Finite Aperture Optics
This module takes the concepts of pupils and resolution that we have discussed in the
previous modules and works through how to apply them to our first-order optical design
systems. We start with a description of how to find the system pupils and windows, then
move on to a discussion of how that affects the imaging properties of this system, and finally
return to the Lagrange invariant and its utility in optical system design.
Learning Objectives
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For a given optical system, determine the location of the stops & pupils and the angle
and height of the marginal and chief rays as they travel through the system.
Calculate the depth of focus and maximum number of resolvable spots allowed by a
given optical system.
Design an afocal Keplerian telescope given a set of specifications.
Since neither window is conjugate to the object or image, we will see later that it suffers from
"vignetting" which is to say loss of light at the edge of the field. In other words, the field stop
is out of focus at the object, causing a gradual darkening to about 50% transmission at the
edge of the field. It is thus generally better practice to put the exit window at the image.
Lyot stop’s purpose: So this is an example of where exit pupils can be something you
want to design, you want to put them in a particular place because they may be something
you want to use to improve the performance of your system.
If the diameter of the aperture stop is D_AS, the diameter of the field stop is D_FS and the
focal lengths of the first and second lenses are f1 and f2, what is the angular field of view of
this imaging system?
Numerical aperture describes the property of cones of light getting off of the object or
to the image. It's not a property of lenses. Now despite that, lenses are often labeled with
numerical aperture. For example, microscope objectives might have 0.5 NA listed on
them. What that means is, if used as designed, with an object that radiates at very large
cone of rays, then you will get an NA of 0.5 off of the object and on into your
microscope. However, you could have an object, let's say, that emits a smaller cone of
light. At that point, you are using that lens at maybe 0.3 NA. The most common mistake I
see students using is when trying to understand the resolution or the MTF (Modulation
Transfer Function, whichis a measurement of the optical performance of a lens.) of an
optical system, they simply use the NA written on the side of the lenses, not the NA as
the lens is being used. And that's the concept I want to make sure we have understood
here.
If I went off through lots of this was a microscope, and this is some small object, then
somewhere, an image space, I've formed an image I'd have a very similar diagram on the
backwards. But presumably, my image would be much bigger because you like
microscopes have large magnification and therefore, because angular magnification is the
inverse of transverse magnification, this alpha in image space, the angle that the marginal
ray made with the axis, would be much smaller than the equivalent angle in object space.
So the numerical aperture in image space would be smaller. That would in turn mean that
the point spread function in the image space would be bigger than the point spread
𝜆
function here in the object space (Since we have the Airy disk 𝑟0 = 0.6 𝑁𝐴 and NA_obj
>NA_img, this guess is right.).
Either using the expressions on the right or simply through intuition given the effective
numerical apertures in object and image space, what are the diffraction limited spot sizes, r0,
in object and image space for the imaging system shown on the left in which the object is at
infinity and the image is a the focal plane? Use the variable FN for the lens F# = f/D and
wavelength λ.
Pupil matching
The idea here is that you might be cascading multiple optical systems packaged or
designed independently. And how should you put them together? How should you think
about how they interact? And now these concepts of windows and pupils are really
important. Because basically you want to look at the exit windows and pupils of the first
system, and the entrance windows and pupils of the second system. And that tells you
how the system interacts. And note that those can often be not physical stops. Remember,
they're the images of the aperture stop and the field stop. And so they can be planes
hanging in space of a particular size. So this is a concept that if you didn't know this class
you'd never even think to look at, you wouldn't know those concepts were there.
When the aperture stop is at the back focal plane, something a bit odd happens. Where is
the entrance pupil? Do a quick graphical ray trace to find out, remembering that images
are found where rays from the object intersect.
You should find that the rays from any point on the aperture stop are parallel in object
space and thus intersect at negative or positive infinity. It seems really awkward to have
an entrance pupil at infinity, but actually it results in a very nice property. Consider how
cones of rays from two different object points are limited by an entrance pupil at infinity.
Then consider a system with the aperture stop at (for example) the first lens. What do
you notice is different about the way light is accepted from the object?
Let's remind ourselves of what information is represented by the marginal and chief rays.
Which of the following tables describes the quantities bounded by the chief and marginal
rays in the object and aperture planes?
The marginal ray has zero height in the object plane and the chief ray has zero height in
the aperture stop by definition. So they must each limit the angle in those two planes.
The chief ray defines the field at the object and the marginal ray hits the edge of the
aperture stop, again by definition. So they must limit the size of the light bundle in those
two planes.
Apply the form of H at an object to the Gaussian beam, using the waist ray as the chief
and the divergence ray as the marginal. Remembering that the constants of order unity in
these expressions depend on the definition of resolution used (this uses Rayleigh) and the
shape of the beam, how many degrees of freedom (spots) are represented by a single
Gaussian beam?
Answer: 1
If we do the calculation for an Airy disk, we get exactly one. The point is that the
diffraction limited focus (Gaussian, Airy or otherwise) represents a single spot and the
two rays of the Gaussian beam are just the chief and marginal rays for an optical system
which transmits just one spot.
Spots are something we count (integers), so "2/pi" doesn't quite make sense. This is
approximately one and the reason its not exactly one is that we measure the Airy disk and
the Gaussian beam slightly differently - one to first null, the other to the 1/e point. This
leads to small changes in the constants in equations like that above.