Calculating, NOT measuring, aperture of compound lenses

[Emmanuel BIGLER]

From Struan:
There is a halfway house between using Oslo and guessing.

.. and in some cases no Oslo nor any guess are required to figure out the diameter of the entance pupil, for example like on fig.13 here ...
http://www.galerie-photo.com/pupille...que.html#fig13

.. when the iris is located in front of the glass like in Wollaston's meniscus and in a good ol' convertible view camera lens with the front group removed.
But our reader wants no iris at all, so this remark is irrelevant, unfortunately.

Regarding the determination of the pupils in a barrel lens with no iris, I remember a lecture I attended a long time ago, and the process was tought to us as follows, very close to what Struan said.
You start by finding the exit pupil by looking backward from the focal point, finding among all images of lens element mounts seen from behind the compound lens, which one appears to be the smallest in terms of apparent angle and not in terms of actual size of the image.
The image of all lens mounts which appears to have the smallest angle as seen from the focal point is the exit pupil.
The entrance pupil is then the image of this lens mount as seen from the front of the lens.
My guess, but I'm not 100% sure, is that it is equivalent to look at all images of lens mounts seen through all glasses from a point of view located on the optical axis, far away ahead of the lens, in front, and selecting the image which looks the smallest in terms of angular size.
So this is exactly what Struan said, with the additional condition that the point of view should be far away and that the size is the angular size of aperture images seen from this point of view.

[Struan Gray]

I love the way having no glass in your lens makes the calculations easier :-)

I did this once when playing with a fantasy of being Abelardo Morell and projecting the world onto the walls of a room. The idea was to make a long focal length lens by mounting positive and negative singlets at each end of a pair of nested sliding tubes, adjusting the focal length by moving the tubes in and out. I wanted to know how much it would matter which lens was in front. In this case, finding the virtual image of the rear lens mount as seen from through the front lens is a simple enough exercise.

Once you get more than a couple of elements or groups, the calculations get tedious enough that I would prefer to learn Oslo. The problem is that real multiplet lenses quickly become 'thick', and even if you cheat and use thin lens formulea to find the combined focal length, specifying the 'distance' to the lens mount acting as aperture gets a bit fraught.

Of course, you can always do the thing graphically too, which would get you in the ballpark.

[Jockos]

Hej Jockos

There is a halfway house between using Oslo and guessing.

As Dan said, the entrance pupil is the aperture as seen from the front of the lens. When you have no explicit aperture stop, the aperture is set by whichever element appears to be smallest when viewed through all the elements between it and the object space out in front of the lens. Provided you don't have too many elements and they are not too far apart, you can step through the lens, finding the virtual images of the mounts for each element, and choosing the smallest of them for the final figure.

If that doesn't seem obvious to you, it might be better to spend your time learning Oslo. Or tinker and measure.

Hej,

Isn't what you're talking about now exactly what I mentioned in my first post, or are you implying there's an equation that I can use for calculating these virtual images? I do understand that the entrance pupil and the focal length can get me the aperture value (N = f/E), but how can I calculate the entrance?

[Dan Fromm]

Jockos, you're working in a vacuum. Pick a design that has the desired maximum aperture. If you're going to get the student version of OSLO, it will have to be one with few elements to keep the number of surfaces under student OSLO's limit. 10, I think, including the diaphragm you don't want. Scale it to give the desired focal length. Start procuring/grinding lens elements.

www.dioptrique.info is, among other things, a library of lens' prescriptions. Since you want an f/2.8 lens, look for f/2.8ers there. There's no need to find a new layout that gives f/2.8 with decent image quality when you can look one up.

One caution. Most of the prescriptions that Eric collected use old glasses that are no longer available. You'll have to find modern equivalents and reoptimize.

[Emmanuel BIGLER]

From Struan in Sweden:

The idea was to make a long focal length lens by mounting positive and negative singlets at each end of a pair of nested sliding tubes, adjusting the focal length by moving the tubes in and out. I wanted to know how much it would matter which lens was in front. In this case, finding the virtual image of the rear lens mount as seen from through the front lens is a simple enough exercise.


Great suggestion, Struan.

Since we are here in a virtual world, let's do a paraxial design of such a compound lens virtually, as a Gedankenexperient.

