MCQ
Magnification produced by astronomical telescope for normal adjustment is $10$ and length of telescope is $1.1\, m$. The magnification when the image is formed at least distance of distinct vision $(D = 25\, cm)$ is
- ✓$14$
- B$6$
- C$16$
- D$18$
here $f_{\circ}=100 \mathrm{\,cm}$ and $f_{\mathrm{e}}=10 \mathrm{\,cm}$
Final magnification $=f_{\circ}\left(\frac{1}{D}+\frac{1}{f_{e}}\right)=14$
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