Question:
A small telescope has an objective lens of focal length 140cm and an eyepiece of focal length 5.0cm. What is the magnifying power of the telescope for viewing distant objects when (a) the telescope is in normal adjustment (i.e., when the final image is at infinity)? (b) the final image is formed at the least distance of distinct vision (25cm)?
Answer:
Given that,
f0 = 140 cm, fl = 5 cm.
Magnifying power = ?
a) In normal adjustment,
magnifying power = f0 / - fl
= 140/-5
= -28 is the magnification
b) When final image is formed at least distance of distinct vision then,
maginification = (-f0 / fl) [1 + fl/D]
= -140/5 [1 + 5/25]
= -33.6 is the magnification.
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