Lens viewing angle - a self-assigned project..

I'll be interested to apply my so far math/optics skills I'm gaining during my current academic study to certain self-assihned project that is aimed to design an optimal lens hood for my LF lenses I currently own (Fuji-W 150mm/5.6 and Grandagon-N 90mm/6.8). The market offerings don't excite me much (each has its own annoying drawbacks), so I thought to try to make one by myself (actually with he help of my relative who is very skilled metal worker and not only). First of course I need to proceed with necessary calculations. To make the things simple at the first, I will only take into account the actual media area (4x5) not taking into account actual excessive lens coverage (image circle), but the hood shall be flexible enough to accomodate moderate movements manually and preferrably accomodate 1-2 square filters in front of the lens (slots for Cokin, Lee or Hitech sized)
So, first I assume there is no movements applied (camera is at the respective detents), the lens is projecting straight onto the film area.
The viewing angle can be calculated easily by given focal lenght and of course it is constant. (i.e. for instance 150mm lens is expected to produce roughly 43 degrees of viewing angle horizontally and approx. 37 degrees vertically). There are two questions so far Im asking myself:

1. What is actual, precise size of 4x5 active film area ? (I guess is must be somewhat smaller then just plain 4" by 5"). Cannot measure myself since have yet shot single frame - still in the process of learning LF basics and have had no time to go to the shop for some 4x5 film to purchase)
2. What is the point the viewing angle is measured from ? (i.e. where is located the top of "viewing angle cone") Shall I take it located at len's front principal focusing plane, len's rear principal focusing plane or perhaps from the film plane once focused at infinity ?

I noticed most LF manufacturers provide full data for their respective lenses including optical construction and physical dimensions, so once I'll know the necessary calculating parameters and will have particular len's manufacturer data, I will be able to proceed with calculations (unless you see a major mistake in my approach).

So, if you feel you can contribute answering my questions above, I'll be grateful.

Alex

The precise image area on film will depend on the design of your filmholders. You could remove the darkslide of one of your filmholders and measure.

Indeed, simple and straight forward. How didn't I think about it myself :-0

Alexz said:

.......
The viewing angle can be calculated easily by given focal lenght and of course it is constant. (i.e. for instance 150mm lens is expected to produce roughly 43 degrees of viewing angle horizontally and approx. 37 degrees vertically). There are two questions so far Im asking myself:

.....

2. What is the point the viewing angle is measured from ? (i.e. where is located the top of "viewing angle cone") Shall I take it located at len's front principal focusing plane, len's rear principal focusing plane or perhaps from the film plane once focused at infinity ?

.......

Alex

Click to expand...

The answer to question #2, is the center of the entrance pupil. The entrance pupil is the image of the aperture as seen from the front of the lens. If a datasheet has info about the entrance pupil, it may use the symbol En. In general, the pupils are not at the same locations as the principal planes.

See pages 122-123 of Applied Photographic Optics by Sidney Ray.

If the datasheet doesn't give the position, there are techniques to measure the position, e.g., you can use another camera to focus on the entrance pupil of the lens, then figure out the position you focused on. You will also find alternative directions on some webpages that tell how to do panaromic photography by rotating the entire camerea. (Many of these webpages incorrectly refer to the nodal point, which is correct for swing-lens cameras.)

It's probably getting carried away to worry about this for a lens hood. Just take the center of the lens and allow some margin.

Thanks a lot Michael, that makes sense. I'm aware entrence (and exit) pupils aren't usually at the principal planes but rather a projected image of the physical aperture by the front and rear optical parts of the lens. But as far as I remember, entrence pupil can actually be positioned either outside or inside the lens and in the letter case it will probaly be difficult to take such measurements. Am I wrong ?

I think if you look at a typical LF lens, you will see that the pupils appear to be within the lens. Perhaps this isn't true for a true telephoto, but most LF lenses have quasi-symmetrical designs that don't do radical things like place the pupils outside of the lens. Neither method for measuring the location of the entrance pupil (focusing on it, or rotating the camera to find the axis that eliminates parallax shifts) depends on the pupil being inside the lens.

