Re: Apparent Distance

Ioannis wrote:

Ο <anteperkovic@xxxxxx> έγραψε στο μήνυμα
news:1114779645.671559.5180@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

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


For the Moon, a = 1/2 degree, and OB ~ 380,000 km.
gives a maximum magnification of 180x. However, it is well known that

Why the apparent discrepancy? 

Because the telescopes do not really "decreased distance". Instead, the
magnify.


I am aware of that. *However*, with magnification there is associated the
notion of "decreased distance". There is a distance for which, if you moved
the object THERE, the object would subtend to the same angle as the one seen
through a telescope with magnification M. So magnification produces a
*virtual* decrease of distance, which is what I am attempting to define
mathematically.

The problem stems from the fact that OA != OB when AB becomes significant compared to the lengths OB and OA. The small angle approximation still holds for the magnification of fine detail very close to B. But it does not hold at all when OA ~ AB.

sin(x) = tan(x) = x  for x << 1

But when x ~ 1 all bets are off!

What is the difference?

Well, the sky seen throught the telescope has circumference not just
360 degrees but M*360 degrees, so objects _can_ have M*a higher then
Pi/2.

I have no idea what you mean with the above. 

If anyone has another explanation, please backup your assertions with math,
to minimize confusion.

You are applying the small angle approximation to very large angles. There is no one value of magnification m that accurately describes the effect of moving a significantly extended object closer by a factor of two. Try it with a newspaper and you will see what I mean.

Regards,
Martin Brown
.

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