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
    $18$

Answer

Correct option: A.
$14$
a
$f_{0} / f_{e}=10, \quad f_{\circ}+f_{e}=1.1$

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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