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As with the front of the lens, the effective aperture can be easily converted into a numerical aperture (NA). 

NA_rear = 1/(2*N*((m/P) + 1))                  (where N = f-number of lens, m = magnification, P = pupillary magnification)

Computing a working NA for the rear of the lens isn’t particularly helpful for me as I prefer to use effective aperture. The interesting part is that the working NA for the front and the rear of the lens are tied together mathamatically.

NA_front = NA_rear*m

It’s a very simple association and it is completely independent of pupillary magnification. That means that with any lens of whatever pupillary magnification and f/number, any gains or losses in NA on the back of the lens associated with the pupillary magnification will have an proportional gain or loss in the  NA on the front of the lens. 

This association can also be tranferred  into terms of f/numbers where it is probably more useful to most photographers (microscopists work predominantly in NA terms).

Working f_num = Effective_aperture/m

As an example, if I have a lens set to f/4 with a P=1 and an m=5, the effective aperture is 4*(5+1) or f/24. Applying the above equation gives you a working f/number of  24/5 or f/4.8. That translates into a NA of  1/(2*4.8) or about 0.10. 

Another example is with microscope objectives. The NA listed on the objective is a working NA at the listed magnification. A 4x/0.10 objective has a working f/number at m=4 of  1/(2*0.1) or f/5. The effective aperture is (5*4) or f/20.

The numerical aperture is the same concept as f/number but on the front of the lens. Whereas the f/number determines the image resolution, the numerical aperture determines the object resolution. Numerical aperture is the angle that the aperture makes with the front focal point.

The basic form of numerical aperture is determined by two measurements – the size of the aperture and the distance that the aperture is from the front focal point of the lens. A larger aperture that is closer to the focal point will have a larger numerical aperture (NA) and will result in higher potential object resolution. The NA is closely associated with the f/number of the lens.

NA = 1/(2*f-num)                   (where f-num = focal length/entrance pupil diameter)

and vice-versa:   f-num = 1/(2*NA)

Technically, the numerical aperture is the sine of the half angle that the aperture makes with the front focal point. I have seen various sources suggest that for small NA’s the tangent of the angle is a close approximation of the sine. While that assertion is true, it is superfluous. Why use an approximate tangent when the sine is correct? Maybe somebody can enlighten me. 

At first glance, it would appear that the NA is equal to the tangent of the angle as the NA is equal to half of the entrance pupil diameter divided by the distance from the entrance pupil to the focal point. The first time I looked at a typical diagram, I thought the same: Why sine and not tangent?

The usual way of diagramming the numerical aperture

The answer lies in the fact that the entrance pupil of a well-corrected photographic lens is not just a circle, it is a section from a sphere with a diameter f (focal length) centered at the focal point of the lens (based on a quote from Rudolph Kingslake – a well respected optical engineer so I’ll go with that).

That means that the hypoteneuse of my triangle is equal to the focal length. The vertical line down to the baseline of the triangle from the edge of the entrance pupil (En/2 in length). That means that the NA represents the sine of the angle and not the tangent.

The actual configuration of the numerical aperture

NA = sin(angle a) = (En/2)/focal length          (where En is the entrance pupil diameter)

Like f-number and effective aperture on the backside of the lens, there is NA and effective NA on the front. It is likewise affected by the magnification and the pupillary magnification. We will go into that concept in the next posting.

Having a pupillary magnfication greater than one does provide a potential advantage in resolution at the detector compared to symmetric and telephoto lenses. So, why aren’t all lenses designed this way? It turns out that there are lots of reasons that lenses aren’t or can’t be designed this way.

As you will recall, having a P>1 means that the lens is retrofocus. With most SLR’s this design is necessary with shorter focal length lenses in order to allow infinite focus. The necessary displacement of the principal planes toward the detector characteristic of retrofocus lenses will cause you to have to give up some working distance. This is because the front principal plane also tends to be displaced toward the back of the lens with a retrofocus design. Since object distance is measured from the front principal plane, you will have more lens in front of it and thus less working distance. With non-macro work, this displacement makes little if any difference in the use of the lens. With macro imaging you may find that the object you are trying to photograph is running into the front element of the lens at a relatively low magnification. A good working distance is a highly prized feature for a macro lens.

A telephoto lens design (P<1) is just the opposite. These lenses tend to have extra working distance, but do pay a small penalty in resolution capabilities. Symmetric or nearly symmetric lenses are a good compromise between these competing factors. You may not get the same resolution capabilities inherent to a retrofocus design, but you will retain a reasonable working distance.

