Legacy Notes for the SIE Lens Hans Georg Schaathun
NTNU---Norwegian University of Science and Technology
Ben David Normann
NTNU---Norwegian University of Science and Technology
In Polar Co-ordinates ¶ ∂ ψ R ∂ x = C 0 f f ′ ⋅ ( cos θ ⋅ sinh − 1 ( f ′ f ⋅ ( cos ϕ cos θ + sin ϕ sin θ ) ) − sin θ ⋅ sin − 1 ( f ′ ⋅ ( − cos ϕ sin θ + sin ϕ cos θ ) ) ) ∂ ψ R ∂ y = C 0 f f ′ ⋅ ( sin θ ⋅ sinh − 1 ( f ′ f ⋅ ( cos ϕ cos θ + sin ϕ sin θ ) ) + cos θ ⋅ sin − 1 ( f ′ ⋅ ( − cos ϕ sin θ + sin ϕ cos θ ) ) ) \begin{aligned}
\frac{\partial\psi^{\mathrm{R}}}{\partial x} &=
C_0\frac{\sqrt{f}}{f'}\cdot
\bigg(
\cos\theta\cdot\sinh^{-1}\left(\frac{f'}{f}\cdot(\cos\phi\cos\theta+\sin\phi\sin\theta)\right)
\\&
-\sin\theta\cdot\sin^{-1}\left(f'\cdot(-\cos\phi\sin\theta+\sin\phi\cos\theta)\bigg)
\right)
\\
\frac{\partial\psi^{\mathrm{R}}}{\partial y} &=
C_0\frac{\sqrt{f}}{f'}\cdot
\bigg(
\sin\theta\cdot\sinh^{-1}\left(\frac{f'}{f}\cdot(\cos\phi\cos\theta+\sin\phi\sin\theta)\right)
\\&+\cos\theta\cdot\sin^{-1}\left(f'\cdot(-\cos\phi\sin\theta+\sin\phi\cos\theta)\bigg)
\right)
\end{aligned} ∂ x ∂ ψ R ∂ y ∂ ψ R = C 0 f ′ f ⋅ ( cos θ ⋅ sinh − 1 ( f f ′ ⋅ ( cos ϕ cos θ + sin ϕ sin θ ) ) − sin θ ⋅ sin − 1 ( f ′ ⋅ ( − cos ϕ sin θ + sin ϕ cos θ ) ) ) = C 0 f ′ f ⋅ ( sin θ ⋅ sinh − 1 ( f f ′ ⋅ ( cos ϕ cos θ + sin ϕ sin θ ) ) + cos θ ⋅ sin − 1 ( f ′ ⋅ ( − cos ϕ sin θ + sin ϕ cos θ ) ) ) Since these expression are constant in R R R
∂ f ∂ x = − ( sin ϕ R ) ∂ f ∂ ϕ ∂ f ∂ y = ( cos ϕ R ) ∂ f ∂ ϕ \begin{aligned}
\frac{\partial f}{\partial x} &=
- \big(\frac{\sin\phi}{R}\big)\frac{\partial f}{\partial\phi}
\\
\frac{\partial f}{\partial y} &=
\big(\frac{\cos\phi}{R}\big)\frac{\partial f}{\partial\phi}
\end{aligned} ∂ x ∂ f ∂ y ∂ f = − ( R sin ϕ ) ∂ ϕ ∂ f = ( R cos ϕ ) ∂ ϕ ∂ f where f = ψ x , ψ y f=\psi_x,\psi_y f = ψ x , ψ y
Amplitudes for arbitrary orientation ¶ The deflection is given as the vector ∇ ψ R \nabla\psi^{\mathrm{R}} ∇ ψ R α ⃗ ( x , y ) \vec{\alpha}(x,y) α ( x , y ) θ \theta θ θ \theta θ
We will let ( x , y ) (x,y) ( x , y ) ( x ′ , y ′ ) (x',y') ( x ′ , y ′ )
[ x ′ y ′ ] = [ cos θ sin θ − sin θ cos θ ] ⋅ [ x y ] \begin{align}
\begin{bmatrix} x' \\ y' \end{bmatrix}
&=
\begin{bmatrix}
\cos\theta & \sin\theta \\
-\sin\theta & \cos\theta
\end{bmatrix}
\cdot
\begin{bmatrix} x \\ y \end{bmatrix}
\end{align} [ x ′ y ′ ] = [ cos θ − sin θ sin θ cos θ ] ⋅ [ x y ] In other words, the ( x , y ) (x,y) ( x , y ) ∇ ψ R ( x ′ , y ′ ) \nabla\psi^{\mathrm{R}}(x',y') ∇ ψ R ( x ′ , y ′ ) α ⃗ ( x , y ) \vec{\alpha}(x,y) α ( x , y ) ∇ ψ R \nabla\psi^{\mathrm{R}} ∇ ψ R
α ⃗ ( x , y ) = [ cos θ − sin θ sin θ cos θ ] ⋅ ∇ ψ R ( x ′ , y ′ ) \begin{align}
\vec{\alpha}(x,y)
&=
\begin{bmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix}
\cdot
\nabla\psi^{\mathrm{R}}(x',y')
\end{align} α ( x , y ) = [ cos θ sin θ − sin θ cos θ ] ⋅ ∇ ψ R ( x ′ , y ′ ) This gives
α ⃗ ( x , y ) = C 0 f f ′ ⋅ [ cos θ ⋅ sinh − 1 ( f ′ f ⋅ x cos θ + y sin θ R ) − sin θ ⋅ sin − 1 ( f ′ ⋅ − x sin θ + y cos θ R ) sin θ ⋅ sinh − 1 ( f ′ f ⋅ x cos θ + y sin θ R ) + cos θ ⋅ sin − 1 ( f ′ ⋅ − x sin θ + y cos θ R ) ] \begin{align}
\vec{\alpha}(x,y) =
C_0\frac{\sqrt{f}}{f'}\cdot
\begin{bmatrix}
\cos\theta\cdot\sinh^{-1}\left(\frac{f'}{f}\cdot\frac{x\cos\theta+y\sin\theta}{R}\right)
-\sin\theta\cdot\sin^{-1}\left(f'\cdot\frac{-x\sin\theta+y\cos\theta}{R}\right)
\\
\sin\theta\cdot\sinh^{-1}\left(\frac{f'}{f}\cdot\frac{x\cos\theta+y\sin\theta}{R}\right)
+\cos\theta\cdot\sin^{-1}\left(f'\cdot\frac{-x\sin\theta+y\cos\theta}{R}\right)
\end{bmatrix}
\end{align} α ( x , y ) = C 0 f ′ f ⋅ ⎣ ⎡ cos θ ⋅ sinh − 1 ( f f ′ ⋅ R x c o s θ + y s i n θ ) − sin θ ⋅ sin − 1 ( f ′ ⋅ R − x s i n θ + y c o s θ ) sin θ ⋅ sinh − 1 ( f f ′ ⋅ R x c o s θ + y s i n θ ) + cos θ ⋅ sin − 1 ( f ′ ⋅ R − x s i n θ + y c o s θ ) ⎦ ⎤