MIT2_71S09_gps4 - MASSACHUSETTS INSTITUTE OF TECHNOLOGY...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 r a y p a t h 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 ); 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 [(...
View Full Document

This note was uploaded on 02/24/2012 for the course MECHANICAL 2.710 taught by Professor Sebaekoh during the Spring '09 term at MIT.

Page1 / 8

MIT2_71S09_gps4 - MASSACHUSETTS INSTITUTE OF TECHNOLOGY...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online