (a,0).

  Submitted by Bractals    
Rating: 2.5000 (4 votes)
Solution: (Hide)
      Label the points as follows:

        O - the origin (0,0)
        A - the source of the light (a,0)
        B - the intersection of the first mirror
            and the x-axis (b,0)
        C - the reflection point at the second mirror 
        D - the reflection point at the first mirror

      ACDA is the path the light beam travels.

      Using the sine rule,
     
            |AO|          |AC|         
        ----------- = -----------
         sin(<ACO)     sin(<AOC)       

            |AC|          |AD|
        ----------- = -----------
         sin(<ADC)     sin(<ACD)  

            |AD|          |AB|
        ----------- = -----------
         sin(<ABD)     sin(<ADB)   
   
      If t is the measure of <AOC, then
     
             a           |AC|         
        ------------ = --------
         sin(135-t)     sin(t)       

            |AC|            |AD|
        ------------- = ------------
         sin(360-4t)     sin(2t-90)  

           |AD|          b-a
        --------- = ------------
         sin(45)     sin(2t-90)

      Therefore,

                                      a sin(t)
                                  ---------------
         |AC|     sin(360-4t)        sin(135-t)
        ------ = ------------- = -----------------  
         |AD|     sin(2t-90)       (b-a) sin(45)
                                  ---------------
                                     sin(2t-90)

                           or

        a sin(t) sin(2t-90)2 = (b-a) sin(45) sin(135-t) sin(360-4t)

                           or

        a sin(t) cos(2t)2 = (b-a) ½ [cos(t) + sin(t)] [- sin(4t)]

                           or
                                 
        a cos(2t) = 2(a-b) cos(t) [cos(t) + sin(t)]

                           or

        a [1 + 2 cos(t)sin(t] = 2b cos(t) [cos(t) + sin(t)]

                           or

        a [cos(t) + sin(t)]2 = 2b cos(t) [cos(t) + sin(t)]

                           or

        a [cos(t) + sin(t)] = 2b cos(t)

                           or

        a sin(t) = (2b - a) cos(t) 

      Thus,

                      sin(t)     2b - a
        m = tan(t) = -------- = --------
                      cos(t)        a

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Some ThoughtsTo startJer2006-10-10 11:36:40
Hints/Tipsplaying with Geometer's Sketchpad (spoiler)Charlie2006-10-08 13:04:40
SolutionAnswer?Adeline Redding2006-10-08 13:02:18
Starting thoughtsLarry2006-10-08 11:03:56
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