Let f0 be the focal length of objective and fe is the focal length of eyepiece.
Given here, f0+fe=30.
Angular magnification of telescope is given by m=fef0.
So, 2=fef0⇒f0=2fe
Then, we have f0+2f0=30⇒23f0=30
Or, f0=20cm.
JEE Main 2022 — Physics Optics
In normal adjustment, for a refracting telescope, the distance between objective and eye piece is 30cm. The focal length of the objective, when the angular magnification of the telescope is 2, will be:
Held on 28 Jul 2022 · Verified 6 Jul 2026.
20cm
30cm
10cm
15cm
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