Astrophotography with 4x5

[Helen B]

... Let's see If I've got this straight. I was under the impression that the larger the front element of the lens physically is, the more light from distant stars that can be gathered and condensed. The aperture controls WITHIN the lens (f/2, f/5.6, f/11, etc) in this case only affect non-point-source light -- skyglow, DSOs, nebulas -- anything that isn't a star.

Light that enters the front element may be interrupted by the iris - ie the aperture control. Whether for stars or for sky glow, it is the light that gets through the iris that counts, so the iris affects everything.

In his book, Michael Covington says "When you're photographing stars, what matters is not the f-ratio but the diameter of the lens."

Maybe the diameter of the entrance pupil* would be more correct.

So, let me try to put this in terms of two commonly available 4x5 lenses. If I had a Nikkor 300mm f/9 and a Schneider 300mm f/5.6: given enough exposure time for the two lenses, they could both record the same nebulae; although it will be complete at dramatically different durations of time. However, the f/9, having a small front element, would only allow so many stars to record on the film. The f/5.6, having a much larger front element, will be able to gather the light of fainter stars and record a greater number than the f/9 optic. Nebulas like Orion would appear identical, but the number of stars around it will be different.

Well, you are holding the lens focal length the same so the difference in the amount of light gathered between f/5.6 and f/9 will be the same for the stars and the nebulae (the absolute aperture, as it is being called, varies in the same ratio as the f-number because the focal length is constant). The only difference will be that at f/9 the image of the stars will be larger, and hence dimmer, than the image at f/5.6 if the size of the star image is determined by diffraction. In practice it probably won't be - other lens aberrations are likely to have an effect when the lenses are wide open.

I hope that makes sense.

Best,
Helen

*The entrance pupil is the image of the iris seen through the front of the lens. It is the diameter of the entrance pupil that determines the f-number. The entrance pupil also happens to be the centre of perspective of the lens - the place from which the lens 'sees' the outside world.

 

[dslater]

George, I by no means want to be contrary but now I've just got to get a steady bead on this booger. Let's see If I've got this straight. I was under the impression that the larger the front element of the lens physically is, the more light from distant stars that can be gathered and condensed. The aperture controls WITHIN the lens (f/2, f/5.6, f/11, etc) in this case only affect non-point-source light -- skyglow, DSOs, nebulas -- anything that isn't a star.

Sorry, that is not right - The aperture controls within the lens will affect both stars and extended objects. This is because when you stop a lens down, you are cutting out the light gathered by the edges of the lens and only using light gathered by the center of the lens. For example, suppose you have a f/2 lens. When it is at f/2, you're using all the light gathered by the front element. If you stop it down to 4/f, then you are only going to be using light gathered by a central circle that is 1/2 the diameter of the front element. For example, if you have a 50mm f/2 lens, then it's aperture is 25mm. If you stop it down to f/4, then you're only going to be using the inner 12.5mm diameter circle.

In his book, Michael Covington says "When you're photographing stars, what matters is not the f-ratio but the diameter of the lens."

Got it. But Covington says that's for stars alone. I want to record the nebulae up there too.

Again Covington says, "Remember that most of the really interesting deep-sky objects are extended...Hence f-ratio is usually more important than diameter."

So, let me try to put this in terms of two commonly available 4x5 lenses. If I had a Nikkor 300mm f/9 and a Schneider 300mm f/5.6: given enough exposure time for the two lenses, they could both record the same nebulae; although it will be complete at dramatically different durations of time. However, the f/9, having a small front element, would only allow so many stars to record on the film. The f/5.6, having a much larger front element, will be able to gather the light of fainter stars and record a greater number than the f/9 optic. Nebulas like Orion would appear identical, but the number of stars around it will be different.

Have I got this right now?

Absolute apertures confuse absolutely.

