The advantage of small sensors

Started Feb 10, 2009 | Discussions thread
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no see Contributing Member • Posts: 814
The advantage of small sensors

Since my first trial

small sensors get as many photons as large sensors
I have tried to understand more about the subject.

So this is my second try - hopefully interesting enough for you to read - thanks.

The advantage of small sensors

[1] Douglas A. Kerr: Some Principles of Photographic Optics

[2] Professor Charles S. Johnson, Jr.: Why Is My 50mm Lens Equivalent to 80mm on a 35mm Camera and Why Is There More Depth-of-Field?

[1] Rather, what we are concerned with is how the lens transforms the

luminance (brightness) of a patch on the object into the illuminance delivered to the film to form the corresponding patch of the image. This involves both the amount of light gathered through the entrance pupil from the object patch and the size of the resulting image patch. Because of the latter, the distance from the lens to the film is also involved. E is the illuminance given to a patch on the image.

In [2] the conditions for getting the same image of a scene with different sensor sizes is discussed (same DoF different CoC).

Now it's nice to combine both articles to get more insight into matters:

Lets look at the illuminance E of a patch [see1] for a small sensor (Es) and a big sensor (Eb), then
Es = (Nb/Ns)^2 * Eb (N = f/D 'f/number')

(when DoF is the same and diameter Ds and Db of the entrance pupils are the same [see1 and 2]).

If the object patch is a DoF cell (resolution cell RC), then it's images are the resolution cells on the small sensor (RCs) and the big sensor (RCb).

Now I want to compare my Pentax K10D sensor (10MP, APS-C, pixel pitch 6 microns) against my Panasonic FZ50 (10MP, 1/1.8", pixel pitch 2 microns). Looking for the CoC in [2] you find: CoCb = 19micron and CoCs = 5.9micron. For both we get about 3x3 pixels to cover their CoC.

This means the FZ50 sensor 3x3 pixels get (Nb/Ns)^2 more illumination than the 3x3 pixels of the K10D: Es = (Nb/Ns)^2 * Eb.

Looking for Ns and Nb in [2] we get: Ns = 4.4 and Nb = 14, which results in (Nb/Ns)^2 = 10.

In other words, if you allow a smaller DoF for the K10D, then you have to open the lens for three stops, to get approx. the same light (illuminance) for these 3x3 pixels of the K10D compared to the 3x3 pixels of the FZ50.

Of course you may look at all patches resp. resolution cells to set up the whole image on each sensor (which should be the same [see 2]).

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no see (resp. ][.Kerusker)
we don't see that we don't see (eye's blind spot)

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