First I should mention that Struan's rule for finding which of the images of individual lens element mouts is the entrance pupil, is the right one and the simplest when we consider using the lens in infinity-focus position.
Actually, comparing angular sizes (in degrees or radians) from a very far distant point of view (object at infinity) is strictly equivalent to comparing actual dimensions of the images (in millimeters or inches).
And the rule I was tought, now I understand, can be more general when one wants to know exactly where the pupils are located for any kind of object and image distances, for example at 1:1 ratio or in a microscope lens working at a high (image/object) magnification ratios.

Hence finding the entrance pupil is easy in principle in a simple paraxial lens design with only 2 thin lens elements. The first diameter to look at, is the actual diameter of the first thin lens element; and the second diameter to look at, is the diameter of the image of the 2nd lens element mount, as seen through the first one. The smallest of the two will be our entrance pupil, and its diameter our entrance pupil diameter.

Let us start with 2 thin lens element.
L1 = a positive thin lens, f1= 150 mm.
L2 = a negative thin lens, f2 = -100 mm.
Lets mount the 2dn element at d = 100 mm behind the first one.

A well-known formula, known as Gullstrand's formula in French textbooks, immediately yields the focal length of he compound:
f = (f1 . f2)/(f1 + f2 - d)

With f1= +150 and f2 = -100 this yields a numerical formula where the focal length f and the distance d are expressed in millimeters:

f = 150 / ( -1.5 + 1 + d/100) = 150 / (-0.5 + d/100)

Note that when d=50 mm, the compound's focal length is infinite, the system is afocal and equivalent to a useless Galilean telescope of magnifying power = 1.5.
Galileo Galilei himslef, the great physicist and inventor of this telescope design, used something more powerful than that !

But when d is longer than 50 mm, the resulting compound has a positive focal length, significantly longer than the distance d between lens element.

Now take d = 100 mm, this completes our design:
f1 = +150 mm; d=+100 mm; f2 = 100m.
In French textbooks this air-spaced doublet would be denoted by (1,5 / 1 / -1).

The resulting focal length is: 150 / ( -0.5 + 1) = 300 mm.

Before finding the pupils, let's locate where the principal plane H' can be.

For this, we have to use the usual conjugation formulae in their more general algebraic form, distance are counted positive to the right (the direction where light propagates, at least in French textbooks, no allusion of course to French politics those days just after our local elections).
The principal plane H' is located 300 mm in front of the focal point F' of the compound, which is the image given by the 2nd lens of the first element's focal point F'1.
Let S1 be the centre of the 1st element and S2 the centre of the 2nd, we have, algebraically:

General algebraic formula, the one I know is
V' = V + C
where V = 1/x'; V = 1/x; C = 1/f
x and x' are algebraic distances measured to the centre of the thin lens element.

here
(1 / S2F') = (1 / S2F'1) + (1 / f2)

With numbers this equation yields, in mm,
1 / S2F' = (1/(150-100) ) - 1/100 = 1/50 - 1/100 = 1/100
F' is located 100 mm behind S2, hence H' is floatng in air at (300-100-100) = 100 mm ahead of S1.

The distance between lens element #1 and the focal point is shorter than the focal length: 200 mm instead of 300 mm. This is a true telephoto.

Great! Actually we must confess that we love telephotos for being kind of eccentrics among the family of view camera lenses, where the principal planes are always located close to the shutter for quasi-symmetrical lens designs; this is not fun

Now our telephoto design is well defined.
Start from H'; 100 mm behind H1 => S1 ; 100 mm behind S1 => S2 : and 100 mm behind S2 => F', the focal point of the compound.

Now let's find the entrance pupil for the (infinity->focus) working conditions, following Struan's rule.

Consider for simplicity that both lens elements have the same diameter D.
Image of the 1st element mount = the element mount itself, located at S1, diameter D.

Image P2, seen from the front, of element #2 i.e S2 as seen through element
#1, CAUTION! this is a bit tricky:
If we consider light propagation in the normal direction of lens use
P2 is our (unknown) virtual object to be determined, located in object space
S2 is where the rays actually go after crossing element #1 and is the image of the virtual object P2 through element #1. (yes, this is a bit difficult to understand)

(1/S1S2) = (1 / S1P2) + (1/f1)
1/100 = 1/S1P2 + 1/150 ==> 1/S1P2 = 1/100 - 1/150 = 1/300 ==> S1P2 = 300 mm behind S1.
The magnification ratio for the image of the 2nd mount is : 300/100 = 3

hence lens mount #1 = identical to its image has a diameter D.

image of lens mount #2 as seen from the front = diameter 3 D ===> this is
the biggest size, hence the entrance pupil is simply lens mount #1 as long
as lens mount #2 is at least bigger than 1/3 of lens diameter #1.