Not quite - since both lenses are the same focal length, the difference if f-ratio is due to a difference in aperture. To simplify the math I'm going to assume the 300mm f/9 is a 300mm f/8. since there is one stop difference between the 2, the f/8 lens needs twice as much exposure as the f/5.6 lens ( ignoring reciprocity failure) to record a nebula and have it appear identical. However, the f/8 lens has 1/2 the light gathering power of the f/5.6 lens, so it will record the same number of stars as the f/5.6 lens if it is given twice the exposure.

On the other hand if you have a 150mm f/5.6 lens and a 300mm f/5.6 lens, then with a given exposure time, they'll both give the same image of a nebula, however, since the 300mm lens has twice the aperture of the 150mm lens, it will record stars 4 times as faint as the 150mm lens. Of course to get the same field of view, you would have to use the 300mm lens with a negative that is twice as large as the 150mm lens.

I hope this is helpful and clears things up a bit. it is helpful to remember that the f-ratio is defined a (focal length of lens) / (aperture of lens). The only way to change the f-ratio is to either change the focal length or the aperture. If you change the f-ratio by changing the aperture, then both stars and extended objects are affected. If you change the f-ratio by changing the focal length, then only extended objects will be affected.

Dan

 

[Struan Gray]

The apparent location of the iris - the optical path length to the image plane - is in the plane of the exit pupil. Easy? But wait. Should we use the exit pupil diameter and location to determine the diameter of the Airy Disk, or should we use the iris diameter and the exit pupil location?

Can't resist....

You find the exit pupil by taking rays from the edge of the diaphragm and seeing where the rear elements form an image of them. It seems to me that the angular spread of rays diffracted off the edge of the aperture will be bent to new angles in exactly the same way, and so the diameter of the Airy disc will be determined by the apparent position and size of the aperture, i.e., by the position and size of the exit pupil.

In a thin, symmetric lens focussed at infinity, the radius of the Airy disc is given by:

r = F tan(arcsin(1.22lambda/D))

Where D is the diameter of the aperture and F is the focal length. Physicists will recognise this as the formula for Fraunhofer diffraction off a circular aperture of diameter D, a distance F from the measuring screen. At small angles, where tan(theta)=sin(theta)=theta, this reduces to the more usually given:

r = 1.22lambda*F/D = 1.22lambda*f

where f is the focal ratio or f-number.

If you now consider diffraction from the exit pupil of a thick, asymmetrical lens of the same focal length and f-number two changes take place. The exit pupil changes diameter compared to the thin lens case, and it changes position.

The f-number is set by the entrance pupil diameter, so a lens with the same focal length and f-number has an entrance pupil diameter, D, which is F/f just as with a thin lens. The exit pupil is different from the entrance pupil by a factor P, the pupillary magnification, and its diameter will be DP.

The position of the exit pupil can be found by remembering that the entrance and exit pupils are conjugate points. That is, their positions obey the normal equation for Gaussian optics, 1/d(entrance)+1/d(exit)=1/f. Remembering that the distances are measured from the principal planes, and that the rear principal plane is a distance F away from the focus screen at infinity focus, leads to the simple fact that the exit pupil is at a distance FP from the focal plane.

Thus, r', the radius of the Airy disc becomes:

r' = FP tan(arcsin(1.22lambda/DP))

and for small angles, r' = r, i.e. the Airy disc is unchanged.

So the only question is what counts as a small angle. In practice, everything: diffraction angles are so small that you are more likely to have to worry about the onset of Fresnel diffraction from wide apertures than you are about changes in the size of the Airy disc from asymmetry in the lens.

To take an extreme example such as the old 360 mm Tele-Arton, which has P roughly equal to 0.5, the change in the diameter of the Airy disc for green light compared to a simple, thin lens is less than the roundoff error in my calculator, even at f32 - i.e. less than a ten-thousandth of one percent. For retrofocus wide angles the effect is even less.

