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Asked in GSEB Board March 2022 · Telescopes
Given: a telescope with objective focal length fₒ and eyepiece focal length fₑ, in normal adjustment.
Idea: magnifying power compares the angle the final image subtends at the eye with the angle the object itself subtends.
A distant object subtending an angle β is imaged by the objective in its focal plane, with size h = fₒ β.
The eyepiece is used as a magnifier on that image with the final image at infinity, so the image subtends α = h/(fₑ) at the eye.
m = α/β = (h/fₑ)/(h/fₒ) = (fₒ)/(fₑ)
So a long-focus objective with a short-focus eyepiece gives the greatest magnification.
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