Is my thinking about equivalence right?

Started 3 months ago | Questions thread
kalisti Contributing Member • Posts: 618
Re: Is my thinking about equivalence right?

RobBobW wrote:

Mark Ransom wrote:

RobBobW wrote:

Mark Ransom wrote:

RobBobW wrote:

- the argument about total light is pointless as what is important is light density. Yes FF will bring in 4 times the light, but FF also has 4 times the surface area of sensor to illuminate, so it is a wash. Faster lenses bring in more light per unit area, period.

This is simply false. The reason apertures are measured by F-stops is because this equalizes the light density per unit area between lenses of different characteristics. A lens at F/1.2 will produce the same light density, no matter the focal length of the lens or the size of the sensor behind it. This means that a sensor with 4x the area really will collect 4x the light, when measured over the whole image, as long as the F-stops/T-stops of the lenses are the same.

Mark, you just said the exact same thing as I did. A given unit area of film or sensor does not care how much film or sensor is around it in order to record the amount of light hitting it. Light intensity/density is what is important. Otherwise we would be using different exposures with different sized sensors. A person can use a hand held light meter reading to determine the correct exposure regardless of the film or sensor format being used. Yes more total light hits the larger piece of film for a given exposure, but that is only because more total light is needed to achieve the required light density per unit area.

I think where you lost me is when you said "it is a wash". FF lenses don't automatically bring in 4x the light, but the larger sensor is able to capture more of what is provided. To get equivalence you need to light up the smaller sensor 4x brighter, which requires an F-stop 2 stops lower/faster.

Sorry, but that is completely incorrect. If the light intensity per unit area is the same, the exposures will be the same. Total amount of light only come into play as the numerator of the ratio to the area of the sensor. The author of your linked article states it correctly once then gets it mixed up later on. What I meant by “it is a wash” is that more light is required to illuminate a full frame sensor to the desired light intensity. The only relevance of total light is in obtaining the desired intensity.

The reason this matters is the nature of noise. The majority of noise in today's cameras is from photon shot noise, which is a property of the light itself and not of the lens or sensor or any other camera electronics. The only way to reduce shot noise is to collect more light. Whether you do this with a larger sensor, a larger aperture, or a slower shutter speed is immaterial.

But this is also a function of the size of the individual light sensor pixels. The more pixels per unit area, be it from sensor size or sensor resolution, the more noise, all other things being equal.

Did you read the link I provided? Pixel size matters a lot less than you think it does. Because in the end noise per pixel doesn't matter as much as noise per unit area of the picture.

Sorry, I did not realize there was a hot link in your text I have seen and read it now. The author has some good points, but is confused on a few issues and has several misleading statements resulting from not clearly stating what he meant. So he contradicts himself or states the opposite of what he says later on.

I think its just crossed wires, both correct, same exposure is as you said, mark said "To get equivalence" which would mean (in my book at least) equal noise(ish), which you would need a greater intensity for MFT.

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