The concept of etendue starts with the idea that it is really hard to stuff a lot of light into a small hole. Etendue is the number that tells us how big the hole is. With that basic idea, let’s start diving into the details.
The amount of light is quantified as energy and is normally measured in Watts (or lumens for visual systems). It is called the radiant flux and is represented by the symbol Φ (pronounced fie, rhymes with pie). The total flux is related to the etendue by:
Φ = L A Ω
Where:
- L is the radiance (or luminance in visible systems). This is an intrinsic property of the source — it is a value you have to look up.
- A is the physical area of the emitting surface.
- Ω (pronounced oh MAY gah) is the solid angle into which the light is emitted, measured in steradians. It is the 3D equivalent of the angles we learned in geometry. If the beam of light is a cone, its solid angle is given by: Ω = 2π sin²θ Where θ (theta, pronounced thay-ta, with a soft th as in thin) is the half-angle of the cone (angle between the center and the edge).
- The product, A*Ω is the etendue.
Etendue is intimately related to conservation of energy. If we transmit light from a source to an image, unless light is absorbed, reflected or scattered out of the beam, it will get to the image. Similarly, radiance/luminance is conserved. And if both energy and radiance/luminance are conserved, etendue must be.
From Source to Image: Hemisphere to F/1
An image will be helpful here (see Figure 1 below):
We begin with a source area A1 emitting into a hemisphere (not possible to draw so let’s imagine that all the light from the source out to a hemisphere is collected by the optical system). This means that we will have a half-angle θ1 = 90°, which corresponds to a solid angle of: \[
\Omega_1 = 2\pi \sin^2 \theta_1 = 2\pi (1)^2 = 2\pi
\]
At the output, the optical system compresses that light into an f/1 beam, meaning that θ2 = 30°. And,
\[
\Omega_2 = 2\pi \sin^2 \theta_2 = 2\pi \left( \frac{1}{2} \right)^2 = \frac{\pi}{2}
\]
Applying conservation of etendue, we see that: \[
2\pi A_1 = \frac{\pi}{2} A_2 \quad \Rightarrow \quad A_2 = 4A_1
\]
\[
\Omega_2 = \frac{\Omega_1}{4}
\]
This tells us that, because of the conservation of energy, if you reduce the output cone angle with respect to the input cone angle, you must also increase the output area. That’s the tradeoff. You can’t stuff a lot of light into a small hole!
The Lagrange Invariant: Etendue in a Plane
While etendue is based on areas and solid angles, the Lagrange invariant (aka optical invariant) expresses a similar concept in terms of ray heights and angles in a plane. It is also a conserved quantity that captures the relationships between ray height and ray angle throughout an optical system.
The formal expression is: \[
H = \bar{y} n u – y n \bar{u}
\]
Where:
- H = the Lagrange invariant
- y = height of the marginal ray
- \( \bar{y} \)= height of the chief ray
- u = angle (in radians) of the marginal ray with respect to the optical axis
- \( \bar{u} \)= angle (in radians) of the chief ray
- n = refractive index at the surface
This invariant, H, stays constant across all surfaces in the system, assuming rotational symmetry.
In Figure 2 below shows the optical invariant at key points in an optical system:
At the aperture stop, the chief ray height \( \bar{y} \) is zero by definition (it intersects the optical axis there), so the invariant becomes: \( H = -y n \bar{u} \)
At the image plane, the marginal ray height y is zero (since it passes through the image point on-axis), giving: \( H = \bar{y} n u \)
Since H is conserved, we can equate both expressions: \( – y n \bar{u} = \bar{y} n u \)
This clearly shows that if you try to reduce the marginal ray height y (say, by shrinking the aperture), you must proportionally increase the field angle \( \bar{u} \) to get the same amount of light through, or vice versa. The same applies across field height and marginal ray angle.
As an example, customers occasionally request a lens with a large entrance pupil (y) and a wide field of view (u bar) that will work with a small image sensor (y bar) at a slow F/# (u). This is impossible. You simply cannot shrink beam size and image size at the same time without violating the laws of physics.
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