I am just starting to looking at the application of wavefront abberommtery to measure abberations in the human eye. I am confused over the pupil conjugate planes, do you use relay lenses to image the "Real (IRIS)" pupil straight on to the Shack Hartmann lenslets? like in the diagram below. Shack Hartmann Wavefront Sensor

If this is the case then why does another book state that all you are interested in is the "effective pupil" see below. Effective Pupil size

Are the pupil conjugates referring to the conjugates of A) the real pupil(Iris), B) The effective pupil or (C) the exit pupil of the eye (which i suppose is just a conjugate of the the "real" pupil).

My own understanding is we must have to image the "Effective pupil" as this will include abberations after the cornea, if we sucessfully image "The real pupil" on the Shack Hartmann this will not include aberrations from the cornea?


2 Answers 2


The answer to your question depends on what you want to study. But first, let's agree on terms: by "optically conjugate" I mean that the deformable mirror/shack-hartmann wavefront sensor is imaging some plane. In case of astronomical AO, the SH WFS is conjugated with the telescope pupil (or, more precisely, with the primary mirror of the telescope). A good presentation about the clarification of this terminology (and generally informative) is here or here [pages 16-20] (on my Dropbox).

Back to your question. The answer (if I understood your question correctly) can be formulated as: the conjugation plane depends on what you are studying. You may want to study the retina, or the iris, or something else. But ultimately, at least one of the source of the aberrations will be the tear-film of the cornea.

As to your last question, I think that the aberrations from the cornea will be in either way (present in the wavefront).

P.S.> You are referring to a "diagram below", but there is nothing attached/linked in your post. These illustrations would be helpful.

Disclaimer: my area is astronomical AO, where the conjugation plane is obviously the primary mirror of the telescope - this is where you want to cancel aberrations.

  • $\begingroup$ Well I want to look at the aberrations of the eye as a whole. So as soon the light leaves the cornea/tear film. In a "perfect" eye the wave front will be flat and the rays collimated. Sorry the "see below" refers to the second image which is a screen shot of a book. I'm confused as the whether the pupil plane mentioned in much of the literature refers to the "effective pupil" as seen in image two, or the "exit pupil" of the eye, I suspect it may be the former. $\endgroup$
    – BigAl
    Commented Oct 23, 2014 at 9:48
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    $\begingroup$ Alec, after your clarifications, I think that the "Pupil plane" refers to the effective pupil (as in the image 2). The hint (for me) is in the caption of the picture 2, saying that "the optics of the entire eye is simplified as an equivalent lens". So yes, I think that you are right: the pupil plane is almost certainly the "effective pupil" as in Fig.2. $\endgroup$
    – virens
    Commented Oct 30, 2014 at 10:53
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    $\begingroup$ I'm pretty sure the ophthalmologists follow you astronomical guys in all the standard optical terminology. It'd be a bit weird if they didn't. I once designed a system for imaging the lens in the eye and didn't notice any weird usages amongst the ophthalmologists I was talking to. $\endgroup$ Commented Nov 5, 2014 at 9:23

As in Virens's answer "A conjugate to B" simply means A is an image surface when B is an object surface and contrariwise (I use the word "surface" here because it is almost always curved to some degree, even though to first order we model these surfaces as planes).

There is no way that this setup can measure anything other than the total aberration of the total eye system, including cornea, lens and aqueous humour. It is quite analogous to a double-pass interferometric test of a lens system as wontedly done with a Fizeau or Twyman-Green interferometer. The light passes through all the eye's components and there is no way of telling which part of the total wavefront aberration is induced by which part of the eye. If you think about the above in terms of what I describe below, it should become clear that you need to image the effective pupil as drawn in your diagram at the plane of the lenslets. In fact, strictly speaking, the talk of pupils in your document is not quite right and what you need to do is image the principal plane of the eye's optics onto the lenslet plane. However, this principal plane is almost always very near to the pupil and, as I discuss at the end, small errors in any of this discussion will not make a great deal of difference.

Now onto conjugates and pupils. I don't believe losing a great deal of sleep about what is conjugate to exactly what is warranted here. Changing the definition slightly is equivalent to adding defocus terms to your wavefront sensor measurement. Rather, I personally find in situations like this to to think of nominally collimating lenses ("infinity conjugate" objective) and to think of them as Fourier transformers. So first an excursion into the idea of thinking of a collimator as a Fourier transformer: it may help you with your thinking about the various pupils in the system.

Collimators as Fourier Transformers, Principal Planes and Pupils

Let's begin with the eye. The probing beam is a collimated beam, and the eye's focussing mechanisms will try to focus it: thus the eye will naturally configure itself to be near to an infinity conjugate objective schematically shown below.

