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)/(alpha)) where y is the vertical distance above the starting position. Show that the path taken by light from (0,0) to (2X,0) is the arc of a circle centeblack at (X,alpha). Note that the origin is chosen to be well above the ground, so that y<0 is allowable.

Ishaan Booker 2022-07-22 Answered
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 < 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.
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Answers (1)

Steven Bates
Answered 2022-07-23 Author has 15 answers
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's principle one obtains the ode
1 v 1 + y 2 = const
Here one solves the "boundary value problem"
{ 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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