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JEE Mains · Physics · STD 12 - Ray optics and optical instruments
  • JEE Mains 2014
  • Medium
  • MCQ

The focal lengths of objective lens and eye lens of a Galilean telescope are respectively \(30\, cm\) and \(3.0\, cm\). telescope produces virtual, erect image of an object situated far away from it at least distance of distinct vision from the eye lens. In this condition, the magnifying power of the Galilean telescope should be

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Answer & Solution

Correct Answer

(D) \(+8.8\)

Step-by-step Solution

Detailed explanation

Given, Focal length of objective, \(f_{0}=30\, \mathrm{cm}\)

focal length of eye lens, \(f_{e}=3.0 \,\mathrm{cm}\)

Magnifying power, \(M=?\)

Magnifying power of the Galilean telescope,

\(M_{D}=\frac{f_{0}}{f_{e}}\left(1-\frac{f_{e}}{D}\right)\)

\( = \frac{{30}}{3}\left( {1 - \frac{3}{{25}}} \right)\)       \([\because D = 25\,{\text{cm}}]\)

\(=10 \times \frac{22}{25}=8.8\, \mathrm{cm}\)

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