Infinity conjugate objective

The objective transforms an on-axis point source to an on-axis, collimated beam, i.e. a plane wave. A point a lateral distance $x$ away from the optical axis gets transformed into a collimated beam that is tilted, with tilt angle $x/f$, where $f$ is the effective focal length. A tilted plane wave at the output implies a scalar field of $\exp\left(i\,k\,\frac{x}{f}\,X\right)$, where $X$ is the distance from the optical axis on the lens output side. Thus, an object represented by a superposition $\psi(x,\,y)$ of point sources induces at the output side the field $\Psi(X,\,Y)$ where $(X,\,Y)$ are the lateral co-ordinates on the output side:

$$\Psi(X,\,Y)=\int_{\mathbb{R}^2} \,\psi(x,\,y)\,\exp\left(i\,\frac{k}{f}\,(x\,X + y\,Y)\right)\,{\rm d}x\,{\rm d}y$$

So our collimator is in fact a Fourier transformer. Pretty neat. BUT: there is an implicit assumption in the above. We need to take the output plane to be at the unique axial position where the collimated beam is transformed by a PURE rotation about the optical axis point at that axial position when the object point source shifts sideways in the object plane from the optical axis At all other axial positions, the transformation is a rotation TOGETHER WITH a lateral translation. The axial position where the transformation is a pure rotation is for most practical purposes the exit pupil of the lens. It is the principal plane (see the Wikipedia page on cardinal points) nearest to the output.

Hopefully my sketch below may make these ideas clearer.

Principal Planes

I've shown a generic imaging system as a box. The principal planes are $P_1$ and $P_2$; these are conjugate to one another (in the sense that rays converging to a point on one of these planes will be imaged to rays converging to a point on the other) and these are almost always the system's entrance and exit pupils, respectively or very near to them. They almost always lie a little inside the physical optical system; a notable exception to this rule of thumb is the Kepler telescope. Note that rays $QR$ and $ST$ are always parallel. So, as the point source moves from $O$ to $Q$, the output collimated beam will rotate precisely about the point $S$ through the angle defined by $\tan\theta = x/f$. Conversely, if a plane wave is incident on the output side and fills the exit pupil, and if this plane wave tilts through angle $\theta$, the focus at $O$ shifts to a focus at $Q$, so the ray bundle converging on $O$ rotates about the point $S$ (approximately: this statement would imply that the focal surface is spherical with radius $f$. It tends to be, but is not always so and indeed the design of field flattening systems aims to iron out this curvature).

The same system also becomes a collimator the other way around if the point source is placed on a plane a distance $f$ (the SAME $f$ as above) to the right of the principal plane $P_2$: a collimated beam will be output from $P_1$ and will rotate about point $R$ as the point source moves laterally on the plane a distance $f$ to the right of the principal plane $P_2$.

The Wavefront Aberration Measurement Setup

So if the eye is perfect, it will focus the collimated beam on the retina, and the resulting point source on the retina from the light scattered by the retina will return as a collimated beam. The aberration on the returned collimated beam is twice that imparted by the eye's optical system, and this is what you are aiming to measure with the Hartmann Shack sensor. Now, what you need is for this collimated beam to be:

  1. Still a collimated beam when it reaches the Hartmann Shack sensor and

  2. Have the right beam diameter to fill the Hartmann Shack sensor and

  3. If the beam tilts at all, i.e. rotates about the eye's pupul (approximately the principal plane) it will rotate about the plane where the sensor lenslets are.

So the optics in between the eye and sensor will be ideally a Keplerian telescope, or equivalent thereto, so that the above conditions will be fulfilled (actually, you need to design for 2 by choosing the right angular magnification of your Keplerian telescope).

If your system is slightly off, and you're not quite imaging the pupils as discussed above, you will get an extra PURE defocus term in your inferred aberration. Most softwares for driving interferometers and wavefront sensors can easily be configured to take this out. The other aberration terms will be unaffected. So inaccuracies in what is exactly conjugate to what will not make any difference to your measurement unless they are big enough that the beam diameter of at the sensor becomes bigger than the sensor itself (so you're getting an overly optimistic idea of the eye's aberration because you're only measuring part of the beam) or if the beam diameter becomes much smaller than the sensor, so that only a few lenslets are involved in the measurement. This problem can be fixed by including an adjustable beam expander.

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    $\begingroup$ Nice explanation but i think the abberometer is not strictly a double pass system, the incoming beam is typically of small width and passes through the center of the eyes optics. $\endgroup$
    – BigAl
    Commented Nov 5, 2014 at 16:37
  • $\begingroup$ @AlecKingsnorth I think you may be right: a 0.5mm beam will not be probing much of the eye's optics at once. However, I don't understand how a higher NA reflected field arises from the retina: if the retina is a Lambertian scatterer, you're going to see speckle and the measured aberrations are going to me more indicative of the retina's surface properties than of the eye's optics. So this can't be how it works. I'd have thought that one would like to have the iris as open as possible (approx 7mm diameter hole) and to fill this region with a collimated beam in a double pass test ... $\endgroup$ Commented Nov 5, 2014 at 23:04
  • $\begingroup$ ... to get the best data on the eye's optics. Maybe not the full 7mm, but more than 0.5mm. I think that the iris can only shut to about 1mm diameter anyway, so more of the eye's optics than that traversed by a 0.5mm wide beam is being used. $\endgroup$ Commented Nov 5, 2014 at 23:05

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