[dtmateojr]

You are probably referencing the notion of "f-stop" (or "f-number") which is defined by "focal_length / aperture_diameter". So decreasing the focal length while keeping the aperture diameter constant, would result in a lower f-ratio.

A lower f-ratio will lead many people, including you, to believe that the lens is now "faster"/"brighter".

However, that is only true as far as "exposure" (i.e., light intensity) is concerned.

A bit of thought (see my post why more samples of a signal reduce noise), however, reveals that light intensity is not essential for image quality (noise levels) but that the overall amount of light received is.

As teleconverters and speed boosters (when used as format adapters) demonstrate, you can change focal length and f-ratio back and forth, without affecting image quality (real world imperfections of the devices not considered) as these devices do not touch the overall amount of light gathered. So while a "speed boosted" lens may look faster ("oh, it turned from an f/11 to an f/5.6!") or a teleconverted lens may look slower ("oh, it turned from an f/5.6 to an f/11!"), the noise levels will not be affected by these conversions at all.

P.S.: A speed boosted lens indeed gets "faster" when you compare the lens with and without speed booster on the crop sensor. Normally, the crop sensor throws away a lot of light (signal) and thus makes the lens slower. An f/2.8 FF lens used on APS-C is only equivalent to an f/4 lens on FF. The speed booster condenses the normally lost light into the APS-C image circle so now all of the light gathered by the lens is used, making it perform at f/2.8 as it would on an FF sensor.

You are indeed a follower of JJ. Instead of explaining how it works you gave me a definition. Everybody knows that f-stop is focal length over aperture. Nobody is arguing with that. Question remains: why would decreasing the focal length while keeping aperture constant make a lens brighter (faster)?

[kh1234567890]

It can be mathematically proven that you can take the same photograph with different format cameras and it can also be empirically demonstrated. Have you looked at the real world demonstrations by DPReview?

Unless that girl has involuntarily elastic nose and ears, those shots are not the same ...

[dtmateojr]

You are indeed a follower of JJ. Instead of explaining how it works you gave me a definition. Everybody knows that f-stop is focal length over aperture. Nobody is arguing with that. Question remains: why would decreasing the focal length while keeping aperture constant make a lens brighter (faster)?

Still no answer? So you raised the concept of a speed booster to refute my arguments when you do not even understand how it works. ROFL! You and Joseph James are geniuses.

[Class A]

Everybody knows that f-stop is focal length over aperture. Nobody is arguing with that. Question remains: why would decreasing the focal length while keeping aperture constant make a lens brighter (faster)?

What is your question?

There are three possibilities:

1. You are asking "Why does decreasing focal length while keeping aperture constant decrease the smallest f-ratio of a lens?" Answer: That's why I provided the definition of "f-ratio". You can easily see that by decreasing the numerator of the quotient, the result becomes smaller as well.
2. You are asking "Why does decreasing focal length while keeping aperture constant make a lens gather more light?" Answer: It does not. When a speed booster is used to reduce the size of the image circle of the lens so that all the light it already gathered before actually hits a crop sensor then of course this changes the FOV of the lens and the amount of light that is recorded, but it does not change the amount of the light the lens gathers.
3. You don't believe that decreasing the focal length while keeping aperture constant makes a lens brighter (faster) and are challenging the idea.

Which of the above is it?
Or is it a different question altogether?

[ScooterMaxi Jim]

At the risk of summing up what the original poster wants to know, let's keep the bare facts in mind.

For purposes of proper exposure, the aperture governs the intensity of light hitting the sensor no matter what the crop factor. So, shooting at ISO 100 with the 17-50 at f/2.8 your exposure on FF or APSc will be the same. For a static scene, you probably will want to factor in the narrower FoV to ensure against visible camera shake by setting the shutter speed higher because you are enlarging the APSc image more to reach the same proportion as a FF. However, the basic exposure principles remain the same.

Now, f-stop is not the perfect measure of how much light is hitting the sensor plane; t-stop (transmission) is the proper measure. However, we typically are not aware of the extent a particular lens loses transmission. As a general rule, though, the fewer elements a lens has, the less light loss. I suspect that the 17-50 f/2.8 at 21mm brings about the same light intensity as a 21 f/3.2 Ltd. due to the latter's simpler design and high-quality coatings. Also keep in mind that all sensors to varying degrees lose some of the ability to gather light at especially large apertures. Past around f/2 this becomes especially apparent. The comparative sensor light gathering between a 50 f/1.4 and a 50 f/1.2 is hardly noticeable (although the effects on image appearance and DoF are easily recognizable).

If the OP wanted to know about the depth of field effect of different sensor sizes (sometimes referred to as circle of confusion), then the equivalence discussion certainly is appropriate. However, we might want to be more cognizant of the knowledge level of the OP, who by now most likely has departed the thread in frustration.

[dtmateojr]

What is your question?



There are three possibilities:

1. You are asking "Why does decreasing focal length while keeping aperture constant decrease the smallest f-ratio of a lens?" Answer: That's why I provided the definition of "f-ratio". You can easily see that by decreasing the numerator of the quotient, the result becomes smaller as well.
   
2. You are asking "Why does decreasing focal length while keeping aperture constant make a lens gather more light?" Answer: It does not. When a speed booster is used to reduce the size of the image circle of the lens so that all the light it already gathered before actually hits a crop sensor then of course this changes the FOV of the lens and the amount of light that is recorded, but it does not change the amount of the light the lens gathers.
   
3. You don't believe that decreasing the focal length while keeping aperture constant makes a lens brighter (faster) and are challenging the idea.
   

Which of the above is it?

Or is it a different question altogether?

Again, nobody asked for a definition. EVERYBODY knows the definition. The question is: do you understand WHY? Do you understand HOW focal length affects light? Because if you do understand, you would not have used a focal reducer to refute my arguments. It SUPPORTS the fact that equivalence is WRONG. I'm simply making you realise for yourself that you are very very wrong.

So explain HOW it works, not what it is.

[normhead]

Nobody is arguing with that. Question remains: why would decreasing the focal length while keeping aperture constant make a lens brighter (faster)?

That depends on if keeping the Aperture constant means, keeping i the same size, or if it means keeping it at say, ƒ4. Both could be considered keeping the Aperture constant. Not that I care, or have a horse in this race just saying. But obviously a one 1 cm opening on a 50mm lens lets in more light than a 1 cm opening on a 200mm lens. It's collecting light from a much wider angle.

[dtmateojr]

That depends on if keeping the Aperture constant means, keeping i the same size, or if it means keeping it at say, ƒ4. Both could be considered keeping the Aperture constant. Not that I care, or have a horse in this race just saying. But obviously a one 1 cm opening on a 50mm lens lets in more light than a 1 cm opening on a 200mm lens. It's collecting light from a much wider angle.

Yes. That's one way of looking at it, by FoV. FoV is directly related to focal length. They are inseparable. Another way of looking at it is using a property of focal length: magnification or the projection of the subject into the sensor. It's NOT about the image circle. A 35mm Nikon DX has a smaller image circle than the 35mm Nikon FX but you can use both of them on DX cameras because they have the same LIGHT PER UNIT AREA. You could probably use the DX on a full frame mirrorless with a shorter flange distance (not sure -- you may have problems with infinity focus). That's all that matters. Total light alone is totally missing the point.

Point is, decreasing focal length does brighten an image. It lets MORE light per unit area. Heck, it lets more light in, period.

[audiobomber]

So I am to understand that when I'm using my tamron 17-50mm 2.8 on my k5 it is equal to a 25-70mm lens now does this affect the aperture also or is it still considered an f2.8 ?

It is always a 17-50mm f2.8 lens because those are lens properties. What you are asking about though, I believe is how does it change on a FF camera.

Full-frame equivalence means same DOF, same FOV, same noise, same diffraction, same perspective as an equivalent FF lens. Perspective is dependent on subject distance, so keep it the same for each system. Adjust focal length and f-number by the crop factor. Use ISO to equalize brightness. Here are a few equivalents:

Nikon V1, 1" sensor, 100mm f/2.0, crop factor 2.7X
@ 10 ft. subject distance, DOF is 0.13 ft.
@ 100 ft. subject distance, DOF is 13.4 ft.

Pentax K-50, APS-C sensor, 180mm f/3.6, crop factor 1.5X
@ 10 ft. subject distance, DOF is 0.13 ft.
@ 100 ft. subject distance, DOF is 13.4 ft.

Nikon D800, FF sensor, 270mm f/5.6, crop factor 1X
@ 10 ft. subject distance, DOF is 0.13 ft.
@ 100 ft. subject distance, DOF is 14.1 ft. (The DOF calculator would not allow f5.4, which explains the slight difference in DOF at 100 feet).

