[Chaos-l] Burgess Finders?
D Gary Grady
DGary.Grady at gte.net
Tue Aug 21 17:42:22 EDT 2007
Michael Hrivnak wrote:
> If I have my math right, I think objects in a perfect 8x50 should be about 1.56x brighter than in a perfect 6x30.
I believe you are absolutely correct with respect to extended objects
such as nebulae but not with respect to stars. It may seem
counterintuitive, but while nebulae and naked-eye planets should appear
about 1.56 times as bright, stars should appear about 2.78 times as bright.
Here's one way to look at it:
Roughly speaking, the apparent brightness of an object on the focal
plane is a function of light per unit area.
If you increase the objective diameter by a factor of 50/30, you've
increased the area by a factor of (5/3)2, so all else being equal you
should be capturing (about) 2.78 times as many photons per second.
However, if you at the same time you increase magnification, you're
spreading those photons over a larger area on the focal plane. Going
from 6x to 8x for example will make an image -- of, say, M31 -- 4/3
times as big across, or 16/9 in area. So the light per unit area is only
9/16 as much as it would be without the magnification.
Taken together, the larger objective and the greater magnification will
make an image of a nebula (25/9)x(9/16) = 25/16 = 1.5625 as bright in an
8x50 finder as it would appear in a 6x30.
*However*, suppose what you're looking at is a star rather than an
extended object. In that case (assuming good optics) the image is for
practical purposes a point at both 6x and 8x (or 1000x for that matter).
So greater magnification doesn't spread the photons over more area (or
more light receptors in your eye), and hence magnification doesn't alter
apparent brightness for stars. Hence stars will appear 2.78 times as
bright in an 8x50 viewfinder in comparison with a 6x30.
A related matter of interest in astrophotography of objects in the solar
system: The exposure needed for a planet or comet visible in the image
as an extended object is related to the planet's distance from the Sun
but not to the planet's distance from the telescope, for the same reason
that when you photograph Aunt Minnie, the exposure you use has nothing
to do with Aunt Minnie's distance from the camera. But when the object's
image is a dot, exposure depends on both.
D Gary Grady
Durham NC USA
dgary at mindspring.com
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