core25 min studyLesson 203 in Physics

Paraxial rays and ABCD matrices

Build first-order ray transport from a declared reduced-angle vector, signed distances, and ordered matrix multiplication.

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What you’ll understand

  • Construct translation and refraction matrices for the vector (y,n theta).
  • Multiply optical elements in physical order.
  • Use the B equals zero condition to identify conjugate planes.

A paraxial optical system maps the height and direction of an entering ray linearly to its outgoing height and direction. Matrix notation turns a long sequence of surfaces into a compact calculation, but only after the ray coordinates and signs are fixed. Different books place angle and height in different orders or use an unreduced angle. We will use one convention consistently, derive its elementary matrices, and test the resulting map with both dimensions and a directly traced ray.

Declare the vector before using a matrix

Let z increase along light propagation from left to right, let y increase upward, and let the small angle θ=dy/dz\theta=dy/dz be positive for an upward-going ray. Our column vector is (y,u)T(y,u)^{\mathsf T} with u=nθu=n\theta, called the reduced angle. Height is in metres and reduced angle is dimensionless. The approximation replaces sine and tangent by their angles in radians and requires small ray angles and small surface slopes throughout the system.

Within a uniform region of index n, propagation through positive axial distance d changes height by dθ=du/nd\theta=du/n and leaves direction unchanged. Thus the upper-right matrix entry has units of length. At a curved interface take radius R positive when the center of curvature lies to the right of the vertex. Linearizing Snell’s law about that local surface normal changes reduced angle by minus surface power times height, while height is continuous.

(y2u2)=(ABCD)(y1u1),T(d,n)=(1d/n01),S=(10−Φ1),Φ=n2−n1R\begin{pmatrix}y_2\\u_2\end{pmatrix}=\begin{pmatrix}A&B\\C&D\end{pmatrix}\begin{pmatrix}y_1\\u_1\end{pmatrix},\qquad T(d,n)=\begin{pmatrix}1&d/n\\0&1\end{pmatrix},\qquad S=\begin{pmatrix}1&0\\-\Phi&1\end{pmatrix},\quad\Phi=\frac{n_2-n_1}{R}
A and D are dimensionless, B is in metres, C and surface power Phi are in inverse metres. S connects the reduced angles immediately before and after one spherical interface.↔ Scroll wide equations horizontally.

Order and determinant are physical checks

If a ray first meets element one and then element two, the total matrix is M2M1M_2M_1: the rightmost factor acts first on a column. Interchanging factors usually changes the system because propagation before bending differs from propagation after bending. A converging thin lens in air has positive focal length f and lower-left entry −1/f-1/f. An initially parallel ray above the axis therefore leaves with a negative slope, as physical convergence requires.

Both elementary reduced-angle matrices have determinant one, so their product does too for ideal first-order transport without ray rejection. That fact expresses preservation of a suitable phase-space area, not preservation of every individual ray height or angle. If a reference uses (y,θ)(y,\theta) instead, refraction can give determinant n1/n2n_1/n_2. Mixing that vector with reduced-angle matrices is a common error that can hide in air-only examples and fail immediately at a dielectric boundary.

Mtotal=MN⋯M2M1,L(f)=(10−1/f1) (air),AD−BC=1M_{\mathrm{total}}=M_N\cdots M_2M_1,\qquad L(f)=\begin{pmatrix}1&0\\-1/f&1\end{pmatrix}\ \text{(air)},\qquad AD-BC=1
The determinant statement uses the ordered reduced-angle vector. L is a thin paraxial lens with f in metres, and all matrices connect explicitly chosen reference planes.↔ Scroll wide equations horizontally.

Imaging removes dependence on launch angle

All rays from one point in an object plane share its height but can have different launch angles. At an image plane they must return to a common height. The equation y2=Ay1+Bu1y_2=Ay_1+Bu_1 shows that first-order imaging therefore requires B to vanish. At those conjugate planes A is the transverse magnification. A negative A means inversion, while the determinant condition gives D=1/AD=1/A and connects angular spread to magnification.

