derive an expression for the magnifying power of an astronomical telescope in normal adjustment?

Asked by shijurlytepz | 12th Oct, 2021, 06:56: PM

Expert Answer:

Astronomical telescope that is used to provide angular magnification of distanr object is shown in figure.
It has an objective lens of long focal length and greater diameter.
Diameter of eye piece and focal length are smaller than that of objective lens.
Light from a distant object enters the objective and a real image is formed
in the tube at its second focal point. The eyepiece magnifies this image
producing a final inverted image. Distance between objective and eyepiece that is called as tube length is
sum of focal lengths of objective and eye-piece length so that inside the tube focal points of both the lenses coincide.
 
The magnifying power m is the ratio of the angle β subtended at the eye by image to the angle α
that is subtended by object at the lens.

m = ( β / α )
 
β ≈ tanβ = h / fe 
 
where h is image height and fe is focal length of eye piece
α ≈ tanα = h / fo
where fo is foacl length of objective lens
 
hence , we get  magnifying power m as
 
begin mathsize 14px style m space equals space fraction numerator h divided by f subscript e over denominator h divided by space f subscript o end fraction space equals space f subscript o over f subscript e end style

Answered by Thiyagarajan K | 12th Oct, 2021, 10:21: PM