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MIT2_71S09_gps4 - MASSACHUSETTS INSTITUTE OF TECHNOLOGY...

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MASSACHUSETTS INSTITUTE OF TECHNOLOGY 2.710 Optics Spring ’09 Problem Set #4 Posted March 16, 2009 — Due Wednesday, April 1, 2009 1. Knocking down one dimension: the Screen Hamiltonian In class, we derived the set of Hamiltonian equations in 3D space, where the position vector q = ( q x , q y , q z ) and momentum vector p = ( p x , p y , p z ) are parameterized by the variable s . The “time” variable s indexes location along the ray in the sense that as we advance s from 0 to s 1 to s 2 to s 3 and so on, we progress along successive locations on the ray path (see the diagram below.) Using this parameterization, we arrived at the set of Hamiltonian equations d q d s = ∂H p , d p d s = - ∂H q , where H = | p | - n ( q ) is the Hamiltonian, and it is conserved because it physically expresses momentum conservation for the ray. In this problem, we will settle on the choice H = 0 for the value of the (conserved) Hamiltonian. plane input plane ray path p output q plane xz z y x ( ) z z ( ) [ ] (0) (0) q p s s s z z z x y 1 3 2 1 2 3 The Hamiltonian equations are a 6 × 6 set of coupled ordinary differential equa- tions. In this problem we will show how this can be simplified to a 4 × 4 set, essentially knocking down one spatial dimension. We begin by the observation that s seems to be somewhat superfluous: instead of it, we should be able to parameterize with respect to a spatial variable, say z . Graphically, this is also shown in the figure, where s 1 , s 2 , . . . have been projected to their corresponding z 1 , z 2 , . . . representing locations on the optical axis. Now the ray is parameterized as [ q ( z ) , p ( z )] instead of [ q ( s ) , p ( s )]. Formally, we do this by eliminating s from the set of equations x = q x ( s ) , y = q y ( s ) , z = q z ( s );
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i.e. , solve the third equation for s ( z ) and substitute into the first two equations, which then become x = q x ( z ) , and y = q y ( z ) . We seek to derive a new set of ordinary differential equations to describe the trajectory [ q ( z ) , p ( z )]. a) Using the chain rule, show that the following equations are true, d q x = p x , d p x | p | ∂n , d z p z d z = p z ∂q x d q y p y d p y p ∂n = , = | | , d z p z d z p z ∂q y d p z p ∂n d z = | p z | ∂q z . b) Use the Hamiltonian conservation principle to rewrite these as d q x d z = p x p z , d p x d z = n p z ∂n ∂q x , d q y d z = p y p z , d p y d z = n p z ∂n ∂q y , d p z d z = n p z ∂n ∂q z ; and show that p z itself, which so pervasively appears in all the denominators, is given by p z = n 2 ( q ) p 2 x + p 2 y . (Technically, both solutions p z = ± . are allowed, but we choose the “+” sign due to the convention that light rays should propagate toward the right hand side, i.e. towards positive z .) c) Now set h q x , q y , z ; p x , p y ≡ − p z q x , q y , z ; p x , p y = n 2 ( q ) p 2 x + p 2 , y and show that the “lateral” dynamical variables [( q x , q y ) ( p x , p y )] obey the 4 × 4 set of Hamiltonian equations (in terms of z ) d q x ∂h d p x ∂h d z = ∂p x , d z = ∂q x , d q y ∂h d p y ∂h d z = ∂p y , d z = ∂q y .
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