 

[Sparky]

Helen - thanks so much for your herculean efforts!! I feel meek that I don't feel there's so much to reply to (well, guess you shut me up pretty well, didn't you!). Just absorbing it all still. I suppose that means I think I'm on the same page with you - I don't really disagree with any of your content, as the thought process is pretty transparent. The only thing you said that I might take exception with is the coincidence of nodal point and plane of iris - which, while for LF lenses - I know designers try to put them in the same place, more or less - but I'd think it the rare occasion when that can happen with a lens for a 35mm system - especially some of the older, long nikkors, etc... where it's just a triplet at the front of the tube - and the aperture is a good way down from that! However- that's not a retrofocus lens... but well, anyway.... I guess that's that.

 

[Struan Gray]

Just for grins, I fired up my double-precision analysis program. Here is a graphical plot showing the percentage change in the diameter of the Airy disc for different levels of asymmetry. A symmetric lens has P=1, telephotos are to the left, retrofocus lenses are to the right. I have plotted the absolute value, which makes them easier to compare - retrofocus lenses shrink the diffraction spot a bit, telephotos increase it.

The fun thing is that the focal length drops out of the percentage, so the change due to asymmetry becomes due to P and the size of the entrance pupil alone, which in a thin lens is the same as the size of the iris. I have plotted this for D=10 mm, which is about right for Helen's 35 mm f3.5 wide open and the old LF telephoto I mentioned at f32.

It all looks swimmingly fine, until you notice that 10^-6 on the scale. Even a lens as lopsided as the Tele-Arton has a negligable effect.

 

[Helen B]

"Can't resist.... "

I'm glad that you didn't resist.

"It seems to me that the angular spread of rays diffracted off the edge of the aperture will be bent to new angles in exactly the same way, ..."

That's a safe assumption, no problem.

"...and so the diameter of the Airy disc will be determined by the apparent position..."

OK so far.

" ...and size of the aperture, i.e., by the position and size of the exit pupil."

There's the assumption (that the diameter of the exit pupil is the correct diameter to use) that I thought needed further proof, and is the one that is most questioned. Once that assumption is made the rest falls easily, as Struan shows. It is not a difficult assumption to prove - it is fairly obvious and some people could consider it trivial - so I may as well add it.

The rear elements create a virtual image of the iris. If the rear elements are considered as a simple lens then, for small angles:

Magnification, m = Dx/Ds (Dx is diameter of exit pupil, Ds is the diameter of the iris or 'aperture stop')

The magnification is the inverse of the angle ratio. The ratio between the angle the Airy Disk from the iris makes at the plane of the simple lens and the angle the same disk makes at the virtual image of the iris (the exit pupil) follows this rule:

m = alphaS / alphaX (where alphaS is the angular diameter of the Airy Disk made by the real iris and alphaX is the angle subtended by the real Airy Disk at the plane of the exit pupil.

Therefore the apparent angle of diffraction from the exit pupil is

alphaX = alphaS * Ds / Dx

We know that

alphaS = 1.22lambda / Ds (this is based on the real aperture, and does not rely on assumptions, other than those used in the original equation)

therefore

alphaX = (1.22lambda / Ds) * (Ds / Dx)

which becomes

alphaX = 1.22lambda / Dx

and that is what we wanted.

Best,
Helen

 

[Struan Gray]

Our paths and posts crossed. I am glad we agree: usually when I discuss asymmetric lenses I get things wrong

I'm thinking of starting a new group: f63.9999999999999999.

 

[George]

George, I by no means want to be contrary but now I've just got to get a steady bead on this booger. Let's see If I've got this straight. I was under the impression that ...

