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Ray Optics and Optical Instruments
Astronomical Telescope
4 questions from 4 papers, each with its answer.
- questions
- 4
- papers
- 4
- years asked, 2017–2026
- 2 of 10
- of Ray Optics and Optical Instruments
- 4.49%
Appears in
Concepts tested
- Astronomical telescope
- Image of objective formed by eyepiece
- Magnifying power of a telescope for a distant object
- Normal adjustment
All Questions
2024 1 question
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A small telescope has an objective of focal length 140 cm and an eye piece of focal length 5.0 cm. The magnifying power of telescope for viewing a distant object is:Magnifying power of a telescope (normal adjustment) $m = \frac{f_o}{f_e}$ $m = \frac{140}{5} = 28$
2022 Phase 2 1 question
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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 m and 2 cm respectively. Consider the following statements about the telescope: (a) The distance between the objective and eyepiece is 20.02 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 objective The correct statements are:In normal adjustment the final image is at infinity and is inverted. $m=-\frac{f_o}{f_e}=-\frac{20}{2\times10^{-2}}=-1000$. Tube length $L=f_o+f_e=20+0.02=20.02$ m. The aperture of the eyepiece is smaller than that of the objective. So (a), (b) and (d) are correct; (c) is wrong since the image is inverted.
2015 AIPMT-II 1 question
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In an astronomical telescope in normal adjustment a straight black line of length L is drawn on inside part of objective lens. The eye-piece forms a real image of this line. The length of this image is I. The magnification of the telescope is:
In normal adjustment the objective is at distance $u = -(f_0 + f_e)$ from the eyepiece.
Magnification by the eyepiece: $\frac{I}{L} = \frac{f_e}{f_e + u} = \frac{f_e}{f_e - (f_0 + f_e)} \Rightarrow \frac{I}{L} = \frac{f_e}{f_0}$ (in magnitude)
$M = \frac{f_0}{f_e} = \frac{L}{I}$
2012 AIPMT-PRE 1 question
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The magnifying power of a telescope is 9. When it is adjusted for parallel rays the distance between the objective and eyepiece is 20 cm. The focal lengths of lenses areIn normal adjustment: $f_o + f_e = 20$ cm ...(i)
Magnifying power: $\frac{f_o}{f_e} = 9 \Rightarrow f_o = 9f_e$ ...(ii)
From (i) and (ii): $10f_e = 20 \Rightarrow f_e = 2$ cm
$f_o = 18$ cm
