Lens effects on photons

[Audionut]

Topic split from http://www.magiclantern.fm/forum/index.php?topic=10111.msg131472#msg131472

In the first graph, you can see the 5D3 and the 60D are just as good. That is, if you would use a 80mm/2.8 lens on 5D3 and a 50mm/1.8 on 60D, you would get the same results (same framing, same low-light performance at high ISOs).

I know you used these lens as example, but you must also consider lens diameter.  That is, two 300mm/2.8 lenses with one having a diameter of 50mm, and the other having a diameter of 100mm, the larger lens will let through four times the light.
Rodger Clarke makes small mention of that here.

I believe that DxO measures ISO without a lens attached, so I only mention the above for completeness.

Pixel pitch seems to play some role also.

[a1ex]

That is, two 300mm/2.8 lenses with one having a diameter of 50mm, and the other having a diameter of 100mm, the larger lens will let through four times the light.
Rodger Clarke makes small mention of that here.

Can you quote the relevant paragraph from Roger Clark? I couldn't find it.

[dmilligan]

two 300mm/2.8 lenses with one having a diameter of 50mm, and the other having a diameter of 100mm, the larger lens will let through four times the light.

Wouldn't a 300mm f/2.8 lens, by definition, have a diameter of 107.14mm? A 300mm lens with a diameter of 50mm, would have an f-number of f/6, right?

[Audionut]

Improving Photon Collection

2) Bigger sensor with correspondingly larger lens. The larger lens, even at the same f/ratio, collects more light, thus delivering more light per angular area than a smaller lens of the same f/ratio. Given two sensors, one twice as large are the other, and with a lens twice the focal length, but the same f/ratio on both cameras, the 2x bigger lens delivers 4 times the light for the same field of view. This is the basic reason why small sensor P&S cameras have poor low light performance compared to larger sensor DSLRs. For more on this topic, see: Does Pixel Size Matter?

Think about telescopes, bigger is better, right?  I once read a really good article on the effects of lens diameter on light throughput, but I can't recall that article.  I'll try increasing my google skills today while thinking about downgrading my 6D for something with a metabones adapter.   


@dmilligan.  f-stop is a function of focal length, not lens diameter.

http://www.lavideofilmmaker.com/cinematography/f-stops-focal-length-lens-aperture.html

f-stop = focal length / diameter of lens opening

edit:  The diameter in this instance, being the iris (aperture blades), not the front lens element.

[dmilligan]

@dmilligan.  f-stop is a function of focal length, not lens diameter.

How could the size of the entrance pupil be larger than the diameter of the lens? It can only be the same or smaller (via aperture blades) => a 300mm f/2.8 lens could never have lens diameter of 50mm, that's impossible. If you're "wide open" or have no aperture blades (i.e. telescope), then entrance pupil size and the diameter of lens are the same thing.

[Audionut]

It can only be the same or smaller (via aperture blades) => a 300mm f/2.8 lens could never have lens diameter of 50mm, that's impossible.

Oh, so the numbers for the example, were a poor choice?    

If you're "wide open" or have no aperture blades (i.e. telescope), then entrance pupil size and the diameter of lens are the same thing.

I'm not sure I understand you're meaning here.  Take two telescopes both with a 1.25" eyepiece, one telescope is 4", and the other is 12".  We both know which telescope collects more light, even though that light gets squeezed through the same output.


http://www.clarkvision.com/articles/low.light.photography.and.f-ratios/
http://www.clarkvision.com/articles/exposure.and.upgrades/

Conclusions

The lens is key to collecting light. Always choose the lens with the largest diameter aperture to collect to most light. This is independent of f/ratio, as we saw in part 2 of this series.

35mm/1.4 = 25mm diameter aperture
85mm/1.4 = 60mm diameter aperture

Of course, this takes into account the aperture of the lens, and not simply the front lens diameter.

[dmilligan]

If you have two telescopes that are 4" and 12", and they have the same f ratio, they will necessarily have different focal lengths. The amount of light delivered to the 1.25" eyepiece will be the same. F ratio tells you the amount of light delivered per unit area to the focal plane. So objects in the eyepiece will be the same apparent brightness between the two telescopes (they will just be much more magnified in the 12" scope since it will have a larger focal length).

