Re: Field of View and Magnfication calculations

On Apr 18, 3:05 pm, Eugene <eugenhug...@xxxxxxxxx> wrote:
On Apr 18, 12:30 pm, wsnel...@xxxxxxxxxxx wrote:



On Apr 17, 12:51 pm, Dave Typinski <möb...@xxxxxxxxxxxxx> wrote:

wsnel...@xxxxxxxxxxx wrote:

In this example we are only concerned with what the front surface of
the objective sees compared with what the observer were to see if he
looked at the ruler from the location of the objective. The objective
could be a Newtonian primary mirror which would be _more_ distant from
the object than is the eyepiece.

Aha, good point.

But still, it's not clear to me why the only distance that counts is
the one between the object and the objective lens (or primary mirror)..

I guess I'm going to have to go learn about it by actually reading a
textbook and doing some exercises.

...the horror, the horror...

It is possible to directly measure the object distance, but the focal
length and image distance must often be calculated or determined via
indirect means. If you treat the telescope as a "black box" or a
"thin lens" the calculations become easier.- Hide quoted text -

- Show quoted text -

Hi wsnel,

In getting the angle subtended by the object, is it
dependent or independent from the system
configuration where the target could be a few
meters away and the eyepiece pulled back or
target at infinity, does any of this affect the
accuracy of the angle subtended calculations
or totally independent, meaning, as long as
one can get the angel subtended and know
the eyepiece degrees like 15mm TV Plossl,
one can always get the magnification by
dividing 50 by the angle subtended irregardless
of the nature of the system whether an object
is focused a few meters away or at infinity?

Distortion in the eyepiece and the objective can affect the results,
but for small angles the thin lens formula would still be
approximately correct.


.

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