Re: Field of View and Magnfication calculations

On Apr 20, 8:07 pm, wsnel...@xxxxxxxxxxx wrote:
On Apr 19, 8:52 pm, Eugene <eugenhug...@xxxxxxxxx> wrote:





On Apr 20, 8:21 am, Dave Typinski <möb...@xxxxxxxxxxxxx> wrote:

Eugene <eugenhug...@xxxxxxxxx> wrote:

What has the thin lens formula got to do with it. I'm
inquiry about the formula for small angle where

theta subtended by object = arctan (object size/distance)

then apparent field of eyepiece like TV Plossl 50 degrees
divided by the theta is equal to the magnification.

Does this magnification calculations based on the theta
and Plossl field accurate when the object is 4 meters
away, 8 meters away, at infinity?

The calculation is accurate everywhere because for any given object of
interest--say, a meter stick--theta will change depending on the
object's distance from the observer.
--
Dave

But in using a compound system with primary and secondary
mirrors. Sometimes the focal length changes because you
are adjusting the rear visual back distance from the
secondary to focus close. Or is the calculation entirely
independent of the focal length of the system??

The focal length of a compound telescope will change if you change the
distance between the primary and secondary in order to focus the
scope.  Usually the primary mirror is what moves but some small Maks
and large RCs move the secondary in order to focus.  For the purposes
of this exercise only the eyepiece should be moved in order to focus.

The focal length of the system is defined by the image distance for an
object at infinity, in which case:

1/infinity + 1/image distance = 1/focal length

where we see that image distance and focal length are now equal.- Hide quoted text -

- Show quoted text -

I know. If one change the distance between eyepiece
focal plane and secondary by moving the primary,
the focal length of the system changes. No problem
with that. But in determining the magnification, can
one simply uses the following method:

theta subtended by object/2 = arctan (1/2 object size/distance)

then apparent field of eyepiece like TV Plossl 50 degrees
divided by the theta is equal to the magnification.

Now the main question is, would the value of the above
solutions equals the actual system focal length divided by
the eyepiece focal length?

E
.

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