Online Depth of Field Calculator

Now let's look at noise equivalence using DXOMark:

Nikon V1, 100mm, 1/300s, f/2.0, ISO 100; SNR is ~35dB
Pentax K-50, 180mm, 1/300s, f/3.6, ISO 350; SNR ~35dB
Nikon D800, 270mm, 1/300s, f5.6, ISO 640; SNR ~35dB

Nikon D800 versus Pentax K-50 versus Nikon 1 V1 - Side by side camera comparison - DxOMark

In summary, photos taken with the above setups will be equivalent in FOV, DOF, perspective, diffraction and noise, i.e. equivalent photos. However, a lens designed for APS-C like your 17-50 will vignette strongly on a FF body.

[dtmateojr]

It is always a 17-50mm f2.8 lens because those are lens properties. What you are asking about though, I believe is how does it change on a FF camera.

Equivalence for a lens means same DOF, same FOV, same noise, same diffraction, same perspective. Perspective is dependent on subject distance, so keep it the same for each system. Adjust focal length and f-number by the crop factor. Use ISO to equalize brightness. Here are a few equivalents:

Nikon V1, 1" sensor, 100mm f/2.0, crop factor 2.7X
@ 10 ft. subject distance, DOF is 0.13 ft.
@ 100 ft. subject distance, DOF is 13.4 ft.

Pentax K-50, APS-C sensor, 180mm f/3.6, crop factor 1.5X
@ 10 ft. subject distance, DOF is 0.13 ft.
@ 100 ft. subject distance, DOF is 13.4 ft.

Nikon D800, FF sensor, 270mm f/5.6, crop factor 1X
@ 10 ft. subject distance, DOF is 0.13 ft.
@ 100 ft. subject distance, DOF is 14.1 ft. (The DOF calculator would not allow f5.4, which explains the slight difference in DOF at 100 feet).

Online Depth of Field Calculator

Now let's look at noise equivalence using DXOMark:

Nikon V1, 100mm, 1/300s, f/2.0, ISO 100; SNR is ~35dB
Pentax K-50, 180mm, 1/300s, f/3.6, ISO 350; SNR ~35dB
Nikon D800, 270mm, 1/300s, f5.6, ISO 640; SNR ~35dB

Nikon D800 versus Pentax K-50 versus Nikon 1 V1 - Side by side camera comparison - DxOMark

In summary, photos taken with the above setups will be equivalent in FOV, DOF, perspective, diffraction and noise, i.e. equivalent photos. However, a lens designed for APS-C like your 17-50 will vignette strongly on a FF body.

Did you look at the interpolated SNR or the real measured SNR?

[Class A]

At the risk of summing up what the original poster wants to know, let's keep the bare facts in mind.

I'm sure you had good intentions and tried to help the OP in a more direct manner.

The trouble is that the deeper discussions that ensued in the thread are the result of people knowing what they are talking about trying to dispel the myth that you should convert focal length in order to understand which FF lens would take the same images like a certain lens on APS-C, and that DOF is affected somehow as well, but that an f/2.8 lens will remain a f/2.8 lens for the purposes of exposure.

While technically accurate, stating that the exposure aspect does not change is very misleading, because in order to take an image on a FF camera that looks the same like one taken on an APS-C camera, we have to use less exposure.

The important image parameters DOF and SNR (noise levels) are determined by the total amount of light used which is proportional to exposure x sensor-area.

That's why stating that "f/2.8 = f/2.8“ is entirely unhelpful. The correct equivalent f-ratio the OP needs to consider is f/4.

Please read the DPReview article.

P.S.: You were right about t-stops, but this discussion was not what the OP was after.

[dtmateojr]

I'm sure you had good intentions and tried to help the OP in a more direct manner.

The trouble is that the deeper discussions that ensued in the thread are the result of people knowing what they are talking about trying to dispel the myth that you should convert focal lenght in order to understand which FF lens would take the same images like a certain lens on APS-C, and that DOF is affected somehow as well, but that an f/2.8 lens will remain a f/2.8 lens for the purposes of exposure.

While technically accurate, stating that the exposure aspect does not change is very misleading, because in order to take an image on a FF camera that looks the same like one taken on an APS-C camera, we have to use less exposure.

The important image parameters DOF and SNR (noise levels) are determined by the total amount of light used which is proportional to exposure x sensor-area.

That's why stating that "f/2.8 = f/2.8“ is entirely unhelpful. The correct equivalent f-ratio the OP needs to consider is f/4.

Please read the DPReview article.

P.S.: You were right about t-stops, but this discussion was not what the OP was after.

