Three Proofs — Question 6

Section 34-8 · Physics pages 5702–5703

Question 6 Ray-transfer matrices recover imaging and magnification in one calculation Represent a paraxial ray by the column (y,u), where u is its slope. Free propagation is T(l) = ( 1 l 0 1 ) and a thin lens is L(f) = ( 1 0 −1/f 1 ) . The object and image planes lie distances p before and q after the lens. F F ′ Schematic; labels, not drawing lengths, specify the data. (a) Multiply the matrices for propagation from object plane to image plane. (b) Impose independence of final height from initial slope and recover both the lens equation and signed magnification.
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Question 6 - Solution Ray-transfer matrices recover imaging and magnification in one calculation F F ′ (a) Method and calculation ( yI uI ) = T(q)L(f)T(p) ( yO uO ) . T(q)L(f)T(p) = ( 1 −q/f p + q −pq/f −1/f 1 −p/f ) . (b) Interpretation and further calculation B = p + q −pq/f = 0 ⇒1/p + 1/q = 1/f . m = A = 1 −q/f = −q/p . Physical check Imaging is the condition B=0: rays launched from one point at different slopes must end at the same image height.

Question 6

Ray-transfer matrices recover imaging and magnification in one calculation

Represent a paraxial ray by the column (y,u), where u is its slope. Free propagation is T(l)=(1l01)T(l)=\left(\begin{smallmatrix}1&l\\0&1\end{smallmatrix}\right) and a thin lens is L(f)=(10−1/f1)L(f)=\left(\begin{smallmatrix}1&0\\-1/f&1\end{smallmatrix}\right). The object and image planes lie distances p before and q after the lens.

Diagram shown in the original SVG worksheet.

  1. Multiply the matrices for propagation from object plane to image plane.

  2. Impose independence of final height from initial slope and recover both the lens equation and signed magnification.

Question 6 - Solution

Ray-transfer matrices recover imaging and magnification in one calculation

Diagram shown in the original SVG worksheet.

(a) Method and calculation

(yIuI)=T(q)L(f)T(p)(yOuO).\binom{y_I}{u_I}=T(q)L(f)T(p)\binom{y_O}{u_O}.T(q)L(f)T(p)=(1−q/fp+q−pq/f−1/f1−p/f).\boxed{T(q)L(f)T(p)=\begin{pmatrix}1-q/f&p+q-pq/f\\-1/f&1-p/f\end{pmatrix}}.

(b) Interpretation and further calculation

B=p+q−pq/f=0⇒1/p+1/q=1/f.B=p+q-pq/f=0\Rightarrow\boxed{1/p+1/q=1/f}.m=A=1−q/f=−q/p.\boxed{m=A=1-q/f=-q/p}.

Physical check

Imaging is the condition B=0: rays launched from one point at different slopes must end at the same image height.