core26 min studyLesson 206 in Physics

Optical invariants and étendue

Connect ray phase-space preservation with the trade between beam area, angular spread, and passive radiance limits.

On this page

What you’ll understand

  • Calculate a two-ray paraxial invariant.
  • Evaluate étendue for a uniformly filled acceptance cone.
  • Apply reduced-radiance conservation with explicit passive-system assumptions.

A lens can concentrate a wide beam onto a smaller spot, but the beam generally emerges with a broader range of directions. Optical invariants quantify this trade. They also explain why an ideal passive concentrator cannot funnel every ray from an extended source into an arbitrarily small area and narrow cone simultaneously. We will connect a simple determinant calculation to étendue, then distinguish increased irradiance at a focus from an increase in the radiance carried by an individual phase-space channel.

Two rays reveal an invariant

Take two paraxial rays at the same plane, with coordinates (y1,u1)(y_1,u_1) and (y2,u2)(y_2,u_2) where u=nθu=n\theta. Their signed parallelogram area in reduced ray coordinates is H=y1u2−y2u1H=y_1u_2-y_2u_1. Any first-order optical map with determinant one preserves this quantity because the determinant of the pair of output columns equals the map determinant times the input determinant. H has units of length because reduced angle is dimensionless.

With a marginal and a chief ray chosen appropriately, H becomes the familiar Lagrange invariant relating object height and numerical aperture to corresponding image-side quantities. Its sign depends on ray ordering, but its constancy does not. The conclusion requires admitted rays and centered paraxial transport. It does not imply that an aperture transmits every possible pair, and it does not convert a first-order determinant identity into a universal theorem for arbitrary scattering or nonlinear media.

H=y1u2−y2u1=n(y1θ2−y2θ1),Hout=(AD−BC)Hin=HinH=y_1u_2-y_2u_1=n(y_1\theta_2-y_2\theta_1),\qquad H_{\mathrm{out}}=(AD-BC)H_{\mathrm{in}}=H_{\mathrm{in}}
The last equality uses the reduced-angle convention and an ideal paraxial map with unit determinant. H is a signed two-ray invariant, not a measure of total optical power.↔ Scroll wide equations horizontally.

Étendue counts spatial and angular acceptance

For a three-dimensional bundle define étendue by integrating projected area and solid angle with an index-squared weight. The projection cosine accounts for a surface being seen obliquely by a direction. In a uniform medium, a planar area A accepting a full circular cone of half-angle alpha has étendue πAn2sin⁡2α\pi A n^2\sin^2\alpha. This exact cone integral follows by integrating cosine times sine over polar angle and full azimuth.

Writing numerical aperture as nsin⁡αn\sin\alpha makes the same result G=πA NA2G=\pi A\,\mathrm{NA}^2. It assumes every point of the selected area admits that same cone; arbitrary position-direction correlations require the integral instead. Solid angle is dimensionless in SI but conventionally labeled steradians, so étendue is reported in square metres steradians. A smaller spot can preserve G by admitting a larger cone. When the available numerical aperture is exhausted, further area reduction must reject part of the original bundle.

G=∬n2cos⁡θ dA dΩ,Gcone=πAn2sin⁡2α=πA NA2,Ln2=constant along a lossless rayG=\iint n^2\cos\theta\,dA\,d\Omega,\qquad G_{\mathrm{cone}}=\pi A n^2\sin^2\alpha=\pi A\,\mathrm{NA}^2,\qquad \frac{L}{n^2}=\text{constant along a lossless ray}
L is radiance in watts per square metre per steradian for a fixed spectral band and polarization accounting. Reduced radiance is conserved in passive lossless ray transport without emission or gain.↔ Scroll wide equations horizontally.