You conclude as if you never read anything that was discussed in this thread. Total light alone does not make sense in photography.

[GeneV]

"...police said he fell down an elevator shaft. Onto some bullets." ~The Bowler, Mystery Men

Would a doctor's autopsy read "Cause of Death : Gunman"? Or would the bullet and the resulting damage be mentioned as cause?

Actually something of both, since ME reports usually include "homicide" "suicide" or "accidental" in the cause as well as the mechanics of death. OK, enough drift.

[Class A]

Again, nobody asked for a definition. EVERYBODY knows the definition.

I did not provide a definition, I asked you a question.

Instead of answering my question, i.e., instead of clarifying what you really want to know from me, you repeated your statement about why using formulas is not the same as understanding something. We cannot make progress like this. If you debate properly, you will be cornered into realising that you are basing your conclusions on fundamental misconceptions. I'm foolish enough to think that I may succeed in letting you realise your fundamental mistakes, despite the fact that entire threads (here and here) were dedicated to make you realise your misconceptions, in part by very knowledgeable people (one of them is responsible for the very useful sensorgen website). FYI, I never participated in these threads, something you seem to assume.

So if we don't want to create a third thread like that, you'll either have to answer questions or I'll have to give up. I had given up earlier when I said it was time for me to leave the thread but then I thought of two aspects that may help some people to understand what is going on. It is my hope that someone yet uninfluenced stumbling over this thread will read my posts and hence not fall victim to the misinformation spread by people who do not understand that exposure must not be equal across formats in order to achieve images that look the same.

The question is: do you understand WHY?

I do. And the result the definition gave me "f-ratio becomes smaller" which is exactly in line with what you stated later:

"It lets MORE light per unit area." -- dtmateojr

So in my first answer to your question I already stated the same fact by saying that the f-ratio decreases. A decrease in f-ratio, as you know, increases the intensity of light.

So why did you have to repeat your question? What you should have done instead is to read my explanation as to why an increase in the intensity of light does not matter (when you decrease the format at the same time).

I already explained that for image quality the main factor is the number of photons captured because the more photons we capture, the lower the shot noise. Everyone knows that through the fact that longer shutter speeds or wider apertures reduce the noise in an image.

Now if you capture the photons gathered by a lens, it does not matter what registration distance a lens has. You can capture all photons, e.g., 50mm behind the lens with a sensor that is 50mmx50mm large, or you can do it 100mm behind the lens, with a sensor that is 100mmx100 large. We have to increase the sensor size as the light emanating from the lens will spread.

Do you understand the difference between "registration distance" and "focal length"?
It seems you don't.

Do you agree that we capture the same photons either way?

If you do, you'll realise that we are taking the same photo, either way. We are just using a different representation in-between, before we show the sensor data to the viewer.

Do you agree that the exposure (light / area) is different between the 50mmx50mm sensor and the 100mmx100mm sensor?

If you do, you'll realise that the exposure can be different but the image captured will be the same.

In fact, the larger the format, the lower the exposure must be.

If we kept the exposure constant when changing formats, i.e., if we said a 17-50/2.8 lens corresponds to a 26-75/2.8 lens on FF, then we would expose the larger format image so that we can print it larger than the small format image without additional noise showing. The latter is not necessary (and incorrect) for an equivalent image. Just because we are using a larger format, does not mean we want to print bigger. Hence we don't need to achieve the same exposure. We would be paying the price of less DOF anyhow, if we used the aperture to achieve the same exposure (i.e., if we used the same f-ratio on both formats).

Do you understand HOW focal length affects light?

Yes, and the definition of f-ratio I provided was in accordance with that understanding.

Because if you do understand, you would not have used a focal reducer to refute my arguments. It SUPPORTS the fact that equivalence is WRONG.

How so?
Please elaborate.

Please, before you answer, read my post here carefully and think about the irrelevance of the lens to sensor distance for capturing an image for how the image will look like, but that exposure depends on the lens to sensor distance (as intensity, i.e., light per unit area decreases the further away the sensor is from the lens). From this, it follows that exposure is not a good parameter to consider when you are concerned how an image will look like (in terms of DOF and noise) across formats.

[dtmateojr]

Do you agree that the exposure (light / area) is different between the 50mmx50mm sensor and the 100mmx100mm sensor?



In fact, the larger the format, the lower the exposure must be.

ROFL! Somebody does not understand the term RATIO.
Is 1/2 less than 2/4? I'll give you a hint that any child would understand: the numerator is the incoming light and the denominator is the sensor area. LULZ!!!