# Fast lenses, and High ISO

Started Jul 18, 2014 | Discussions thread
Re: Exposure and brightness.

Great Bustard wrote:

bobn2 wrote:

Great Bustard wrote:

EinsteinsGhost wrote:

Great Bustard wrote:

EinsteinsGhost wrote:

Great Bustard wrote:

EinsteinsGhost wrote:

You're wrong about that, unless you mean that it is also wrongly used for DOF based arguments. Fast is about speed (faster the lens, shorter the exposure time for identical conditions).

What shutter speed can you use on mFT at f/2 that you cannot use on FF at f/4?

Obviously, you didn't try to understand the point made. Let me make it simpler:

Scene brightness: 9 EV

ISO: 100

With these conditions, f/2 will have a faster shutter speed of about 1/4000s. An aperture of f/4 will give you 1/2000s. A larger or smaller sensor will not change that.

Obviously, you do not understand ISO. Why would a FF photographer feel compelled to shoot the same ISO as the mFT photographer?

I don't see a reason for a FF photographer feeling compelled to shoot an ISO 100 shot at ISO 3200.

Neither do I:

http://www.josephjamesphotography.com/equivalence/#purpose

If one system can take a photo that another system cannot, and that results in a "better" photo, then, of course, we would do so.

I see both photographers to shoot at the lowest possible ISO and in this case, it will be the base ISO.

It's fair to say that if f/2 meters at 1/4000, and thus f/4 would meter for 1/1000, then, sure, 1/1000 is more than likely to be "fast enough" in almost all circumstances. What happens with f/2 meters at 1/400? Will f/4 at 1/100 be "fast enough"?

Please tell us about the exposure differences here as they relate to the visual properties of the photo:

• 50mm f/2 1/100 ISO 400 on mFT
• 100mm f/2 1/100 ISO 400 on FF
• 100mm f/4 1/100 ISO 1600 on FF

Same exposure on first and second (f/2, 1/100s, ISO 400). Lower exposure in third by two stops.

That is correct! However, you forgot to tell us how this relates to the visual properties of the photo.

The visual property that an exposure is all about is brightness of the scene. For same brightness, same exposure is expected. With higher ISO (third bullet above), you're doing just that, increasing brightness by two stop to compensate for reduced exposure by two stops.

So what we see here, really, is that exposure is merely part of the equation:

• Exposure (photons / mm²) = Sensor Illuminance (photons / mm² / s) · Time (s)

Not exactly. Luminous exposure, which is what we use in photography, is measured in luminous flux times time, which is luminous energy. Not quite the same thing as 'photons / mm². If we were radiologists using radiometric exposure, then we'd be measuring that in W/m², and it would become clearer that it is a power density, and integrated over time it becomes an energy density, which can then be related to a photon count, given some assumed distribution of the photon energies. The relationship between radiometric and luminous exposure is that luminous exposure is weighted by the luminosity function, that is, includes only visible energy. So, while our cameras do in fact work as photon counters, that is an approximation tio what they should be doing photographically (film cameras are also photon counters, they just count photons in pairs and their QE is rather low).

We discussed this a while back:

http://www.dpreview.com/forums/post/39448295

So are you saying, for example, that one billion red photons falling on the sensor does not result in the same exposure as one billion blue photons falling on the sensor? If a room were illuminated by a 100W red LED, would the camera meter differently than if it the room were illuminated by a 100W blue LED?

It should. Exposure is, as you know, measured in lux seconds. Read about what the 'lux' is here: http://en.wikipedia.org/wiki/Lux. Specifically it is lumens per square meter. Lumens here: http://en.wikipedia.org/wiki/Lumen_(unit)

Luminous flux differs from power (radiant flux) in that luminous flux measurements reflect the varying sensitivity of the human eye to different wavelengths of light, while radiant flux measurements indicate the total power of all electromagnetic waves emitted, independent of the eye's ability to perceive it.

The number of candelas or lumens from a source also depends on its spectrum, via the nominal response of the human eye as represented in the luminosity function.

http://en.wikipedia.org/wiki/Luminosity_function

Photopic (black) and scotopic (green) luminosity functions.[c 1] The photopic includes the CIE 1931 standard[c 2] (solid), the Juddâ€“Vos 1978 modified data[c 3] (dashed), and the Sharpe, Stockman, Jagla & Jägle 2005 data[c 4] (dotted). The horizontal axis is wavelength in nm.

So, indeed green and red (or blue) light is differently weighted with respect to exposure.

• Brightness (photons / mm²) = Exposure (photons / mm²) · Amplification (unitless)

No, no, no. You don't get 'brightness' by 'amplifying' exposure. The luminance of a viewed image depends on what you're using to view it. The luminosity of your TV,monitor or projector or the strength of the light that you're using to view a print. It also continues to provide light as long as you view it, it's luminous energy is not limited by the luminous energy in the exposure. So, not 'amplification' at all. This is the root of the 'ISO' misunderstanding. So, the output of`the photo isn't a luminance. In film days it was a 'density', in digital days it's a file value denoting a grey scale. It simply represents the value from black to whiter than white that we want this exposure to represent (white is set a bit grey so we can have convincing light sources and specular reflections in our photos, if what was white, they'd just look white). Generally rather than 'amplification' it is a 'mapping', which maps the set of exposures which the sensor measures to a set of grey scale values which will include gamma correction and very likely a film like S- curve.

When I talk about "brightness", I was thinking in terms of the nominal values for the color channels in the image file. For example, (200, 200, 200) is "brighter" than (100, 100, 100). Now, I'm sure the mapping is non-linear, so I'm not saying that (200, 200, 200) is twice as bright as (100, 100, 100). But that's what I meant by "brightness". (Note that "amplification" does not necessarily mean "gain", although an analog gain could be part of the amplification).

The word 'amplification' implies that it is linear (unless you qualify that it isn't) and the proposition that is is unitless implies that you are getting the same thing out as goes in. That is false. The photographic reproduction chain falls short of the viewers eye. It isn't a case of reproducing the light that was incident on the sensor - that is the mistake of thinking that makes people believe that somehow ISO is a magic light amplifier. Best to steer clear from any such misleading terminology. There is no 'amplification' needed - because the output from the photographic process is the analogue of what was in film days a 'density' - simply a filter value from letting no light through (or reflecting no light) to letting some arbitrary amount greater than zero through. Film ISOs are defined in terms of exposure to produce a given density and digital ISOs in terms of exposure to produce a given file value. It absolutely is not either 'gain' or 'amplification'. As I said, once you start thinking that way, confusion results, so best not to propagate such a confusing

For example, if we took a photo of a white wall at f/4 1/100 ISO 400 on mFT and f/8 1/100 ISO 1600 on FF, and the ISOs were calibrated the same, then if a pixel for mFT read (100, 100, 100), it would read the same on FF. On the other hand, if the FF photo were taken at f/8 1/100 ISO 400, then the nominal values for the color channels in the image file would be lower -- "less bright".

Yes, because ISO 1600 maps one quarter of the exposure to 12.7% (or 100% depending on the ISO you want to use) than does ISO 400. There is no 'amplification', just a different mapping, or 'scaling' if you like. The important thing is what's coming out isn't light, so the idea that there is some unitless 'gain' or 'amplification' in there is dead wrong.

-- hide signature --

Bob

Complain
Post ()
Keyboard shortcuts: