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lungovw
14-Jul-2016, 03:28
Dear all,

Can't figure that. I know the formula is f = Focal Length / Entrance pupil, but, it is not at all intuitive for me that if you have light from a point at infinity falling on two lenses A and B, both with the same entrance pupil, both converging that light to points: fa at 150mm from the lens and and fb at 300 mm from the lens, for me, it is the same amount of light on both images. How come that lens A is one f stop brighter and will need half of the exposure time compared to B?

152827

Ray Heath
14-Jul-2016, 03:38
Because, in your diagram, can you see that the camera body for the 300 mm lens will need to be, physically, twice as long to bring the image into focus at the film plane.

So, the same amount of light will need to travel twice as far.

In my practice of building simple lens sliding box cameras I've come to see focal length as not just the determinant of angle of view but also the determinant of how far the light needs to travel within the camera body to arrive in focus at the film plane.

lungovw
14-Jul-2016, 03:49
Thank Ray for this insight. I understand that light becomes weaker with distance when it is diverging, when it is spreading onto surfaces that are bigger with the square of the distance. But in the case of lenses light is converging and should not, for this reason, should not lose strength for travelling longer distance. I assume it does, I just don't understand how.

Pfsor
14-Jul-2016, 04:32
Dear all,

Can't figure that. I know the formula is f = Focal Length / Entrance pupil, but, it is not at all intuitive for me that if you have light from a point at infinity falling on two lenses A and B, both with the same entrance pupil, both converging that light to points: fa at 150mm from the lens and and fb at 300 mm from the lens, for me, it is the same amount of light on both images. How come that lens A is one f stop brighter and will need half of the exposure time compared to B?

152827

If you have a pinpoint light source at infinity (such as star light) in such a case it is the actual physical size of the aperture (same in your case?) and not the f number that determines the exposure. A fact well known to astro-photographers. For a pinpoint light source at infinity if the physical aperture size is the same in your lenses the exposure is the same too.

ic-racer
14-Jul-2016, 06:15
The angle of view in A is greater (you don't show this so the diagram is not correct) and it collects more light to a smaller area making the image brighter.

To clarify your original post title, focal length has no effect on aperture size, they are independent. Focal length affects F-number.

rdenney
14-Jul-2016, 06:24
Dear all,

Can't figure that. I know the formula is f = Focal Length / Entrance pupil, but, it is not at all intuitive for me that if you have light from a point at infinity falling on two lenses A and B, both with the same entrance pupil, both converging that light to points: fa at 150mm from the lens and and fb at 300 mm from the lens, for me, it is the same amount of light on both images. How come that lens A is one f stop brighter and will need half of the exposure time compared to B?

152827

Your illustration shows what appear to be collimated light passing through a lens and converging to a point, which seems unrealistic to me. You have point objects in the distance that are rendered as point objects on the film plane, and areas of light in the distance that are rendered as areas of light on the film plane. The usual diagram showing resolution of distant point objects shows divergence from the object to the aperture, and convergence behind it, but even that is useful more for showing how a lens resolves than how it captures light.

Consider this: The exit pupil is what the film sees. That exit pupil is controlled by the aperture, and is always smaller than the film. Thus, the light is spreading from the exit pupil to the film, and is thus subject to the inverse square law. The better illustration is therefore rays converging at the lens, crossing at the rear nodal point, and then diverging to the film, the opposite of your diagram.

Rick "all models are false, though some are useful" Denney

lungovw
14-Jul-2016, 06:29
Great, thanks Pfsor, considering what you said I started to figure it out. If instead of pinpoints we imagine minimal surfaces, it is a fact that shorter focal lengths will project smaller images than the longer ones. So the same amount of light, captured by the same entrance pupil, will concentrate in smaller surfaces on film and in this way will be actually brighter. Could we say that? Any thoughts?

N Dhananjay
14-Jul-2016, 06:41
Do not think of it as light growing 'tired' because it has to travel a longer distance. Think of it as that a bunch of light rays from an object falling all over the surface of the lens. They get bent to converge towards the focal point on the film plane. So what the film plane sees is a bunch of rays emanating from the aperture of the lens. Now if the same aperture is further away, it is going to look smaller and consequently dimmer. Think of looking at the disc of a torchlight from 1 ft away vs 10 ft away. It is the light collecting ability that is affected by the apparent diminishing of the size. That is why the moon which reflects a pale light from the sun (but is closer to us) looks brighter than a star which is boiling cauldron of energy which is much further away.

Cheers, DJ

lungovw
14-Jul-2016, 07:24
Ic-Racer, the diagram does not consider an angle of view, just one point shedding light onto the lens. Exactly as Pfsor put it, like a pinpoint star. I think, if I understood you correctly, that is not right to say that a wide angle lens collects more light than a telephoto lens. For a wide angle using a very small film size works as a telephoto lens. Agree that it would be better to put is as "How Focal length affects F-number".

Jac@stafford.net
14-Jul-2016, 07:59
For a wide angle using a very small film size works as a telephoto lens.

Technically, it is a long lens. Telephoto lenses are shorter physically than their effective focal length.

ic-racer
15-Jul-2016, 07:42
Ok is see, the diagram shows a collimated light source.

Emmanuel BIGLER
15-Jul-2016, 11:54
Hello from France !

Before looking at the collimated beam as shown in the first post, let us examine this simple comparison of 2 photographic situations.

