{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

MIT2_71S09_gsol4

# MIT2_71S09_gsol4 - MASSACHUSETTS INSTITUTE OF TECHNOLOGY...

This preview shows pages 1–4. Sign up to view the full content.

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

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

View Full Document
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
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 = 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 .

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 18

MIT2_71S09_gsol4 - MASSACHUSETTS INSTITUTE OF TECHNOLOGY...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online