I honestly don’t know how all of these issues factor into lens design. I would guess that with a typical SLR lens is designed either retrofocus or telephoto out of necessity. The differences in resolution and working distance mean very little at non-macro focus distances. At high magnification, these factors become much more important and have to be considered in the design process. The question is: do you want working distance or improved resolution?

A small point to end this entry. The resolution difference between lenses of various pupillary magnification is really pretty small. The improvement can be easily shown mathematically, but in the real world the difference is generally minor. That’s why many choose to ignore this factor.

We have previously discussed the concept of effective aperture and have found that it is related to the size of the aperture and the distance the aperture is from the detector. A small aperture that is close to the detector is equivalent to a larger aperture farther away – i.e. as long as they make the same angle with the detector they will have the same effective aperture.

We have been talking about the “aperture” previously for this discussion, but the real work of the effective aperture is done at the exit pupil. As previously defined, the exit pupil the apparent size and position of the aperture as seen through the lens. This all means that the pupillary magnification have an effect on the effective aperture, although predominantly with close-up and high magnification.

To set the stage for this discussion, you need to know how the pupil moves in relation to the rear principal plane as the pupillary magnification is changed. With a symmetric lens (P=1), the entrance and the exit pupils are equal in size. So, a 100mm f/4 lens will have a 25mm entrance pupil and a 25mm exit pupil. In this case the entrance and the exit pupils will be positioned at the principal planes. The effective aperture at infinite focus will be focal length divided by the exit pupil – 100/25 or f/4. 

For non-symmetric lenses (P<>1) with the same f/number, the exit pupil will travel along this same cone forward or backward depending upon the pupillary magnification, its size and position proportional to the pupillary magnification. This concept is easier to see in a diagram. A point that took me a while to figure out is that the diameter of the cone as it crosses the rear principal plane will always be the same same size as the entrance pupil, no matter what the pupillary magnification is.

The size and position of the exit pupil in relation to the rear principal plane is proportional to the pupillary magnification

At infinity focus (focus at the rear focal point of the lens), the effective aperture is always the same as the aperture setting. The exit pupil may be only half the distance from the focal point with a P=1/2 lens, but the size is also halved and the angle it makes with the detector is unchanged. At any focus closer than infinity, the front aperture setting and the effective aperture will start to diverge as you would expect with the effective aperture. The difference is that the resulting effective aperture will vary depending upon the pupillary magnification. Again, this is best seen with a diagram.

The same apertures as above for a 100mm f/4 lens. Effective apertures at 1:1 magnification in blue

The formula that defines this relationship of the effective aperture to the pupillary magnification is:

N’ = N*((m/P) + 1)                        (where N’ = effective aperture, N = lens aperture setting, m = magnification, and P = pupillary magnification)

The long and the short of it is that with a P>1 you will lose less aperture as the magnification is increased and the opposite for a P<1. A larger effective aperture (smaller number) means more potential image resolution. Sounds like all high magnification lenses should have a P>1 – well maybe not as we will discuss the plusses and minuses of this situation in the next installment.

The aperture is the main determinant of potential lens resolution. I use potential because the imperfections in lens design and manufacture will also strongly affect the actual resolution of the lens. So, assuming a perfect lens, the aperture is the main determinant of resolution in most imaging situations.

The relative size of the aperture in relation to the detector determines the size of the Airy disc. This encompasses two quantities: 1) The size of the aperture and  2) The distance that the aperture is from the detector. A large aperture that is close to the detector will produce the highest potential resolution.

The easiest way for me to think about the relative size of the aperture is in terms of the angle that the aperture makes with the detector. A large aperture that is close to the detector will allow light from a larger number of incident angles to hit the detector. That same aperture farther away will lessen that variety of angles.

The angle that the aperture makes with the detector determines the amount of diffraction. A large angle will produce less diffraction and a small angle will produce more. More diffraction means a larger Airy disc.

the realtive size of the aperture can be expressed as the effective aperture. The aperture setting on a lens is only applicable to infinity focus. That means an f/8 aperture focusing at the focal length of the lens. Two f/8 lenses will have the same amount of diffraction at inifnity focus regardless of the focal length since the calculation of f/number already compensates for different distance from the aperture to the detector.

As the lens is focused closer than infinity, the aperture will move away from the detector. This increased distance will lessen the angle that the aperture makes with the detector and will thus increase the diffraction and the size of the Airy disc. The distance that the aperture is from the detector will be m+1.