So if we want a f/3 lens of 300 mm focal length, the entrance pupil should have a diameter of 300/3 = 100 mm.
hence D = 100 mm for a f/3 lens.

a positive thin lens with f=+150 and D=100 mm is a big one and its minimum thickess at the centre is about 4 mm.


The attached drawing summarises what we have calculated here.

Note that this pupil position is specific to this design with no f-stop inside the lens!
In real telephotos, the f-stop is located at some distance behind the entrance positive group, and the entrance pupil is often located behing the iris itself, sometimes close to the focal plane!

And note that this is a paraxial calculation, nothing is know regarding aberrations!

[Struan Gray]

Isn't what you're talking about now exactly what I mentioned in my first post, or are you implying there's an equation that I can use for calculating these virtual images? I do understand that the entrance pupil and the focal length can get me the aperture value (N = f/E), but how can I calculate the entrance?

Hej, och Glad Påsk!

In your first post you seemed to be assuming that you could take the mount diameter as the size of the entrance pupil. The vital point is that this is the wrong diameter - you need to take the diameter of the hole as seen looking through the front of the lens. Any glass which makes the hole look larger will allow it to gather more light, increasing the f-number. In a compound lens with several elements, the overall light-gathering ability is limited by the mount diameter which looks the smallest.

Imagine a lens made up of three simple elements. You need to decide which of the three mount diameters looks the smallest when seen from the front of the lens.

The mount for the first element has no glass between the world and itself. The mount diameter is the relevant diameter. This is the simplest case.

The mount for second element is seen through the first element, which either magnifies or diminishes the apparent size of the mount, depending on whether the first element is a positive or negative lens, and the distance between the two elements. In this case, you can use the thin lens formula (sometimes called the Gauss' lens law) to find the position and size of the virtual image of the mount.

The mount for the third element is seen through the combined lens formed by the first and second elements together. Thus your first step is to find out the focal length of the combination (using Gullstrand's equation). Your second step is to work out where the combined lens is (how far in front of the third element is can be said to be) and hat's where you need to get into finding principle planes and their positions. Once you have done that, you use the lens formula again to find the position and size of the virtual image of the lens mount for the third element.

Now you have a list of three candidate apertures, one for each element. The f-number is the smallest of these, divided by the focal length of the whole lens.

Conceptually, none of this is hard. It just takes repeated applications of Gullstrand's formula and the lens equation. You need to be careful to keep track of plus and minus lengths properly, but otherwise, the main problem is tedium. Which is why computers are better at this kind of calculation than humans :-)

Emmanuel has given you a nice worked example (thanks Emmanuel!), but it might not be for the type of lens you were thinking of building.

Note that unless you are building very complex lenses, or very odd ones with long tube lengths, none of this angels-dancing-on-pinheads stuff is really necessary. The mount diameter divided by the focal length will get you within a stop or so. But if you really need f2.8 and not f2.2, or f4.5, you - or your computer - need to do the math.

[Dan Fromm]

Struan, we're all working in a vacuum. I just read the thread from the start. The OP never mentioned the coverage he needs. If he wants a 180 mm lens that covers 4x5 he'll need a design that covers at least 45 degrees. And he wants f/2.8.

I took a look at Eric's library, as I suggested the OP do in post #14 in this discussion. Until proven otherwise he's dreaming. 45 degrees and f/2.8 come close to defining the null set. I don't see how he can make what he wants to be starting with off-the-shelf lens elements. I find it hard to believe that he's going to design a lens, procure and grind glass and mount his new elements.

[Struan Gray]

https://xkcd.com/356/

[Jockos]

Struan, Emmanuel,

Thank you very much for your help. I've been running the calculation on some candidate lenses, and got very close to the f2.8 goal!

If the camera has a working shutter, I'll get the elements and try it out. If it works as a PVC pipe prototype, I've made a printable CAD design for a barrel integrated in a board, which should fit the Speed Graphic.
Lots of if's right now, but it's been an interesting Easter project!

[Struan Gray]

Good luck, and have fun.

If the lens doesn't work out, there's always the 7" Aero Ektar.