On Tue, 18 Mar 2003 16:13:59 GMT, "ralford" <ralford@bigfoot.com> wrote:

>Thanks for the references - we'll talk about math later :)

>

>The crucial point I had not identified is the well known convention that a

>lens is classified (wide, normal, tele, etc.) by the ratio of its

>focal-length to the film/ccd diagonal. That combined with, the hopefully,

>well known fact that light intensity falls off as 1/r^2 we can do the

>following:

>

>"the same type" leads to

>

> f1/d1 = f2/d2 (1)

>

>where the "ascii subscripts" indicate two different film/ccd sizes, f is the

>focal length, and d is a linear dimension, the diagonal is fine for this

>discussion.

>

>Let E be the intensity (I is too close to 1 for text). The inverse power

>law implies that the intensity for a lens is proportional to 1/f^2, thus

>the intensity ratio at the focal point for two lenses, is

>

> E1/E2 = [ (1/f1) / (1/f2) ]^2 = (f2/f1)^2 (2)

>

>from (1), f2= f1*(d1/d2),

>

>E1/E2 = [ (f1*d1/d2) / f1) ]^2 = (d1/d2)^2

>

>And the intensity does indeed change for different image areas as we all

>expect. The fact that is varies as the square emphasizes the effect.

>

>I'm happy now :)

But, I'm not....;-)

It is all really much simpler than all of this,

really....;-) If you are comparing two CCDs for

relative sensitivity, based on size alone, the

whole lens issue is irrelevant (you could stop

right there...;-), since all relevant aspects

for the lenses used, if properly specified, are

the *same* for the two different CCD sizes, and

therefore the lenses can be dropped from the

comparison equation. It does not matter if the

lens for one is a tele, and the lens for the other

is a WA (assuming even illumination), nor if one

lens is bigger in diameter, or whatever. If the

lenses are correctly focused on the same (distant)

subject, with the same reflectivity and illumination,

and both lenses are set at the same aperture (and

have the same efficiency), and both cover their

respective CCDs, and are used at the same relative

aperture, then their contribution is the same for

both CCDs - they both pass the same "unit intensity"

of light to the CCDs, with the larger CCD obviously

receiving more "units". Better to simply remove

these confusing-but-equal-in-effect lenses, and see

the obvious directly: the two CCDs, under the same

even illumination, with the same type of construction

(including sensor type and number, but not sensor

size, which would change proportionally with chip

size), will not be equally sensitive; the larger

will be more sensitive...

I'm not sure why this is so difficult...;-)