CBSE 2024 · Region 5 · Set 1 · Q19 · 2 marks
A telescope has an objective lens of focal length $\displaystyle 150$ cm and an eyepiece of focal length $\displaystyle 5$ cm . Calculate its magnifying power in normal adjustment and the distance of the image formed by the objective.
Marking-scheme solution
$\displaystyle m=\frac{f_{0}}{f_{e}}$
$\displaystyle =\frac{150}{5}=30$
$\displaystyle \frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\displaystyle \frac{1}{150}=\frac{1}{v}-\frac{1}{\infty}$
$\displaystyle v=150 \mathrm{~cm}$
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CBSE Class 12 Physics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.