An astronomical refracting telescope is being used by an observer to observe planets in normal adjustment.…

An astronomical refracting telescope is being used by an observer to observe planets in normal adjustment. The focal lengths of the objective and eye piece used in the construction of the telescope are $20 \mathrm{~m}$ and $2 \mathrm{~cm}$ respectively. Consider the following statements about the telescope: (a) The distance between the objective and eye piece is $20.02 \mathrm{~m}$ (b) The magnification of the telescope is $(-) 1000$ (c) The image of the planet is erect and diminished (d) The aperture of eye piece is smaller than that of objectie The correct statements are:
  1. (a), (b) and (c)
  2. (b), (c) and (d)
  3. (c), (d) and (a)
  4. (a), (b) and (d)

Solution

Given,
$\begin{aligned}
& f_e=2 \mathrm{~cm}, \text { and } \\
& f_o=20 \mathrm{~m}=2000 \mathrm{~cm}
\end{aligned}$
For normal adjustment,
$\begin{aligned}
\text {Magnification power } & =\frac{-f o}{f e}=\frac{-2000}{2} \\
& =-1000 \\
\text {Length of telescope } & =f o+f e \\
& =2000+2 \\
& =2002 \mathrm{~cm} \\
& =20.02 \mathrm{~m}
\end{aligned}$
The image formed is inverted and magnified, and the aperture of objective is smaller than eye piece of the telescope.
$\left(f_o~ \& ~f_e\right.$ are focal lengths of objective & eye piece respectively.)

Asked in: NEET 2022 (Phase 2)

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