The velocity of light above a hot surface decreases with the height from that surface. The velocity is given by v = v 0 ( 1 − y α ) where y is the vertical distance above the starting position. Show that the path taken by light from (0,0) to ( 2 X , 0 ) is the arc of a circle centeblack at ( X , α ). Note that the origin is chosen to be well above the ground, so that y &lt; 0 is allowable.Attempt at a solution: After working through from the velocity relation given in the question, we get the following ODE: 1 v 0 ( 1 + y α ) 1 + y 2 = − C where C is a constant.

Question

The velocity of light above a hot surface decreases with the height from that surface. The velocity is given by   v  =      v    0    (            1      −      y        α    ) where   y is the vertical distance above the starting position. Show that the path taken by light from (0,0) to   (  2  X  ,  0  ) is the arc of a circle centeblack at   (  X  ,  α  ). Note that the origin is chosen to be well above the ground, so that   y  &amp;lt;  0 is allowable.Attempt at a solution: After working through from the velocity relation given in the question, we get the following ODE:             1                        v          0                (                              1            +            y                    α                )                  1          +                      y            2                                =  −  C where   C is a constant.

Steven Bates · Accepted Answer

The question concerns the light path in a vertically inhomogeneous medium with the velocity   v  (  y  ) supposed linearly dependent on the height   y  (  x  ).Here  v  =      v    0        (    1    −          y      α        )  From Fermat&#039;s principle one obtains the ode      1          v              1        +                  y                      ′            2                                =  constHere one solves the &quot;boundary value problem&quot;      {                                        y            ′                    =                                                                      α                  2                                −                                  C                  2                                                  v                  0                  2                                                (                y                −                α                                  )                  2                                                            C                                  v                  0                                                (                y                −                α                )                                                                                                                  y          (          0          )          =          0                                                          y          (          2          X          )          =          0                        The ode is solved by separation of variables remembering that   (      u        )    ′    =      u    ′        /    (  2      u    ).One obtains            α      2        −          C      2              v      0      2            (    y    −    α          )      2        =  C      v    0    (  x  +  B  )with      {                            B          =          −          X                                                          C          =                                    α                                                v                  0                                                                      X                    2                                    +                                      α                    2                                                                                              Finally the equation of the circle  (  x  −  X      )    2    +  (  y  −  α      )    2    =      X    2    +      α    2

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