MIT2_71S09_gsol4

MIT2_71S09_gsol4 - MASSACHUSETTS INSTITUTE OF TECHNOLOGY...

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MASSACHUSETTS INSTITUTE OF TECHNOLOGY 2.710 Optics Spring &09 Solutions to Problem Set #4 Due Wednesday, April 1, 2009 Problem 1: Knocking down one dimension: the Screen Hamiltonian a) The main goal of this problem is to show how the 6 & 6 set of Hamiltonian equa- tions can be simpli&ed to a 4 & 4 set of ordinary di/erential equations known as Screen Hamiltonian equations . The Screen Hamiltonian equations describe the evolution of the intersection of the ray path with the screens, that are perpendicular to the optical axis, as z advances. The geometry of this problem is shown in Figure 1. We begin by writing in an explicit form the set of Hamiltonian equations, dq x = @H ds ; dp x @p x = ds ± @H ; @q x dq y = @H ds ; dp y @p y = ds ± @H ; @q y dq z = @H ds ; dp z @p z = ds ± @H (1) ; @q z where the conserved Hamiltonian is, H = n ( q ) ± q p 2 x + p 2 y + p 2 z = 0 : (2) As shown in Figure 1, di/erent points in the ray trajectory ( s 1 ; s 2 ; s 3 ; ²²² ) have been projected to their corresponding axial coordinate ( z 1 ; z 2 ; z 3 ; ) changing the parame- terization of the ray from [ q ( s ) ; p ( s )] to [ q ( z ) ; p ( z )] : We then apply the chain rule when taking the derivatives of the components of the position vector with respect to z and use the results of equations 1 and 2, dq x dx = dz dx = dz ds ds @ = dz & H @ @p x ±& p z ( @H ± 3) dq ) x p x = dz ; p z dq y dy = dz dy = dz ds ds @ = dz & H @ y p z ( ± 4) dq ) y p y = dz ; p z dq z dz = dz = 1 : (5) dz Similarly, we take the derivatives of the components of the momentum vector with 1
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respect to z , dp x dp x = dz ds ds @ = dz & H @ @q x ±& p z ( @H ± 6) dp ) x p = j j dz & @n p z ; x dp y dp y = dz ds ds @ = dz & H @ y p z ( ± 7) dp ) y p = j j dz & p z ; y dp z dp z = dz ds ds @ = dz & H @ z p z ( ± 8) dp ) z p = j j dz & p z : z b) From the Hamiltonian conservation principle of equation 2, we see that j p j = n ( q ) , so equations 3 to 8 become, dq x = p x dz ; dp x p z = dz & n @n p z ; @q x dq y = p y dz ; dp y p z = dz & n @n p z ; @q y dp z = dz & n @n p z (9) : @q z Solving for p z from equation 2, p z = q n ( q ) 2 & ² p 2 x + p 2 y ³ : (10) c) We now set, h ( q x ; q y ; z ; p x ; p y ) ± & p z ( q x ; q y ; z ; p x ; p y ) = & q n ( q ) 2 & ² p 2 x + p 2 y ³ ; (11) and use it to elliminate p z , that appears in the equation set 9, dq x p x = dz q @h = n ( q ) 2 & ² p 2 x + p 2 y ³ ; (12) @p x dq y p y = dz q = n ( q ) 2 & ² p 2 x + p 2 y ³ ; y dp x n = dz & q n ( q ) 2 & ² p 2 x + p 2 y ³ = x & ; x dp y n = dz & q n ( q ) 2 & ² p 2 x + p 2 y ³ = y & : y 2
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Figure 1: Screen Hamiltonian. We also have the additional equation, dp z n = dz & @n h : (13) @z d) The equation set 12, is a proper set of Hamiltonian equations with h as the Screen Hamiltonian ; therefore, h is conserved in this 2D space. However, equation 13 shows that the Screen Hamiltonian is not conserved in general, unless @n=@z = 0 , where the index is invariant along the optical axis. It is okay for the Screen Hamiltonian to not be conserved because the lateral momentum, ( p x ; p y ) ; is not generally conserved; however, the 3D momentum, ( p x ; p y ; p z ) ; must be conserved. This is also known as the Phase Matching Condition .
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MIT2_71S09_gsol4 - MASSACHUSETTS INSTITUTE OF TECHNOLOGY...

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