Well Danny, as others have undertaken the task to clear this mess up, it seems that I'm coming with a cross after the funeral...
Anyway, we take it easy. You know that the f-ratio (5.6; 2.8 etc.) governs the ability of the lens to see the sky glow or the nebulae. The lens diameter is there to tell us what magnitude stars will be photographed and how bright they will be.
Now if you want to take a picture of a greater number of stars with two different lenses, just calculate the lens diameter (Focal length / f-ratio) and compare the numbers. The bigger number is the winner for stars.
If you want to know how sensitive the lenses are for the the sky glow or - for your bigger pleasure, the nebulae - just compare their f-ratio numbers. The bigger number (smaller aperture in use) let's say 2.8 is the one that will see more (more than 5.6) of both the sky glow and the nebulae. Juggle with both ways to see what the lens will see... and forget all the other stuff. Leave something for the horse - it has a bigger head! I don't know, did I forget anything?
Heck, reading the answers I got so entangled in all the concepts and names that I hardly know whom I'm answering to and what about

 

[George]

...
If you want to know how sensitive the lenses are for the the sky glow or - for your bigger pleasure, the nebulae - just compare their f-ratio numbers. The bigger number (smaller aperture in use) let's say 2.8 is the one that will see more (more than 5.6) of both the sky glow and the nebulae....
I don't know, did I forget anything?
Heck, reading the answers I got so entangled in all the concepts and names that I hardly know whom I'm answering to and what about

Ha, ha . I managed to say nonsense there too! Let's try again - for the sky glow and the nebulae, the bigger number of the f/ratio (smaller aperture in use) will see LESS of the sky glow and nebulae. Thus a f/5.6 lens is more resistant to the sky glow and unfortunately the nebulae too, than a f/2.8 lens. Not so sure how I will see that tomorrow though...

 

[snaggs]

Back to the picture making stuff.. Danny.. I currently have a 240mm and I know that once I start doing this.. whilst 240mm will be great for rich field.. I'm going to want a slightly tighter crop of orion.. probably even tighter than yours..

I was thinking the Schneider Tele-Xenar 400mm f/5.6.. is that going to be a worthwhile difference? The objective will certainly be larger.

Daniel.

 

[dslater]

Hi Danny,
Just wanted to say I love that pic of Orion with the 300mm f/5.6. I had a 300mm f/5.6 Caltar that I sold to buy a fuji 300mm f/9 - the caltar was just too big and heavy for my Linhof 4x5. Now I wish I had kept it for this purpose .

 

[konakoa]

Daniel, Yes. No. Maybe. Honestly, after reading everything here I don't know what to think anymore. But If I had to guess I'd say sure. I just dove in and started shooting with the kit I had. There comes a point where you just have to go for it and try it. IMHO for your folding field camera that lens would be a good fit. It'd be nice for landscapes too.

I haven't used one, and I only know a little about the telephoto lens designs from what I've read about them. All I know is that they have a different design that puts the optical center of the lens somewhere out in front of the lens itself; that lens tilts and swings are a little different and you won't have as big of a image circle to play with. I'm afraid you'll just have to go for it and see -- then post back here what you find for the rest of us to mooch off of.

dslater, I feel your pain. I nearly got a 300 f/9 myself - but at the last moment I went the step up to the 300 f/5.6 just because of my interest in astrophotography. It was dumb luck that I went for the 5.6 as it works so well for this purpose.

 

[konakoa]

By the way, I don't mean to be daft to the incredibly detailed postings here on why this stuff works the way it does. The formulas and theories posted are really interesting; they're just a bit difficult to comprehend.

It feels a bit like when we were all learning lens apertures for the first time. "f/16 is a bigger number than f/2, so that's what I want in low light!" .... not quite, yet with time it gets more clear.

 

[25asa]

My head hurts after reading all this stuff.

Would my f/2.5 Aero Ektar work?

 

[George]

It surely would, it has been used by astrophotographers for that purpose. But the sky glow could be a problem (depending where you take pictures) with long exposures as the f/2.5 is quite open. Just go for it.

 

[Raphael]

hi all,

My head hurts after reading all this stuff.

Would my f/2.5 Aero Ektar work?

Surely, but I know people who made an astrograph with an Aero Ektar, and they used to stop it down to f/4 or f/5.6, because they got strong astigmatism, i.e not-round shaped stars on the film.
Maybe I have a link about that somewhere...

In addition, advancements in optical design and making do that a f/2.5 lens of this age old doesn't perform as well as a "modern" f/2.5 lens.

best regards,

Raphael