Re: Stupid question about magnification

canopus56 wrote:

> The problem with extrapolating this short-distance single lens
> relationship to celestial distances (i.e. - what does the telescope's
> objective alone do?) is that the single thin lens equation breaks down
> after about 30 times the distance of the focal length. Where the object
> distance become effectively infinite - 1/o - simply becomes "zero" and
> the thin lens equation dissipates into a meaningless -
>
> 1/f = 1/i
>
> Similarly, o - the distance to a celestial object - moves to an
> effective infinite distance, then -
>
> M = -i/o
>
> - becomes a meaningless "zero" dimensionless power.

Not really. First off, you slipped from apparent (visual) magnification
we are
talking about with telescopes, to optical magnification, given by the
ratio of
(real)image-to-object size. Second, replacing object distance by
"infinity"
is not correct: even the farthest quasar is still at a finite distance
from us.
Consequently, optical magnification doesn't become "meaningles" at much
greater
object distances than 30 f.l. The exact formula is M=f/(o-f) -
simplifying to M=f/o
for "o" much greater than "f" - giving for, say, the Sun, at its
average distance of 150 mill km,
M=6.666^(-12) for f=1000mm. Multiplied by the Sun's actual diameter
(1.39 mill km)
it gives you the exact size of its image in the focal plane of a 1000mm
f.l. objective - 9.3mm.
If you'd apply your logic of zero magnification, the image itself would
have zero size.

Sure, it can be also derived from Sun's angular (apparent) size of
little over half a degree.
Note that optical magnification for astronomical objects, while
numerically very small,
is not "meaningless". Apparent magnification results directly from the
image's physical
size, and is just about "meaningless" as the real image itself.

Vlad

.

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