410aex1soln - Exam 1 Phys410 Fall 2011...

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Exam 1 — Phys410 — Fall 2011 Prof. Ted Jacobson www.physics.umd.edu/grt/taj/410a/ jacobson@umd.edu Point values are the numbers in brackets. Total points = 50 . Write clearly and explain your work to be eligible for partial credit. 1. A mirage happens when the air near the ground is hotter than the air above the ground, so that the air density is less near the ground, hence the speed of light is greater. We can use Fermat’s principle of least time to find the path traveled by a light ray under these circumstances. Let the speed of light be c/n ( y ), where the index of refraction has the linear form n ( y ) = n 0 + ky for some constants n 0 and k . Consider a light ray that goes from ( x 1 ,y 1 ) to ( x 2 ,y 2 ), where x is the horizontal coordinate and y is the vertical coordinate. (a) Write an integral over x that expresses the total travel time of the light ray in terms of the given quantities. [3] (b) Convert your integral into an integral over y (don’t worry that there may be two x values for a given y along the full path). This simplifies the next step. [2] (c) Find the differential equation for x ( y ) by imposing the condition that the travel time is minimized. [3] (d) Convert your equation into a differential equation for y ( x ). [2] [The math is quite similar to the soap film and catenary problems we solved. In fact, the solution for y has the form y = a + b cosh( x/b ) for some constants a and b .] (a) If s is the arc length we have ds/dt = c/n , so time = Z dt = Z ds/ ( ds/dt ) = (1 /c ) Z dsn ( y ) = (1 /c ) Z x 2 x 1
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410aex1soln - Exam 1 Phys410 Fall 2011...

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