I finally 'get' equivalence

Started May 22, 2012 | Discussions thread
OP Dorus Senior Member • Posts: 1,065
Re: You're ignoring total exposure. Back to the drawing boards.

Oh, yes, I did mean at same shutter speed and ISO setting. Thanks for pointing that out - it was implicit in my OP but I should have been more clear.


gollywop wrote:

Dorus wrote:

it has taken me long time to get to this insight, and looking at the discussions, I am not the only one

What has always mystified me (don't ask me why, as it is so obvious) is the importance of seeing this concept ONLY in the context of the final product. So let's say that I am going to print an 18x24 picture. If I printed that picture from a 35 mm equivalent Full Frame sensor, at an f-stop of 2.8, that picture will have incorporated 4x more light than if it was taken on a 4/3 or m4/3 sensor at the same f-stop of 2.8. Thus, with identical sensor technologies, the FF picture should have better noise performance and better DR (at least if the media of the final product allows to see the difference between the 2, but that is a separate discussion), since 4x more light was used to generate the picture.

The f-stop is only one part of the exposure; you're forgetting shutter speed and ISO. Let's suppose the ISO is equivalent (which doesn't always mean exactly the same value, depending on manufacturer). If you had an FF sensor with shallow photosites that filled at f2.8 in 1/1000 sec vs. an FT sensor (micro or not) with deep pockets that filled, with the same scene, at f2.8 in 1/250 second, they would both acquire the same total light energy. It's not just the sensor size. Well depth, and other things matter also. If the base noise level of the FT in this case were good, it could well have better noise characteristics than the FF.

Of course, the FF sensors tend also to have deeper wells than the FTs, so this objection is mainly theoretical. But it's more than just relative sensor size.

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