Effective aperture: angle "a" is the largest and "c" the smallest (most diffraction)

The relative size of the aperture can be expressed as the effective aperture. The effective aperture can be expressed as :

N’ = N*(m+1)              (where N’ is the effective aperture, N is the actual aperture setting, m is the magnification)

An f/8 lens will have an effective aperture of f/16 when working at 1:1 magnification. The effective aperture is the most important determinant of diffraction and thus the resolution that the lens is producing at the detector. That means that as the magnification increases, the effective aperture also increases. An effective aperture of f/16 at 1:1 will produce the same amount of diffraction as a real aperture of f/16 at infinity focus.

The astute out there might say ” Hey, since the effective aperture increases as the magnification increases I shouldn’t get any more resolution out of a lens as I increase the magnification.” This is where the (m+1) factor comes into play. At low magnification the “+1″ makes a larger difference in the effective aperture than it does at high magnification.

As an example: If I am working at 1:1 and f/8, my EA (effective aperture) is 16. At 2:1 my effective aperture 24. The magnification has doubled, but my effective aperture has only risen by 50%. that means that I will get more detail out of the image at 2:1 than 1:1. This effect lessens as the magnification rises and eventually you won’t get signficant increased resolution out of a lens by increasing the magnification and is commonly called “empty magnification.”

As the magnification rises, the only effective way to get increased image resolution is to make the aperture larger, thus decreasing the effective aperture. That means opening up the aperture. The problem lies in the fact that the lens aberrations will increase as the aperture is opened and will tend to negate any improvement and even worsen the image beyond a certain aperture setting.

This trade-off between aperture size (diffraction) and aberrations means that a lens will have a sweet spot for resolution and have a sharpest aperture. Most commercial lenses will be sharpest in the f/5.6 to f/8 range. Lenses with larger sharpest apertures tend to be specialty macro lenses (shorter focal length bellows lenses) and microscope objectives.

Next: It’s not really the aperture, it’s the exit pupil.

The Airy disc as described in the previous entry is the smallest point of light that can be focused by a lens at a particular aperture and represents a diffraction pattern. The size of the disc is a direct determinant of image resolution.

The resolution of a lens is defined as the closest that two objects can be to each other while still being separately resolved by the lens. The two objects in question are two Airy discs. As the two discs get closer and closer together they will start to merge and at a certain point the separation between them will be lost. 

The top two are fully separated, the middle are barely separated and near the resolution limit, the bottom two are not resolved

The distance between two Airy discs where they are still considered to be resolved separately is the radius of the disc – also called the Raleigh criterion. A smaller Airy disc means a smaller disc radius and a higher resolution. This distance is somewhat arbitrary as there is still a small amount of contrast remaining in the space between the discs at the Raleigh criterion. If the discs are moved any closer, the remaining contrast between the two objects will completely disappear. The point where all contrast is lost between the adjacent discs is called the Sparrow criterion and is the absolute limit of resolution.

The Raleigh criterior is the most commonly used measure of resolution:

Airy disc radius = 1.22 * N * light wavelength ( N = aperture size, light wavelength is commonly 546 or 550 nm or 0.550 um, a wavelength of green)

The Airy disc diamter is just double the radius.

The Sparrow criterion is expressed as:

D = 0.947 * N * light wavelength (about 80% of the Raleigh criterion)

Diffraction is a topic that you can make as simple or complicated as you wish. I will try to stick to the basics. Diffraction is the apparent bending of light around an object. Diffraction is an extremely important topic in macro photography as it is often the major determinant of image resolution.

Diffraction doesn’t really bend the light as it goes around an object. A good analogy to think about is a wave coming through a gap in a seawall. The wave will hit the gap and it doesn’t just continue in a straight line, the wave will radiate out in all directions from the gap in a circular pattern. Light acts in a very similar way.

waves entering a gap without diffraction

what really happens at the gap

Diffraction in a lens is most prominent at the aperture and the size of the aperture determines the amount of diffraction. When the aperture is small, diffraction is increased. When the aperture is large, diffraction is decreased. As light traverses the aperture all of the waves that get through will start to interact with each other. If the crest of a wave meets another crest, they will add together to form a larger wave. Two troughs will add to make a bigger trough. A crest and a trough will cancel each other out.

All of this “interference” is happening on a dizzying scale on the way to the camera detector. The  pattern that light forms is a well-defined pattern called a “diffraction pattern”. The pattern is a series of light and dark bands related to the interference of the waves. The center band of light in the pattern is the most important and contains most of the light energy. The width of this center band of light is related to the size of the aperture. Small aperture makes the band wider (more diffraction) and a large aperture makes the band more narrow (less diffraction).

typical diffraction pattern associated with an aperture

This center band of light is termed the “Airy disc”. The Airy disc is the smallest size that a point of light can be focused on the detector and is a direct determinant of image resolution.

An Airy disc (homemade, roughly accurate)

The size and position of the pupils is closely tied to the position of the principal planes and the aperture. We have already shown pictures of the entrance and exit pupils with telephoto and retro focus lenses and have shown that they are quite asymmetric. This asymmetry ties back into the position of the principal planes in these lenses.

It is easiest to start by discussing a symmetric lens. A symmetric lens will have an equal distribution of positive and negative elements in the lens. This results in the principal planes being near the center of the lens. Place the aperture at the center of the lens and you have a truly symmetric lens.

If  there are the same lens elements in front and behind the aperture, the entrance pupil and the exit pupil will be the same size. In this situation the pupils and the principal planes will be in the same position. The entrance pupil will be in the same position as the front principal plane and vice versa.

The pupils for a symmetric lens

To add some asymmetry into the lens: A retrofocus lens will have more negative lenses in the front (object side) of the lens than in the rear (detector side). This asymmetry leads to the front and rear principal planes being displaced backward toward the detector. Since there are negative elements in the front of the lens, the aperture will tend to look smaller from the front. From the rear, the aperture will appear larger because of the positive elements (like using a magnifying glass).

The pupils will also change position in relation to the principal planes. The smaller entrance pupil will be out in front of the front principal plane when viewed fron the front – i.e. will move toward the front focal point of the lens. The larger exit pupil, when viewed from the rear of the lens, will move farther away from the viewer and away from the rear focal point of the lens (much easier for me to show this movement with diagrams). Just the opposite occurs with a telephoto lens.

A symmetric lens. P = pupillary magnification, E = Entrance pupil, E' = Exit pupil

The amount that the pupils move in relation to the principal planes is directly related to the pupillary magnification.  Referring to the above diagram: The pupillary magnification is two (the exit pupil is twice as large as the entrance pupil, ie. a retrofocus lens).  This means that the exit pupil will 2x the focal length behind the rear focal point. The entrance pupil will be 1/2x the focal length behind the front focal point. I left the real aperture position out of this diagram for simplicity (if you could call it that), but it is somwhere in front of the two principal planes not far from the entrance pupil.

In the coming installments I will try to start explaining why these factors actually make a difference in macro photography and why they don’t mean diddly for landscape photography. But first, I will have to start getting into aperture related topics like resolution and depth of field to establish definitions for that discussion.

The f/number of a lens is a way to help standardize the function of various lenses. The f/number of a lens helps to determine the exposure needed, the depth of field and the potential maximum resolution of a lens. All of these factors tie directly into the pupils of the lens and the focal length.

The f/number of a lens is the focal length divided by the entrance pupil size. the 200 mm lens in the previous posting has an entrance pupil 50 mm in diameter. that makes the lens a maximum of f/4. A 100 mm f/4 lens would have an entrance pupil of 25 mm in diameter.

f/number  =  f/E            (f = focal length, E = entrance pupil diameter)

 The f-number of a lens is only accurate at infinity focus. Two f/4 lenses will allow about the same amount of light to hit the detector at the same shutter speed, regardless of the focal length. They will not always be equivalent because two lenses will have a different amount of light loss as the light travels through the lens. Light loss in a lens is dependent upon the number of elements and various lens coatings in each lens.

At infnity focus the detector will be one focal length away from the rear principal plane of the lens. A 200 mm lens will be twice as far away as a 100 mm lens but its entrance pupil is twice as large. This will result in the aperture for the two lenses being the same relative size in relation to the detector. The angle that the edges of the aperture makes with the detector will be the same.

200 mm f/4 exit pupil.

100mm f/4 exit pupil - note angle A is the same as the 200

You may have noticed that we haven’t talked about the exit pupil of the lens, only the entrance pupil. This is because at infinity focus, the relative size and distance of the exit pupil will be constant regardless of the pupillary magnification and won’t effect the numbers. This issue will be taken up separately in the next installment.

If I were to stop down the lens to f/8 on my 200 mm lens the entrance pupil will be 25 mm ( 200/8) - half of that at f/4. The area of the pupil will be 1/4 that of f/4 and it will let in 1/4 as much light and will require a shutter speed of 4x as long to get the same amount of light to the detector ( for a similar exposure). The stop half-way between f/4 and f/8 will let in half as much light. That would be f5.6 (not f/6, halfway between the two but the square root of 2 between them).

The main stops you will see on a lens will be f/1, f/1.4, f/2, f/2.8, f/4, f/5.6, f/8, f/11, f/16, f/22, f/32… Each stop will allow half of the light of the next lower number and will lengthen the shutter speed by 2x to get a similar exposure at the detector.