This condition assumes the considered rays pass through the clear apertures; a matrix does not reject an obstructed ray automatically. It also cannot reveal spherical aberration, polarization, diffraction, or the finite brightness of a source. A good workflow first checks the paraxial map, then traces representative aperture rays, and finally asks whether higher-order or wave effects matter. A beautifully multiplied matrix is still only as valid as the linearization used at every surface.

A lens followed by free space

In air a thin converging lens has f=0.20 mf=0.20\,\mathrm m. Find the matrix from immediately before the lens to a plane 0.10 m after it, then propagate the input ray y=0.010 my=0.010\,\mathrm m, u=0.020u=0.020.

  1. The lens matrix has C=−5 m−1C=-5\,\mathrm{m^{-1}}. Since propagation occurs second, the total is T(0.10,1)L(0.20)T(0.10,1)L(0.20), not the reverse product.

  2. Multiplication gives M=(0.50.10−51)M=\left(\begin{smallmatrix}0.5&0.10\\-5&1\end{smallmatrix}\right). Its determinant is 0.5+0.5=10.5+0.5=1, and its entries have the expected units.

  3. The output height is 0.5(0.010)+0.10(0.020)=0.007 m0.5(0.010)+0.10(0.020)=0.007\,\mathrm m. The output reduced angle is −5(0.010)+0.020=−0.030-5(0.010)+0.020=-0.030. Directly, the lens gives slope −0.030 and the subsequent drift lowers height by 0.003 m.

The result

The ray reaches height 7 mm with slope −0.030 radians. Because B is 0.10 m, this pair of reference planes is not an object-image conjugate pair.

Explore the model

Interactive lab

Interactive lab · matrix

A transport shear and its determinant

Use normalized coordinates to study a translation-like shear. Reset, set Diagonal scale a = 1 and Shear b = 1, and predict determinant 1 with a sheared square. For a fresh comparison Reset, set a = 1 and b = 0, and predict the identity. Readouts describe the selected final matrix even during playback.

A transport shear and its determinantThe selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Current readings: Area scale · det A: 1.4; Singular values · σmax, σmin: 1.597, 0.876; Condition number · κ₂(A): 1.823. All parameters are adjustable with the labeled controls below.Matrix A with first row 1.4, 0.6 and second row zero, one; display interpolates between identity and A while final determinant remains 1.4.−4−2024−2−1012x · equal geometric scaleyA = [ 1.4 0.6 ; 0 1 ]Display blend: 0% · final values below
Interpolated geometryOriginal unit squareSelected final transform
Area scale · det A
1.4
Singular values · σmax, σmin
1.597, 0.876
Condition number · κ₂(A)
1.823
t = 0 s

The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forOnly the a = 1 shear resembles normalized free propagation. This triangular family cannot represent a powered lens, a general ABCD system, or real apertures.

Check your understanding

For the declared column vector (y,u), what condition makes two planes conjugate in first-order imaging?

Try it yourself

In a uniform medium with index 1.5, a ray begins at height 2 mm with reduced angle u=−0.030u=-0.030. It propagates 0.15 m. Find its ending height and the determinant of the translation matrix.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Convert reduced angle to geometric slope by dividing by index before calculating the height change.

Compare with the solution

The geometric slope is θ=u/n=−0.020\theta=u/n=-0.020. The height change is 0.15(−0.020)=−0.003 m0.15(-0.020)=-0.003\,\mathrm m, so the final height is 0.002−0.003=−0.001 m0.002-0.003=-0.001\,\mathrm m, or −1 mm. Equivalently B is 0.15/1.5=0.10 m0.15/1.5=0.10\,\mathrm m. The translation determinant is one; propagation preserves reduced-angle phase-space area even though this ray crosses the axis.

Take it with you

  • Specify ray ordering, angular coordinate, and signs before copying an ABCD matrix.
  • The first encountered element is the rightmost matrix factor.
  • B equals zero is the first-order imaging condition between chosen reference planes.

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