Brightness limits require the right quantity

Radiance describes power per projected area per solid angle. Irradiance describes power per receiving area after integration over incident directions. A focusing lens can increase irradiance by delivering a wider accepted angular distribution at the spot while leaving reduced radiance unchanged. Across a refractive-index change, ordinary radiance need not remain numerically constant; the conserved quantity is L/n2L/n^2. Omitting that factor incorrectly rules out familiar immersion-optics benefits or falsely claims a brightness violation.

The passive brightness statement applies to a specified frequency band, polarization channels, and geometrical-optics transport without gain or internal emission. Absorption reduces transported radiance; clipping can reduce the étendue of the surviving subset only by rejecting power. Scattering generally broadens occupied phase space rather than losslessly compressing it. Combining separate wavelengths or orthogonal polarizations changes the channels being counted. Coherent interference and wavelength-scale structures require wave descriptions rather than an unqualified application of an incoherent ray-bundle formula.

A four-to-one area reduction

In air, a uniformly filled bundle occupies 4 mm24\,\mathrm{mm^2} with numerical aperture 0.10. An ideal lossless system reduces its area to 1 mm21\,\mathrm{mm^2}. Find the required output numerical aperture and input étendue.

  1. The input étendue is G=π(4)(0.10)2=0.04π mm2 srG=\pi(4)(0.10)^2=0.04\pi\,\mathrm{mm^2\,sr}. Uniform filling and a circular cone make the simple area-times-NA-squared formula applicable.

  2. Conservation requires 4(0.10)2=1 NAout24(0.10)^2=1\,\mathrm{NA}_{\mathrm{out}}^2, giving NAout=0.20\mathrm{NA}_{\mathrm{out}}=0.20. Area fell by four while angular acceptance squared rose by four.

  3. This output NA is less than the air maximum of one, so it is geometrically possible. An output optic accepting only NA 0.10 could accept at most one quarter of the original uniformly distributed phase-space power at that smaller area.

The result

The ideal output needs NA 0.20, and the bundle étendue remains 0.04 pi square millimetres steradians. Keeping the old cone would require rejecting power.

Explore the model

Interactive lab

Interactive lab · matrix

Shear preserves area without preserving shape

Use the transformed unit square as a phase-space analogy. Reset, set Diagonal scale a = 1 and Shear b = 0, and predict area scale 1. For an independent run Reset, set a = 1 and b = 2, and predict the same determinant despite a strongly sheared shape. Readouts always describe the selected final matrix.

Shear preserves area without preserving shapeThe 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 forThe analogy illustrates area preservation, not a full optical bundle. This positive triangular matrix family does not compute étendue, radiance, aperture clipping, or a general optical invariant.

Check your understanding

How can a passive lossless lens increase irradiance at a focus without violating ray brightness conservation?

Try it yourself

A uniform filled bundle in air has area 2 mm22\,\mathrm{mm^2} and NA 0.30. A receiving channel has area 0.5 mm20.5\,\mathrm{mm^2} and NA 0.20. Find the maximum fraction of input power that can enter that channel for uniform reduced radiance.

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

Give me a hint

Compare receiving and source étendue; common pi factors and index factors cancel.

Compare with the solution

Input étendue is π(2)(0.30)2=0.18π mm2 sr\pi(2)(0.30)^2=0.18\pi\,\mathrm{mm^2\,sr}. Receiver acceptance is π(0.5)(0.20)2=0.02π mm2 sr\pi(0.5)(0.20)^2=0.02\pi\,\mathrm{mm^2\,sr}. Their ratio is 0.02/0.18=1/90.02/0.18=1/9, about 11.1 percent. Thus at most one ninth of the input power can be coupled under the uniform-radiance assumptions. A different spatial or angular source distribution would require integrating its radiance over the actually accepted subset.

Take it with you

  • Reduced-angle unit-determinant maps preserve a signed two-ray invariant.
  • A smaller beam area requires a larger angular acceptance to preserve étendue.
  • Passive ray transport conserves reduced radiance; rejecting rays can reduce accepted étendue only with lost power.

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