Michael, I think you just told me that the apex of the cone of rays that the lens projects on the film is the center of the lens' entrance pupil. Did I misunderstand?

Cheers,

Dan

Yes, thank Micahel, understood. I suspect it would be not that easy to determine its position by these methods, perhaps the more realistic approach would be to assume entrance pupil position to coincide with phyisical aperture location which is often given by the manufacturer's datasheet. Then, making flexable prototype of the hood would allow the fine tuning to adjust one for the actual viewing angle coverage (i.e for the actual entrance pupil location)
Dan, perhaps Michael meant the apex of the cone producing image circle is at the center of exit pupil (the aperture image as produced by rear part of the lens) rather then entrance pupil.

The center of the entrance pupil is relatively easy to pin down analytically. Trouble is, for a lens hood that doesn't vingnette you have to calculate the marginal rays - those that just squeak by the edge of the aperture. Easy enough to do for a particular lens design - if you have the full prescription - but in general it's easiest to measure.

I would put a bright light under my darkcloth so it illuminates the ground glass but not the room. Then place a sheet of paper in front of the lens at the distance where you want your hood to end and trace the resulting shape. This is easier to do if you have a copy stand. You can always expose and process a piece of printing paper if you want a permanent record.

Sounds fair, Struan, thanks (albeit not sure I got your point about marginal rays - do they also participate in actual image formation on the film ?). The experiment you suggested sounds pretty much straight forward I'll certainly consider one, however my goal wasn't really round design but rather a rectangular - that keeps clear just the area that will be actually recorded on the film, i.e. it will cut a part of the side rays that the lens is able to gather, but those rays do not contribute to the image formation on the film. Or probably this isn't a good idea ?

BTW, Michael, you mentioned the book you was referencing to (Applied photographic optics) - thought why not to purchase one for myself ? Checked Amazon - WOW, over 150$ ! So far have never purchased such an expensive books (although acquired more then few professional-related (EE-related math, signal processing, photoshop,...). Tend to think this one really worth such high tag...

If you illuminate the ground glass you will get a rectangular light spot on a piece of paper held in front of the lens, not a round one.

Marginal rays: if you look at the front of your lens the aperture looks roughly circular. All the ray paths from your eye that pass through the aperture will go to make up the projected image of your eye on the ground glass or film. Optics defines a number of rays that are special, mostly because they aid calculation rather than because they are more or less important than any other rays.

The first is the "Chief Ray", which passes through the center of the entrance pupil, that is, through the center of the aperture as is appears to you looking at the font of the lens. This ray describes where the lens sees from: the center of the entrance pupil is the 'position' from which the photo is taken.

The others are the marginal rays, which are rays that go from the object (here, your eye) and just graze past the aperture. These rays define edges of the bundle of light passing through the lens, and are the ones that are worst affected by geometric aberrations.

If your lenshood is not going to vignette, you need to make sure it does not cut off any of the rays. You need to be sure that you can see the whole aperture from every point on the film and, equivalently, from every point in the scene that you are photographing. You need to allow the marginal rays to pass through the shade.

When you adjust a compendium shade (i.e. one that can be moved about) you commonly check through the cut-out corners of the ground glass to make sure that you cannot see any part of the shade: that you see the whole aperture and only the whole aperture. You can also do this from the front, by looking in through the front of the lens and checking that you can see the whole ground glass. This second method is the one I find easiest to do, but there's a gotcha: you must make sure that you can see the corners of the ground glass across the whole aperture, not just in the middle.

Wow, Struan, many thanks. Yes, being at the beginning of my optics study, I see lots of stuff to learn is ahead...

Dan Fromm said:

Michael, I think you just told me that the apex of the cone of rays that the lens projects on the film is the center of the lens' entrance pupil. Did I misunderstand?

Cheers,

Dan

Click to expand...

Yes, you misunderstood. Alexz wants to make a lens hood, which I understand to be in front of the lens, hence the relevant pupil is tne entrance pupil. The apex of the cone that the lens projects to the film is the center of the exit pupil.

