Chapter 1 · Ray optics

Ray transfer matrix analysis

Trace a paraxial ray through spaces and lenses, then use the agreement of many rays to test whether a system forms an image.

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The engineering question

A thin-lens equation predicts an image plane for one ideal lens. How do we trace rays through several propagation distances and optical elements without redrawing the system from scratch each time?

Ray transfer matrices give a compact answer: write the state of one paraxial ray as a two-component vector, then apply one matrix for each physical step in its path.

Model

Paraxial rays in a centered system

At axial coordinate , represent a ray by its transverse height and its small angle from the positive direction.

  • Valid when: the system is centered, one meridional – plane is sufficient, and angles are small enough that .
  • Omitted effects: diffraction, aberrations, finite apertures, and strongly nonparaxial rays.

A ray has a height and a slope

A paraxial ray enters one generic optical element. At its input plane, the ray state r subscript 1 has height x subscript 1 and angle theta subscript 1; at its output plane, the state r subscript 2 has height x subscript 2 and angle theta subscript 2. The element matrix maps the input state to the output state.
Figure 1.3. A single optical element maps the ray state at its input plane to the ray state at its output plane.

Positive is upward from the optical axis. Positive means the ray height increases as it propagates in positive ; a downward-sloping ray has negative .

Two building blocks

For propagation through a positive physical distance , the angle stays fixed while the height changes by :

An ideal thin lens at leaves the ray height unchanged and changes its angle. A converging lens has :

Thus . A ray above the axis is bent downward by a positive lens, as it should be.

Follow the ray, read products from right to left

Suppose a ray propagates a distance to a lens, passes through the lens, and then propagates a distance to a sensor. The physical sequence is written

The first physical operation, , is the rightmost matrix because it acts on the input vector first. Here and are positive path lengths. They are not the signed object and image coordinates: for a real object at and a real sensor at .

The system matrix gives an imaging test

Multiply every step to describe the whole system as one matrix:

The image-plane height is therefore . A focused image plane requires : then every paraxial ray from the same object height has the same final height, regardless of its initial slope. This is the matrix version of imaging.

An image brings many rays to one point

A correctly focused ideal system sends every paraxial ray leaving one object point to the same image-plane height. The rays may have different angles at the object, lens, and sensor, but their final agrees.

Three differently sloped rays leave a point 20 millimetres above the optical axis, travel 150 millimetres to a converging thin lens, and converge at a point 40 millimetres below the axis on a sensor 300 millimetres to the right.
Figure 1.4. For a 100 mm focal-length lens, a point at maps to . Different initial slopes produce the same final height when the sensor is at the image plane.

Explore: move the sensor plane

Start from the Figure 1.4 system, then move the sensor while the rays leave the same object point with different initial slopes. The explorer makes the coefficient observable: the sensor fan collapses only when . The controls use positive physical path lengths, so and .

Worked examples

Build the operators, use their order, then test an image-plane claim.

Guided

Propagate one ray

At one plane, a ray has and . Propagate it in free space. What are its new height and angle?

Hint / setup

Use . The angle does not change in free space.

Full solution

Using millimetres for the height and distance gives and . The negative slope lowered the ray by .

Scaffolded

Focus a finite calibration target

A surgical-imaging camera is checked with a calibration target. Its lens has . A feature at height is to the left of the lens. Find the exact sensor distance and image height in the paraxial thin-lens model. Then show that the sensor height is independent of the ray’s initial slope .

Hint / setup

Use millimetres: and . The target angle is , so it is securely paraxial. Start with , apply , , and , and set the coefficient of to zero.

Full solution

The sensor-height coefficient of is

Imaging requires , so

At that distance, the sensor height is independent of and equals

This agrees with the signed thin-lens result: , , and . The image is inverted. This focus location is exact within the paraxial thin-lens model. Here the sensor must move beyond , an shift relative to the focal length, so is not a good approximation.

Independent

Trace rays from a sample point

A point in a sample is at . A lens with is at , and a sensor is at . Show that rays with different initial slopes all reach the same sensor coordinate. Check against the magnification from Chapter 1.1.

Hint / setup

Write the input vector as . The propagation distances are and .

Full solution

Expanding the height calculation gives and . Therefore

The result does not depend on . From Chapter 1.1, , so , in agreement.

Before you trust the answer

Model selection and engineering checks

  • Are the rays paraxial enough that is a useful approximation?
  • Do the lengths in each propagation matrix use the same unit as the ray height?
  • Does the rightmost matrix describe the first physical operation?
  • Does a centered system preserve an on-axis ray with ?
  • For an image plane, do rays from one object point reach a common output height? If not, is the sensor misplaced, or is the model missing an effect such as aperture clipping or aberration?