# I think the notion of FF = heavier lens may not be true

Started Nov 8, 2013 | Discussions thread
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EinsteinsGhost wrote:

Put 200/2.8 on a FF camera, take note of your exposure variables. Turn crop mode on (I understand Canon can't do it, but Sony and Nikon allow it). Do you expect exposure values to be off by a stop?

The answer I gave before was:

http://www.josephjamesphotography.com/equivalence/index.htm#exposure

This section will answer the following four questions:

• For a given scene, what is the difference in exposure, if any, between f/2.8 1/200 ISO 400 and f/5.6 1/200 ISO 1600?
• What role does the ISO setting play?
• What role does the sensor size play?
• What does any of this have to do with the visual properties of the photo?

As mentioned in the introduction of this essay, the concept of Equivalence is controversial because it replaces the paradigm of exposure, and its agent, f-ratio, with a new paradigm of total light, and its agent, aperture. The first step in explaining this paradigm shift is to define exposure, brightness, and total light.

Now let me flesh it out by quoting more from the link:

Mathematically, we can express these four quantities rather simply:

• Exposure (photons / mm²) = Sensor Illuminance (photons / mm² / s) · Time (s)
• Brightness (photons / mm²) = Exposure (photons / mm²) · Amplification (unitless)
• Total Light (photons) = Exposure (photons / mm²) · Effective Sensor Area (mm²)
• Total Light Collected (electrons) = Total Light (photons) · QE (electrons / photon)

So, we can now answer the questions posed at the beginning of the section:

The exposure (light per area on the sensor) at f/2.8 1/100 ISO 100 is 4x as great as f/5.6 1/100 ISO 400 for a given scene luminance, regardless of the focal length or the sensor size. However, the brightness for the two photos will be the same since the 4x lower exposure is brightened 4x as much by the higher ISO setting. If the sensor that the f/5.6 photo was recorded on has 4x the area as the sensor as the f/2.8 photo (e.g. FF vs mFT), then the same total amount of light will fall on both sensors, which will result in the same noise for equally efficient sensors (discussed in the next section).

If you are still confused, please let us know.

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