To take your example again, a 300mm lens with a maximum aperture of f/2.8, would never have any diameter other than 107mm because any smaller diameter is impossible (as we already established) and any larger of a diameter would be a waste of glass (because you'd have to stop it down to 107mm to get an f ratio of f/2.8 ).

[Audionut]

If you have two telescopes that are 4" and 12", and they have the same f ratio, they will necessarily have different focal lengths.

And if they have the same focal lengths?

and any larger of a diameter would be a waste of glass (because you'd have to stop it down to 107mm to get an f ratio of f/2.8 ).

The diameter has to be large enough to not interfere with the Entrance pupil.  It would not need to be stopped down to get an f ratio.  The f ratio is a function of focal length / entrance pupil.

A larger front element allows more light to be focused.  Of course, it would also be more expensive and heavy.

[dmilligan]

If two telescopes have the same focal lengths and the same focal ratios, then they MUST have the same diameter!!!!!

This is basic optics, I really can't believe you of all people have a misunderstanding about this.

The focal ratio is simply the ratio of the focal length to the effective lens diameter. The entrance pupil merely controls the effective lens diameter, they are the same thing. When you stop down, you are basically throwing away lens diameter. If there is no  controllable "entrance pupil" (like telescopes) then the "entrance pupil" is just the effective lens diameter (aka aperture).

If want to prove me wrong then find me any telescope for sale on the Internet where the focal ratio does not equal the focal length of the scope divided by the telescope's aperture (diameter).

[SpcCb]

Maybe you should see this problem with a different point of view because telescopes (most of them, specially big one) have a secondary mirror, so (light) obstruction.
If you made a test with two telescopes a the same f/d, same focal and the same aperture but with a different obstruction you could see major variation in light energy at the focus plan.
It is a fact who explain why refractors are more luminous than reflectors telescopes.

So maybe you should use 'collecting surface' area instead of f/d or aperture (?).

[Audionut]

Looks like memory hasn't served me well on this instance.  With further research, it's clearly all about aperture diameter.

[Levas]

I don't get this discussion. Bigger front element saying something about how much light it gathers ?
Or is this discussion about T-stops ?
T-value is often very close to F-value, but as far as I know there aren't lenses with lower T-stop then F-stop 
 

[dmilligan]

F-stop just tells you the ratio of focal length to lens diameter, this is (usually) close to, but not equal to the T-stop, the T-stop takes into effect "losses". The f-stop is in effect the "ideal" t-stop for a particular lens.

Losses can come from various things, such as obstructions on telescopes as SpcCb pointed out. They also come from reflections off of the surfaces of lenses and imperfect transmission through glass. So if we wanted to quantify how much actual light reaches the focal plane, we use T-stop.

I see what Rodger Clark was trying to point out. That the larger the aperture, the more light is collected per angular area of the subject (regardless of f ratio), which is true. However, that does not mean that more total photons are collected, because that also depends on the focal length, which effects the FOV (aka the angular size of your subject).

[Audionut]

I see what Rodger Clark was trying to point out. That the larger the aperture, the more light is collected per angular area of the subject (regardless of f ratio), which is true. However, that does not mean that more total photons are collected, because that also depends on the focal length, which effects the FOV (aka the angular size of your subject).

http://www.stanmooreastro.com/f_ratio_myth.htm

True S/N (Object S/N)

All of the above is very interesting but it actually says almost nothing about the relationship of exposure-time, aperture, f-ratio, etc. to “true S/N”!

“True S/N” (or “object S/N”) refers to the actual information content of the image.  Object S/N measures information about the target-object and determines important qualities of the image, such as limiting magnitude and feature contrast and visibility.  Object S/N is primarily determined by object brightness, aperture, and camera QE.  F-ratio itself has virtually no effect on object S/N, except for some potential secondary camera noise effects (discussed below).

Information about an astronomical object (star, galaxy, nebula, features of galaxy or nebula, etc.) is contained in the light that falls onto Earth.  That light consists of a certain number of photons per second per square meter of earth’s surface.

The quality of information from an object depends on how many photons are captured and measured by the instrument.  The number of object photons available to the camera is solely determined by:

• Object flux (photons/second/square-meter)
• Aperture size (square-meters) and efficiency
• Exposure time

Focal length (and thus f-ratio) has absolutely no effect on the number of photons collected and delivered.

A longer focal length, just means we capture the same amount of photons from a smaller FOV.  The number of photons being captured from the subject hasn't changed with focal length.

[dmilligan]

Yes for a particular subject, but when you change the FOV, you may change the "amount of subject", hence you change the total amount of photons collected.

Example: a particular aperture means you can capture a particular number of photons for each star in your scene. Now change the focal length and keep the aperture the same. We'll still capture the same number of photons per star, but we will capture a different number of stars, therefore we will capture a different total number of photons. If we use a longer focal length, we'll capture a smaller FOV, hence fewer stars, hence less total photons. For a shorter focal length the opposite will be true.

[Audionut]

I see what you mean (edit:  and it clarifies my general statement about FL above), but the focal length doesn't affect the SNR of the subject.

It's not the total photon on sensor count that we are interested in, it's the total photon count from the subject of interest.  Sticking with the astro theme, if I'm capturing a nebula, the photon count from stars outside of the nebula are of no interest from an SNR standpoint.

Focal length is important from an detail standpoint, to ensure maximum sensor coverage (resolved detail).

[SpcCb]

The number of object photons available to the camera is solely determined by:

    Object flux (photons/second/square-meter)
    Aperture size (square-meters) and efficiency
    Exposure time


Focal length (and thus f-ratio) has absolutely no effect on the number of photons collected and delivered.

My car is blue, so my car is not blue

Maybe it could help (not absolute but can be use for an approach):

I = (T*pi*B)/{4*(f/d)²*(1+M)²}

I, image illumination in foot-candles
T, transmittance of the lens
B, object luminance in candles/square foot
M, image magnification

So yes, with using some 'perfect' accessories, if you compare two images at different focal length (all other param equals) but cropped to fit the same field it should be close in term of illumination (question of magnification).
But it is cheating a bit on the result samples.. If you compare raw fields of course lower f/d images will get higher illumination.

[IliasG]

Focal length (and thus f-ratio) has absolutely no effect on the number of photons collected and delivered.

Lets say you have a FF sensor and 70-200/f2.8

You shoot at a small target in the sky (say one that you would need 1000mm to cover all the frame) ..
Shoot at 100mm 1sec f/2.8 then 200mm 1sec f/2.8 .. by the summary of
   1. Object flux (photons/second/square-meter)
   2. Aperture size (square-meters) and efficiency
   3. Exposure time

You will have recorded 4X more photons with 200mm f/2.8 because the "2. aperture size (sq-meters)" will be 4X larger 

[Audionut]

It wasn't the focal length that delivered more photons.  100mm f/1.4 will also have the same aperture size.


@SpcCb
The image magnification (M), only plays a role in how the pixels record the photons, right?

[SpcCb]

The image magnification (M), only plays a role in how the pixels record the photons, right?

We are speaking about energy collected in all the field, with no consideration about the recording device, if I'v understood (?).

If the goal is to defined how many energy get each pixels of a sensor, it is quite different.

Do we want to compare with constant pixel size? If yes, it's like there's no magnification between samples, so the energy received is function of the f/d (with constant diameter we should get something function of the focal length).
Or do we want to compare with different f/d and different pixel size? (strange method, by the way) In this case it's like we also change the flux area covered by 1 pixel, so somewhere the energy collected.

[Audionut]

We are speaking about energy collected in all the field, with no consideration about the recording device, if I'v understood (?).

I'd rather speak about energy collected from the subject of interest (with no consideration about the recording device).  We could use an 8mm lens and collect the energy of the entire Milky Way, but that is pointless if we are trying to capture and render Jupiter.
The same with a portrait shot with a shallow DOF.  The energy collected from the out of focus areas is of little interest, since it is blurred, we can just apply noise reduction to these areas, since there are no fine details that need respect.

So the image magnification is useful from an detail standpoint, because of the lens ability to resolve detail, and the recording device ability to record that detail.  But from a photon count standpoint (SNR) (from the subject of interest), image magnification plays no role.

I believe this is an important distinction, since we choose focal length for reasons other then photon count.  If we are limited by shot noise, we would choose a lens with an greater aperture diameter.  If we a limited by resolved detail, we would choose a lens with the correct focal length, and/or, optics that resolve finer detail.  In the first case, we increase the SNR, in the second case, we increase our ability to record fine detail.

If focal length alone (constant aperture size), changes the amount of photons collected from a point source, then I am happy to be corrected.