Apparent Distance

The definition of a telescope's magnification M is standard. Viewing an
object that subtends a *finite* angle to the naked eye (so we are not
talking about star images) with magnification M, creates the illusion of
"decreased distance". Let's call this "apparent distance" under
magnification M.

Here are some of my thoughts on it:
The Magnification M of a telescope is defined as:
b/a, where b is the angle the object subtends to when seen through the
instrument, and a is the angle the object subtends to when seen with the
naked eye.


Following this, define "Apparent Distance" of the object to be the
"perceived distance" by the eye, when the object is seen through such an
instrument at magnification M.


Doing some simple geometry, where the object seen with the naked eye is
AB and the perception through the instrument is A'B':

http://users.forthnet.gr/ath/jgal/astronomy/figs/magnification.GIF

(Observer is situated at O)


If a is the angle the object subtends to when seen naked eye (picture is
symmetric so it's chopped in half), M is the telescope magnification, and
OB is the object's actual distance, the "apparent distance" will be OB', so
then


tan(a) = AB/OB (1)
tan(M*a) = A'B'/OB' (2)


[a < M*a <= A'OB']


Since AB = A'B', it follows from (1) and (2) that:


tan(a)*OB = tan(M*a)*OB', which solved for OB', gives:


OB' = tan(a)*OB/tan(M*a) (3)


Therefore, knowing the object's angle a, the magnification M and its
true distance, the "apparent distance" the object sits at when viewed with
said instrument is given by (3).


A couple of observations:


>From the geometrical picture, clearly:


0 < a < M*a <= Pi/2. (4)


(The object cannot be at an "apparent distance" smaller than 0 to the
observer under any circumstances and it is, say of finite size, so a = 0
only if OB = +oo).

The above though constraints (3) badly. For example:


For the Moon, a = 1/2 degree, and OB ~ 380,000 km.


After doing the appropriate conversions and plugging in the numbers, (4)
gives a maximum magnification of 180x. However, it is well known that one
can use magnifications in excess of x180 for the Moon for example.


Why the apparent discrepancy? Does anyone have any ideas about how this
magnitude can be calculated practically or if my calculations are way off?
For example, for finite sized objects, such as Jupiter or M31, what would be
the "apparent distance" when the corresponding object is viewed at
magnification M?

For reference, here's the sci.math url, where some of these ideas are
discussed.

http://tinyurl.com/bhypq

Thanks much in advance
--
I. N. Galidakis
http://users.forthnet.gr/ath/jgal/
Eventually, _everything_ is understandable

.

Relevant Pages

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