solutionhw1 - Solutions to Homework Set 1 1 Snellius law...

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Solutions to Homework Set 1 1) Snellius’ law : Parts a) and b) are trivial, so I will not write out the solutions. Part c) has two sub-parts: i) Setting the variation of F 1 [ y ] to zero gives d dx n ( x ) y 0 1 + y 0 2 ! = 0 . Now elementary calculus tells us that y 0 = tan θ , where θ is the angle between the light ray and the x -axis. The x -axis is, however, for this geometry, the normal to the planes of constant n ( x ). Thus we have θ = ψ , where ψ is the angle of incidence as it is usually defined in optics. Using the standard trig identities sec 2 θ - tan 2 θ = 1, and sec θ = 1 / cos θ , we reduce our equation to d dx n ( x ) sin ψ = 0 , or n ( x ) sin ψ = const. This last equation is the usual form of Snell’s law. ii) For F 2 [ y ] we can use the first integral to deduce directly that n ( y ) y 0 2 1 + y 0 2 - n ( y ) q 1 + y 0 2 = const., or, collecting terms, n ( y ) 1 + y 0 2 = const. Again using y 0 = tan θ and sec 2 θ - tan 2 θ = 1, this reduces to n ( y ) cos θ = const. In this geometry the angle of incidence is the complement of θ , so cos θ = sin ψ , and again Snell’s law is obtained.
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