I forget the exact formula, but f/ stops are Logarithmic. If you have some Logarithmic graph paper (not sure whether Log-Log or Semi-Log), a straight line at an angle will make a nomograph, which is handy for consulting in the field.

The only aspect I'm still having trouble with is how the effective focal length changes as the lens is moving. I know the lens's labeled focal length never changes, as that is a static number that states the distance from the rear node to the film plane when focused at infinity.

We can use trigonometry to compute the angle of view for any lens. For the 98.6 by 123mm format that has a 157.6 mm diagonal.

For example, if you have a 300mm f5.6 lens that's bellows are extended to 450mm for a portrait, after plugging it into the equation you get a need for 1.17 increase in exposure.

However, specifically for view cameras, we're moving the lens physically closer to the object, often by a significant amount when focusing. This inherently causes the field of view to change, which is different than many modern 35mm lenses, which generally have little movement when focusing, (or mostly front element movement) keeping the field of view relatively static, or at least it contains so little movement that it's hardly noticeable.

Now, focal length is a measurement taken when the lens is imaging an object 1000 meters distant or further. At this distance, light rays from the object arrive at the lens as parallel rays. If the object is closer than infinity (Latin for as far as the eye can see), then the light rays arrive diverging. Because the lens has limit ability to refract (Latin to bend inward), should the object be closer than infinity, the lens to focused image distance will be elongated. For objects closer than infinity, the name changes from focal length to back focus distance.

This all applies to traditional lenses that move as a whole for focussing. Many modern lenses for smaller formats focus by moving only a few elements inside the lens, which is a form of zoom. One would have to do practical tests on such a lens if the angle of view was critical.

The relative amount of movement, as a percentage of focal length is the same with any format. Macro rings or bellows are frequently used on small cameras for macro photography, and the same principles apply as with large format cameras.

Yes, the *effective* focal length—and therefore also the angle of view and f/stop—will typically change with a change of focus.

The only aspect I'm still having trouble with is how the effective focal length changes as the lens is moving. I know the lens's labeled focal length never changes, as that is a static number that states the distance from the rear node to the film plane when focused at infinity.

Interesting! Thank you, that definitely makes sense. So then in theory is the effective focal length equivalent to the distance from the end of a view camera's lens to the film plane?

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If for example, I have a 300mm lens mounted to my 4x5 and focus to infinity, then indeed I would have an effective focal length (angle of view) of 300mm of coverage at infinity.

The area across which your camera can image is known as the field of view or FOV, the larger the FOV the more of your sample you can see.

An "equivalent" focal length lens on 4x5, (150mm) focused at 1:1 will have some pretty dramatic bellows extension (300mm). What does this new field of view look like at this distance relative to how it would look using a 35mm equivalent? Is it the 4x5 300mm equivalent of 80mm on 35mm film? If not, is there an easy equation to determine this?

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However, specifically for view cameras, we're moving the lens physically closer to the object, often by a significant amount when focusing. This inherently causes the field of view to change, which is different than many modern 35mm lenses, which generally have little movement when focusing, (or mostly front element movement) keeping the field of view relatively static, or at least it contains so little movement that it's hardly noticeable.

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The focal length only stays the same if the lens focuses entirely by being moved away from the image plane. True for a view camera but almost never true for a modern camera.

The only aspect I'm still having trouble with is how the effective focal length changes as the lens is moving. I know the lens's labeled focal length never changes, as that is a static number that states the distance from the rear node to the film plane when focused at infinity.

The angle of view normally published is that angle subtended by the diagonal view. Personally, I find this value near worthless. However, TV sets and computer screens are sized by their diagonal measure.

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That would mean then that the effective working focal length / aperture for this lens (originally a 300mm f5.6) when focused in this way is a 450mm f8ish?

Hi there! Thank you for the reply. That makes sense. However, why does the field of view not change in this scenario if the lens is moving further away from the film plane? Specifically view camera lenses, which often move dramatically to focus.

My question is—relative to a 35mm equivalent focal length, how dramatic is this shift? For example, if we took a 50mm lens on a 35mm camera and focused it to 1:1, there is little lens movement when focusing, so the field of view stays relatively static at both infinity and 1:1.

With this assumption, the magnification—otherwise defined as the size of an object on a sensor or film divided by the size of the object in real life—is equal to the lens-film distance divided by the lens-subject distance. The effective focal length then is equal to:

But many small camera lenses—unlike large format lenses—have internal focusing, where internal elements move inside to focus. This has the advantage that the front element doesn't move or rotate, but it complicates the calculations, and that's beyond my knowledge.

