Chapter II: Matrix Methods in Paraxial Optics

Source: Gerrard and Burch, Introduction to Matrix Methods in Optics (1975), Chapter II. Distances below follow the book’s reference-plane convention.

Illustrative problems

Problem 2.1 — Refraction by the end of a plastic rod

For the air-to-plastic surface,

\[\begin{split}P=\frac{n_2-n_1}{r} =\frac{1.56-1}{0.028}=20\ \mathrm{m^{-1}}, \qquad R(P)=\begin{bmatrix}1&0\\-20&1\end{bmatrix}.\end{split}\]

If \(x\) is the image distance inside the plastic, form

\[\begin{split}M=T(x,1.56)R(20)T(0.15,1), \qquad T(t,n)=\begin{bmatrix}1&t/n\\0&1\end{bmatrix}.\end{split}\]

Imaging requires \(B=0\), which gives \(\boxed{x=0.117\ \mathrm m=11.7\ \mathrm{cm}}\). At that distance \(A=-0.5\), so a 2-cm object forms a \(\boxed{1.0\ \mathrm{cm}}\) inverted image. The determinant remains unity.

Problem 2.2 — Imaging through a double-convex glass rod

Each surface has power \(P=(1.6-1)/0.024=25\ \mathrm{m^{-1}}\), and the reduced thickness is \(0.028/1.6=0.0175\ \mathrm m\). The rod matrix is

\[\begin{split}M_s=R(25)T(0.028,1.6)R(25) =\begin{bmatrix} 0.5625&0.0175\\-39.0625&0.5625 \end{bmatrix}.\end{split}\]

Set \(B=0\) in \(T(x,1)M_sT(0.08,1)\). This gives

\[\boxed{x=2.439\ \mathrm{cm}},\qquad m=A=-0.39024.\]

The 2-cm object therefore produces a \(\boxed{0.780\ \mathrm{cm}}\) inverted image beyond the second surface.

Problem 2.3 — Back focal distance of a spherical bead

For a 2-cm-diameter bead of index 1.4, both refracting surfaces have power \(40\ \mathrm{m^{-1}}\). Thus

\[\begin{split}M_s=R(40)T(0.02,1.4)R(40) =\begin{bmatrix} 3/7&1/70\\-400/7&3/7 \end{bmatrix}.\end{split}\]

A parallel input has \(V_1=0\). After an air gap \(x\), its height is \((A+xC)y_1\); setting this to zero yields

\[\boxed{x=-A/C=7.5\ \mathrm{mm}}\]

beyond the bead. Direct multiplication verifies \(AD-BC=1\).

Problem 2.4 — Lantern-slide projection

The image is 20 times the 2-inch slide height. With object and image distances \(u\) and \(v\),

\[\frac vu=20,\qquad u+v=10.5\ \mathrm{ft}.\]

Hence \(u=0.5\ \mathrm{ft}=6\ \mathrm{in}\) and \(v=10\ \mathrm{ft}\). The imaging condition gives

\[\boxed{f=\frac{uv}{u+v}=5.714\ \mathrm{in}}.\]

The lens is therefore 6 inches from the slide; the conjugate distances add back to 10.5 feet.

Problem 2.5 — Positive and negative lens pair

Working in metres, use \(P_1=12.5\ \mathrm{D}\), \(P_2=-8.333\ \mathrm{D}\), and

\[M=T(x)R(P_2)T(0.06)R(P_1)T(0.24).\]

The top-right element is \(B=0.12-x\); hence \(\boxed{x=12\ \mathrm{cm}}\) to the right of the negative lens. At this plane \(A=-1\), so the final image is inverted and has the same 3-cm height as the object.

Problem 2.6 — Longitudinal magnification

For a thin lens of power \(P\), the image condition from \(T(-V)R(P)T(U)\) is

\[U-V+PUV=0, \qquad V=\frac{U}{1-PU}.\]

Differentiation gives

\[\boxed{\frac{dV}{dU}=\frac{1}{(1-PU)^2}} =\left(\frac VU\right)^2=m_T^2.\]

Thus longitudinal magnification is the square of lateral magnification. Its nonnegative sign is consistent with nearby conjugate planes moving in the same longitudinal sense under the book’s signed-distance convention.