The idea is to take a picture of a rectangular gray card located sufficiently far away so that the sharp image focuses in the focal plane.
Sure, you'll get a minimum of discernable details on the gray card in order to focus properly, but this is not important at first.

If the diameter of the entrance pupil is the same in both cases, it is not difficult to admit that the total number of photons emitted by the target per second, photons crossing the entrance pupil of the lens, is the same in both cases.

Now all those photons will hit the image in the focal plane and will be distributed all over the image.
In case #1 we use the 15O mm lens, and imagine that our image size is 4x5 inches (yes, we need a HUGE gray card to totally fill the image format ;) )
In case #2 with the 300 mm lens, the full image size will be 8 x 10 inches, i.e. 2 times bigger in terms of linear image size, its surface is 4 times larger.
Hence the number of photons per second and per square area of the image is 4 times smaller with the 300 mm lens if compared to the 150 mm.
Yes, but if I use a 4x5 inch film back with the 300 mm lens?
Nothing is changed in the previous explanation, the number of photons received per second and per square are is 4 times smaller.

4 times smaller in terms of photons per second and per square area in the image will need an increase of 4X in terms of exposure time, i.e. equivalent to 2 f-stops i.e. a f-number multiplied by 2 = 300/150.


Now let's have a look at the perfectly collimated beam.
For example a beam of light coming from a star.
Imagine, first, that both lenses are diffraction-limited.
It can be shown (and this is not easy to explain) that the image of the star is a diffraction spot.
With the same physical aperture size, the diffraction spot size will be twice as large with the 300 mm with respect with the 150 mm. Again the image area where photons are detected is 2 times larges in linear size, 4 times bigger in terms of total area with the 300 mm lens, but the incoming number of photons per second is the same in both cases.

OK, now a real lens is never diffraction-limited ; what happens if both lenses share the same lens formula, hence, with the same aberration patterns ?
Again the 300 mm lens with show an aberration spot 2 times larger in linear size, 4 times bigger in surface (this, also, is not easy to demonstrate without maths), and, again the number of photons per second and per square area in the aberration spot will be 4 times smaller.

Now let us put everything together, aberrations plus diffraction limiting image sharpness, plus a large-size object like a gray card, nothing is changed, for the same diameter of the entrance pupil, the number of photons detected per second and per square are in the image will be 4 times smaller with the 300 mm lens with respect to the 150 mm, provided that we detect the image in the focale plane in both cases.

The important point here is that the relevant quantity for photographic exposure is the number of photons per second and per square area incoming on film or silicon detector.

The only limit to this situation would be: an image of a star delivered by a diffcation-limited lens, with a diffraction spot smaller that the pixel collection area. In this situation, yes, the total number of photons is relevant, not the number of photons per square area.

Mark Sawyer
15-Jul-2016, 13:02
Imagine instead of that collimated beam, you're photographing a flower at a set distance. That flower reflects the same amount of light into either entrance pupil, but with the longer lens, that same amount of light is spread out over a larger image of the flower at the ground glass. Does that help?

Nodda Duma
15-Jul-2016, 14:22
Emmanuel, perfect explanation. It's all in the irradiance.

Emmanuel BIGLER
15-Jul-2016, 15:45
Well; thinking about it twice, the explanation I gave for the case of aberration-limited focusing of a collimated beam is not correct.
The aberration spot will scale like the focal length if and only if the physical aperture size scales like the focal length, hence at constant f-number between the 150 and the 300 mm lens.
Here we are dealing with the same physical aperture and a f-number multiplied by 2x between the 150 and the 300 mm lens.
In those conditions it is not easy to compare both aberration spots.

And whichever the resolution limit might be, either diffraction- or aberration-limited, when the image spot size is smaller than one pixel, the number of photons per square area becomes irrelevant.

-------------------------------------------

Now for something related, and somewhat off-topic, regarding extended sources vs. point sources.
If you look at the moon either with the naked eye or through an instrument like a pair of binoculars or a telescope, the apparent visual brightness of the moon is almost the same in both cases (except for the transmission factor of glasses which is never 100%).
The reason for the same brightness is that when you look through the instrument, the image of the moon is magnified, say, 10X, but at the same time the number of photons captured by the instrument is multiplied by 100X because the entrance pupil of a 10X well-designed telescope has a diameter 10x bigger than the human eye's pupil.
Hence each part of the image, in terms of number of photons per second and per square area, is the same.

Now if you look at stars, they are point objects and their image is a diffraction spot, and the diffraction spot size is the same for all telescopes featuring the same f-number for the primary mirror (or primary lens, for a refractor telescope).
Hence the bigger the entrance pupil of the telescope, the brighter the image of stars; in this situation the telescope acts as a funnel to catch a number of photons proportionnal to the surface of the entrance pupil, projecting them into a diffraction spot size which is the same for all instruments with the same f-number.
For example all f/10 telescopes yield and image of a star with a diffraction-limited diameter of about 10 microns.
The difference between the small amateur telescope of focal length 1 m and a pupil of 10 cm, vs. a big telescope with 5 m of focal length and 50 cm in pupil diameter is that the big telescope will collect 5x5 = 25 times more photons for each image of a star. Both telescopes feature the same f-number N=10 (relative aperture f/10) and will yield the same diffraction-limited resolution of about 10 microns in the focal plane.