Is this question in relation to my comments about panoramic photography? For swing lens cameras with stationary film, you want to rotate the lens about an axis through its rear nodal point. When you rotate the entire camera to take multiple photos, you want to rotate the camera about an axis through the entrance pupil of the lens.

Thank you Struan, that clarifies more, but still I'm not sure I'm getting the point of marginal rays improtance. The ability to figure entrance pupil analytically is indeed quite possible given full lens specifications (which is unlikely to obtain for a real photo lens unless you're working in particular lens design company), but marginal rays ..
From your explanations I understood marginal rays also participate in actual image formation on the film (i.e. on the GG), am I wrong ? If they do, they must be included into len's viewing angle, right ? (i.e. viewing angle of about 43 degrees for 150mm lens projecting on 4x5 format includes all the rays that take their part in image projecting on GG, including marginal rays). In such case, hood that is based on viewing angle (with apex at the center of entrance pupil) shouldn't produce any visible vignetting.

Or perhaps I didn't get your point and you meant there are rays (margianl) that are actually outside of veiwing angle but still prticipate in image projecting onto GG, so that calculating the hood for viewing angle only, I'll inadvertantly cut off those rays thereby causing visible vignetting on GG (and thus on the film) ? In this case, that mean that taking into account vewing angle for hood calculations is mandatory but not sufficient condition and I have to take into account those marginal rays ...

Can you please enlight me a bit more regarding this issue ?

This is a lot easier to comprehend with a sketch. Here's one for a simple 50 mm f/2 lens.

Oh, great Helen, thank you.
So according to your sketch, the merginal rays are those in bold named "cone of rays entering the lens" right ? I wonder how these rays relate to cone made from the center of the entrance pupil ? (in your sketch the rays are entering through the physical aperture)

This is how I think about it. Imagine trying to take a photo of, say, a cat sitting in front of your lens. On the cat's collar is a small pearl. Light from the room, the sun and any strobes you are using will scatter off the pearl in all directions. The more of that scattered light you can capture, the brighter the focussed image of the pearl will be. The scattered light can be thought of as a hedgehog-like spray of light rays streaming out from the pearl.

In an ideal world you would collect all the rays. That would require a very large lens - one which filled the half-hemisphere of space on your side of the cat. In practice, only microscopists can manage this, because small lenses look large to tiny objects, which is why there are microscope objectives with very fast f-numbers, but few photographic lenses.

Anything that cuts off any of the rays will reduce the f-number of the lens, and dim the image on the ground glass. In the absence of a formal aperture stop, the edges of the glass elements will set the limit, but because the position and shape of the aperture affect how well aberrations are controlled, you should try to ensure that the 'real' aperture is what limits the light passing through the lens and onto your film.

So, in an ideal setup, the spray of rays from the pearl is limited only by the lens' aperture stop. This is often drawn as a cone of rays with its apex at the pearl and its base on the aperture. Strictly, the cone should change shape as it passes through the front elements, but the simple picture is good enough for understanding.

There is a second cone of rays from the exit pupil (the back of the aperture stop) whose apex coincides with the film when the camera is focussed on the cat.

Anything that cuts of any of either cone will cause vignetting, which is just an effete French word for a part of the image being darker than it should be - i.e. not all the rays being collected. A lens shade, a filter holder, a sagging bellows, a finger, a hat, whatever. Anything at all that gets in the way of those two cones will restrict the number of rays that get collected, and so dim the image. In an extreme case, the obstruction will cut off all light, in which case it gets a special name: a field stop.

In this case we're thinking about a lens shade. If you caclulate the angles and opening size of the shade assuming an angle of view given by the simple formula for infinity, objects closer than infinity at the edge of the photo will be vignetted. They will be able to see the center of the lens, but their view of the side of the aperture closest to them will be obscured by the edge of the lens shade. You need to make a shade a bit wider than that needed for infinity to avoid vignetting.

