Re: Stupid question about magnification

Paul Schlyter wrote:
> In article <d4v1t1del4t41meago5oclmcjh8306hi8l@xxxxxxx>,
> Chris L Peterson <clp@xxxxxxxxxxxxxxxxxx> wrote:
<snip>
Paul replied -
> No magic is involved here, and the traditional telescope formulas still
> work. It's only a perhaps somewhat different way to view the issue of
> magnification. And that view is at least somewhat useful - the magnification
> figures assugned separately to the telescope objective and eyepiece are
> not meaningless.

I believe Chris's point (in part) is that magnification of a telescopes
objective alone is, as a matter technical optics terminology,
meaningless because the equation for a single lens's magnification is
meaningless when the imaged object is at infinitive distance.

The term "magnification" can used in the context of magnifying
terresterial objects using a simple single lens system, as in a "3x
hand magnifier" that you can buy at the local 5 and 10. The
magnification of the simple single lens at terresterial distances is
described by the thins lens equation:

1/f = 1/i + 1/o
http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/lenseq.html

where f is the focal ratio of the lens, i is the distance to the image
and o is the distance to the object.

Magnification in this short-terresterial distance and single lens
context is:

M = -i/o
http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/lensdet.html#c2

where i and o are known terresterial distances. For example, I am now
holding a hand magnifier near a coffee cup on my table and am
magnifying its image several times.

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.

You can't have more of an "effective infinite distance" of the object
than say, something like the Orion Nebula, currently overhead. As a
matter of practical visual detection, I understand that an effective
infinite distance starts for objects at about 30 times the focal
length. The thin lens equation for magnification works at smaller
distances where the light is entering the lens radiating from a point
source. At an infinite distance, light effectively enters the
objective lens a bundle or "beam" of parallel light rays. Because
light enters the objective in parallel rays, the thin lens equation
breaks down.

For this reason, I believe that Chris is saying that magnification of
celestial objects at an effective infinite distance does, as a matter
of technical optical terminology, not exist, for a single lens system.
You have an angular size of the true object on side of the lens and
it's linear size at the prime focus on the other size. There is no way
to divide these dissimlar measures to reach the dimensionless scalar of
"power," a.k.a. magnification, when the object is at an infinite
distance.

Though there is no magnification of a celestial object with a single
lens system, as magnification is technically and mathematically
defined, I can take my reading handlens and hold it up at an arm's
length aligned to a nearby mountain. As I move the lens back and forth
between 0 and 1 focal length, magnification does occur. Similarly,
when pointing my 10" newt reflector at the Moon, I typically just
remove the 2" eyepiece, put my eye at prime focus, and find the image
of the Moon reflected in the secondary mirror. That image appears
magnified. As a lay convention, I can express the Moon's magnification
in this simple one lens system in terms of its apparent angular size,
which can be compared to the 1/2 degree true angular size of the Moon,
to compute a dimensionless "power" or magnification. My reflector's
telescope object has a "power," even when pointed at celestial objects.


So, maybe for a lay convention and the single lens system consisting
only of the objective of a telescope, "power" might be considered the
true angular size of the object in the night sky to its angular size
seen by the eye at prime focus.

This of course, should not be confused with true telescopic
magnification, referred to by Chris - the true optical magnification
created by the special case of a two lenses with coincident focal
lengths, i.e. - the traditional Keplerian telescope.

Although probably not proper in terms of technical optics terminology,
IMHO, in lay discussions, it probably is acceptable to talk about the
"power" or magnification of a single lens system alone - including if
you are talking about just the telescope's objective alone. This
assumes that both the participants in the discussion have a grasp of
the differences between magnification in various optical systems, i.e.
-

Magnification in case 1 (single positive lens, object at more than one
focal length but less than effective optical infinity),

Magnification in case 2 (single positive lens, object at less than one
focal length - a hand magnifier), and

Magnification in case 4 (two positive lenses, focal lengths coincident
- a traditional Keplerian telescope viewing a beam of light consisting
of parallel rays from an infinitely distant light source).

Maybe the concept of diopter, used by opticians to describe the power
of single lens eyeglasses, would be helpful here. I understand the
diopter is simply the reciporical of the single lenses focal length.
http://en.wikipedia.org/wiki/Diopter

Warm regards in an overly-heated thread -

Canopus56

P.S. - I'm no optics expert like Chris and the other posters in this
thread.

.

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