Problem 2.7 — Minimum object-to-image distance

For real conjugates of a positive thin lens, \(1/u+1/v=1/f\). Therefore

\[\frac{u+v}{f}=\frac{(u+v)^2}{uv} =\frac uv+2+\frac vu\geq4.\]

Consequently

\[\boxed{u+v\geq4f},\]

with equality only at \(u=v=2f\). This is also the stationary point of the separation found by differentiating with respect to either conjugate.

Problem 2.8 — Cardinal points of a hemispherical lens

Take the entrance plane surface as the first reference plane, translate a reduced distance \(r/n\), and refract at the spherical exit. Reducing the matrix to principal-plane form gives

\[\boxed{f=\frac{r}{n-1}},\qquad \boxed{H_1\text{ lies }r/n\text{ inside the plane face}}, \qquad \boxed{H_2\text{ is at the curved-surface vertex}}.\]

The result tends to infinite focal length as \(n\to1\), and the second principal point remains at the vertex because there is no propagation after the only powered surface.

Problem 2.9 — Cardinal points of a separated positive-negative pair

The two-lens matrix is

\[\begin{split}M=R(-10)T(0.05)R(10) =\begin{bmatrix}0.5&0.05\\-5&1.5\end{bmatrix}.\end{split}\]

Therefore the equivalent focal length is \(\boxed{f=-1/C=20\ \mathrm{cm}}\). Relative to the positive-lens plane, the first focus is \(D/C=-30\ \mathrm{cm}\) and the first principal plane is \((D-1)/C=-10\ \mathrm{cm}\). Relative to the negative-lens plane, the second focus is \(-A/C=+10\ \mathrm{cm}\) and the second principal plane is \((1-A)/C=-10\ \mathrm{cm}\). Each focus is 20 cm from its associated principal plane.

Problem 2.10 — Two-lens eyepiece and chromatic error

Multiplication gives

\[\begin{split}R(P_2)T(t)R(P_1)= \begin{bmatrix} 1-P_1t&t\\ -(P_1+P_2-P_1P_2t)&1-P_2t \end{bmatrix},\end{split}\]

so

\[\boxed{f=\frac{1}{P_1+P_2-P_1P_2t}}.\]

For lenses of the same glass, write \(P_i=(n-1)G_i\) and set \(d(1/f)/dn=0\). The transverse-achromat condition is

\[\boxed{t=\frac12\left(\frac1{P_1}+\frac1{P_2}\right)}.\]

This removes chromatic change of magnification, but \(D=1-P_2t\) still varies with index, so the eyepiece retains longitudinal color.

Problem 2.11 — Cardinal points across unequal exterior indices

In centimetres and inverse centimetres, the thick-lens matrix is

\[\begin{split}M=\begin{bmatrix}0.8&2\\-0.06&1.1\end{bmatrix}, \qquad \det M=1,\end{split}\]

with \(n_1=1\) and \(n_2=1.4\). The cardinal data are

Quantity	Input side, from first surface	Output side, from second surface
Focus	\(F_1=D/C=-18.3\ \mathrm{cm}\)	\(F_2=-n_2A/C=+18.7\ \mathrm{cm}\)
Principal point	\(H_1=(D-1)/C=-1.67\ \mathrm{cm}\)	\(H_2=n_2(1-A)/C=-4.67\ \mathrm{cm}\)
Focal length	\(f_1=-n_1/C=16.7\ \mathrm{cm}\)	\(f_2=-n_2/C=23.3\ \mathrm{cm}\)

The two nodal points coincide at the common center of curvature: 5 cm to the right of the first surface, equivalently 2 cm to the right of the second.

Problem 2.12 — Internally reflected glass sphere

Construct the chain from refraction at the left surface, translation across the sphere, reflection at the right surface, return translation, and final refraction. Moving both reference planes from the left surface to the sphere center simplifies the result to

\[\begin{split}\boxed{M_c= \begin{bmatrix} -1&0\\[2pt] -\dfrac{2(2-n)}{nr}&-1 \end{bmatrix}}.\end{split}\]

The zero \(B\) element shows that the center images onto itself with lateral magnification \(-1\). For \(1<n<2\) the equivalent focal length is positive. At \(n=2\), \(C=0\) and \(M_c=-I\): the bead is afocal and retroreflects each incident ray (within the paraxial and aberration limits).