Vignetting isn't the only thing affected by lens shade cutoff. With some subjects, and some lense, the bokeh of the photo will be radically affected by an obscured aperture. Some people don't care, some use it to artistic effect. Your call.

Don't forget to un-nail the cat.

Alex,

"So according to your sketch, the merginal rays are those in bold named "cone of rays entering the lens" right ? I wonder how these rays relate to cone made from the center of the entrance pupil ? (in your sketch the rays are entering through the physical aperture)"

Yes, no, maybe, it depends. The rays shown in bold are marginal rays according to some people's definition, but other definitions suggest that a marginal ray originates at a point on the axis, not off-axis as shown on my sketch.

My sketch is of a simple lens, so the physical aperture (aperture stop) and entrance pupil are coincident. The entrance pupil is located at the point at which the principal ray would cross the optical axis if it were not deviated by the lens. The principal (chief) ray is the one that passes through the centre of the aperture stop. Therefore in a simple lens, the entrance pupil is the same as the aperture stop.

To avoid the vignetting of objects at infinity with the lens at infinity focus (ie at the widest field of view for a lens without floating elements) the lens hood must not obstruct the prism projected forwards by the entrance pupil (same cross-section as the entrance pupil) at the maximum angle to the axis set by the field stop (ie the film gate, the angle being the half-angle of view of the lens for the format – taking into account the rectangular nature of the field stop if you wish). That’s what my sketch shows – the limits of those prisms. The bold lines are drawn from the principal rays by drawing a parallel line that grazes the edge of the entrance pupil.

If you design a lens hood that does not obstruct the parallel bundle of rays heading towards the entrance pupil, and thence to the edge of the film, from an object at infinity you will also prevent the cut-off of light from a closer object at the same angle. Properly designed, it is objects at infinity, not close, that define the acceptable limits of a lens hood.

Best,
Helen

Helen B said:

Properly designed, it is objects at infinity, not close, that define the acceptable limits of a lens hood.

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Mea Culpa. You're right.

An optimal lens hood has nothing to do with the angle to the edge of your film straight on. If you stop your lens down to f22 or so and look from the back light through to the front, and tilt the lens until the filter ring just enters your line of sight through the aperture, that angle is what you're after. You're after the maximum angle (somehwere between 72 and 75 degrees for your 150mm plasmat) that the lens can deliver to film. Any less will ultimately vignette the film area before you've used your maximum movements. That's also why the old standard compendium lens shade bellows type lens hood that can be adjusted just out of vignette range for many different situations is actually optimal. If you're 4 feet away from someone with the 150 shooting a portrait, that adjustable bellows can be pulled way out before it vignettes. You're using a small portion of the max angle the lens can deliver in that scenario. But if you're photographing the empire state building and have used every bit of maximum rise, that little bellows is pulled way in because you're now using every bit of the 72 degrees the lens can deliver and you don't want your shade to get in the image. Now your 90mm has 105 degrees useable. But you can still use that same little compendium lens shade by pulling it all the way in. That's also why you'll see the corners cut out of the older cameras ground glass. It was so the photog could have a straight shot right through the pupil to see if he was getting a vignette from his shade.

Jim,

Maybe in explanation of why lens movements have not been discussed, only general principles: Alex's original question included "So, first I assume there is no movements applied (camera is at the respective detents), the lens is projecting straight onto the film area."

Best,
Helen

Yes, that's it.
I meant first to gain an understanding of basic approach, then, once being accustomized with basic fundamential knowledge I will consider additional aspects, such as movements, for instance.

Now, it seems I'll be better to get into a good reading of US Military optical design handbook which I found to be excellent resource (thank you Dan for directing me this this source).

Alexz said:

Yes, that's it.
I meant first to gain an understanding of basic approach, then, once being accustomized with basic fundamential knowledge I will consider additional aspects, such as movements, for instance.

Now, it seems I'll be better to get into a good reading of US Military optical design handbook which I found to be excellent resource (thank you Dan for directing me this this source).

Click to expand...

Alex, the bill is in the mail.

hi Helen,

welcome back long time no read where you been?

lee\c