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Thick lenses will change the points from where you measure distances: they will no longer be in the center of lens but some other points depending on lens design (even completely in front of or behind the lens barrel), and small format lenses typically are very thick, and determining these distances can be a problem for the photographer.

Interesting! Thank you, that definitely makes sense. So then in theory is the effective focal length equivalent to the distance from the end of a view camera's lens to the film plane?

Also, In regards to how this also changes the f-stop, is there a formula one can use to determine what the new f-stop value would be at a given distance? Can you use the bellows extension equation for this?

The focal length of a lens is the distance from lens to film when focussed at infinity. At any closer distance, the angle of view is reduced and the f number is no longer accurate.

The 4X5 plate measures 101.6mm by 127mm. Subtracting a 1mm border it becomes 98.6mm height by 123mm with a diagonal of 157.6m.

My question is—relative to a 35mm equivalent focal length, how dramatic is this shift? For example, if we took a 50mm lens on a 35mm camera and focused it to 1:1, there is little lens movement when focusing, so the field of view stays relatively static at both infinity and 1:1.

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Also, In regards to how this also changes the f-stop, is there a formula one can use to determine what the new f-stop value would be at a given distance? Can you use the bellows extension equation for this?

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Or so at 1:1 magnification, the effective focal length is doubled. When objects are at infinity, the magnification is zero, and so effective focal length equals the design focal length.

I’m a bit confused as to how effective focal length wouldn’t change when focusing LF cameras, since the lens itself is moving away from the film plane.

The p factor is usually assumed to be 1 and so that would have no effect on the equation. However, this pupil magnification, which is the apparent aperture width as seen from the back of the lens divided by the width as seen from the front, will change the results if it exists. However, from what I've seen, large format lenses are often symmetric and so have no pupil magnification. Small format lenses with a large pupil magnification won't have a dramatic increase of focal length with close focus, and so this design is common with macro lenses.

For example, if you have a 300mm f5.6 lens that's bellows are extended to 450mm for a portrait, after plugging it into the equation you get a need for 1.17 increase in exposure.

The effective focal length will double, assuming a thin, symmetric lens, which is a pretty good approximation for large format lenses, and usually a poor approximation for small format lenses.

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Here's an exampleThis 51.5mm lens has an angle of view of 45.89 degrees which drops to 39.75 degrees when focused to about 0.15x

For example, if you have a 300mm f5.6 lens that's bellows are extended to 450mm for a portrait, after plugging it into the equation you get a need for 1.17 increase in exposure.

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However, if I focus that lens to 1:1, then my bellows is extended to 600mm. Does this mean that my 300mm lens on my 4x5 then would now have an effective focal length (again, angle of view) of a 600mm lens? If not, why?

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That would mean then that the effective working focal length / aperture for this lens (originally a 300mm f5.6) when focused in this way is a 450mm f8ish?

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My question is—relative to a 35mm equivalent focal length, how dramatic is this shift? For example, if we took a 50mm lens on a 35mm camera and focused it to 1:1, there is little lens movement when focusing, so the field of view stays relatively static at both infinity and 1:1.

Here's one example.This 60mm lens is 60mm with an angle of view of 39.4 degrees at infinity.At 1x it's a 49.43mm focal length and 19.96 degree angle of view.

Hi there! Thank you for the reply. That makes sense. However, why does the field of view not change in this scenario if the lens is moving further away from the film plane? Specifically view camera lenses, which often move dramatically to focus.

Also, In regards to how this also changes the f-stop, is there a formula one can use to determine what the new f-stop value would be at a given distance?

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A model of this phenomenon is a pinhole, which has a focal length equal to the distance from the hole to the film plane. Obviously they don't focus, but moving the pinhole back and forth has the effect of changing the focal length as well as f/stop.

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If the object is imaged at unity 1:1, the object will be 2 focal length ahead of the lens = 600mm back focus distance. We substitute this distance for focal lenght.

An "equivalent" focal length lens on 4x5, (150mm) focused at 1:1 will have some pretty dramatic bellows extension (300mm). What does this new field of view look like at this distance relative to how it would look using a 35mm equivalent? Is it the 4x5 300mm equivalent of 80mm on 35mm film? If not, is there an easy equation to determine this?

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That would mean then that the effective working focal length / aperture for this lens (originally a 300mm f5.6) when focused in this way is a 450mm f8ish?