Re: Image Luminosity vs magnification (long explanation)

From: Bill Ferris (billferris_at_aol.comic)
Date: 08/31/04 

Date: 31 Aug 2004 01:01:34 GMT

Kurt Fisher wrote:
>The amount of light collected by the telescopic, its _light grasp_,
>is the ratio of the area of the objective divided by the area of
>the human eye pupil, or about 7.62 mm at its maximal dilated pupil.
>Using the basic forumla for the area of a circle - A= (D/2)^2 * pi
>gives -
[snip; ignoring formulae with transmission factor]
>Ignore the transmission factor for now, by making it equal to 1 or
>100%), giving:
>
>G = ( (D_obj^2/4 * pi) ) / ( D_exit_pupil^2/4 * pi ) Eq. 2.1
>
>or by cancelling, simply -
>
>G = D_obj^2 / D_exit_pupil^2 Eq. 2.2

I'm assuming "G" is the relative light grasp of an aperture in relation to the
light grasp of the human eye. Is that correct?

In your Eq. 2.1, do you mean to use "exit_pupil" or "eye_pupil?" If you're
calculating the relative light grasp of an aperture with respect to the eye,
then you probably meant "eye_pupil." Exit pupil does, of course, play a role in
determining the relative light grasp of two optical systems. For example, in an
earlier message, you brought up a 10-inch aperture operating at 10X. If we
compare that system to a human eye with a pupil size of 7-mm (just to use a
nice round number), then we see the following:

Eye: aperture equals 7-mm, magnification equals 1X

10-inch Telescope: aperture equals 254-mm, magnification equals 10X

The telescope will have an exit pupil of 25.4-mm, which will reduce its
effective aperture from 254-mm to 70-mm. This is 10X the diameter of the
observer's eye pupil, which translates to 100X the light-grasp. So G=100, if we
ignore transmission factor. In this instrument, an object's apparent magnitude
(apparent total brightness) will be 5 magntiude (100X) greater than the
object's apparent magnitude to the naked eye. However, since the object is
magnified 10X, it will also have 100X the surface area. So, the apparent
surface brightness of the object will be equal in both the telescope and to the
naked eye.

In another message, you wrote:
>For example, in Table 6, the following telescopes roughly might
>produce an image of the same unit brightness:
>
>1) a NexStar 4 at 20x (relative brightness 0.33% in Table 6),
>2) an 8" telescope used at 60 power (relative brightness 0.3% in
>Table 6),
>3) a 10" telescope used at 80 power (relative brightness 0.33%
>in Table 6), and maybe a
>4) a 10-power 50mm binocular (relative brightness somewhere
>between 0.02% and 0.56% in Table 6).

What do you mean by, "same unit brightness?" I'm trying to reconcile how these
setups might be equivalent. If we apply the rule that an extended object will
have the same surface brightness at a given exit pupil in all apertures, then
we can see that the binoculars and Nexstar 4 produce roughly equivalent exit
pupils (5.0-mm and 5.1-mm, respectively). The 8- and 10-inch scopes produce
roughly same-sized exits at 3.3-mm and 3.2-mm, respectively. However, in terms
of the perceived appearance of an extended object, the 10-inch aperture will
kick ***--to put it mildly--on the binocs and 4-inch scope, and should show a
richer, more detailed view than the 8-inch. This last will be a more subtle
difference but, all other factors being equal, the 10-inch will still be
superior.

I've put together an Excel table that quantifies the above examples in terms
common to amateur astronomer. If you'd like, I'd be happy to email you this
file.

Regards,

Bill Ferris
"Cosmic Voyage: The Online Resource for Amateur Astronomers"
URL: http://www.cosmic-voyage.net
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