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
Brief solution
1. Method.
For the air-to-plastic surface,
2. Key step.
3. Answer.
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.
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For the air-to-plastic surface,
If \(x\) is the image distance inside the plastic, form
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
Brief solution
1. Method.
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
2. Key step.
3. Answer.
The 2-cm object therefore produces a \(\boxed{0.780\ \mathrm{cm}}\) inverted image beyond the second surface.
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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
Set \(B=0\) in \(T(x,1)M_sT(0.08,1)\). This gives
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
Brief solution
1. Method.
For a 2-cm-diameter bead of index 1.4, both refracting surfaces have power \(40\ \mathrm{m^{-1}}\). Thus
2. Key step.
3. Answer.
beyond the bead. Direct multiplication verifies \(AD-BC=1\).
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For a 2-cm-diameter bead of index 1.4, both refracting surfaces have power \(40\ \mathrm{m^{-1}}\). Thus
A parallel input has \(V_1=0\). After an air gap \(x\), its height is \((A+xC)y_1\); setting this to zero yields
beyond the bead. Direct multiplication verifies \(AD-BC=1\).
Problem 2.4 — Lantern-slide projection
Brief solution
1. Method.
The image is 20 times the 2-inch slide height. With object and image distances \(u\) and \(v\),
2. Key step.
3. Answer.
The lens is therefore 6 inches from the slide; the conjugate distances add back to 10.5 feet.
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The image is 20 times the 2-inch slide height. With object and image distances \(u\) and \(v\),
Hence \(u=0.5\ \mathrm{ft}=6\ \mathrm{in}\) and \(v=10\ \mathrm{ft}\). The imaging condition gives
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
Brief solution
1. Method.
Working in metres, use \(P_1=12.5\ \mathrm{D}\), \(P_2=-8.333\ \mathrm{D}\), and
2. Key step.
3. Answer.
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.
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Working in metres, use \(P_1=12.5\ \mathrm{D}\), \(P_2=-8.333\ \mathrm{D}\), and
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
Brief solution
1. Method.
For a thin lens of power \(P\), the image condition from \(T(-V)R(P)T(U)\) is
2. Key step.
3. Answer.
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.
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For a thin lens of power \(P\), the image condition from \(T(-V)R(P)T(U)\) is
Differentiation gives
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
Brief solution
1. Method.
For real conjugates of a positive thin lens, \(1/u+1/v=1/f\). Therefore
2. Key step.
3. Answer.
with equality only at \(u=v=2f\). This is also the stationary point of the separation found by differentiating with respect to either conjugate.
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For real conjugates of a positive thin lens, \(1/u+1/v=1/f\). Therefore
Consequently
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
Brief solution
1. Method.
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
2. Answer.
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.
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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
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
Brief solution
1. Method.
The two-lens matrix is
2. Key step.
3. Answer.
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.
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The two-lens matrix is
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
Brief solution
1. Method.
Multiplication gives
2. Key step.
3. Answer.
This removes chromatic change of magnification, but \(D=1-P_2t\) still varies with index, so the eyepiece retains longitudinal color.
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Multiplication gives
so
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
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
Brief solution
1. Method.
In centimetres and inverse centimetres, the thick-lens matrix is
2. Key step.
3. Answer.
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.
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In centimetres and inverse centimetres, the thick-lens matrix is
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
Brief solution
1. Method.
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
2. Answer.
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).
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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
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).