Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Medical
  • An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The separation between the objective and the eye piece is <img alt="36 \mathrm{~cm}" src="https:

An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The separation between the objective and the eye piece is 36 \mathrm{~cm} and the final image is formed at infinity. The focal length \mathrm{f_{0}} of the objective and the length \mathrm{f_{e}} of the eye piece are   

Option: 1

f_{0}=45 \mathrm{~cm}$ and $f_{e}=-9 \mathrm{~cm}


Option: 2

f_{0}=50 \mathrm{~cm}$ and $f_{e}=10 \mathrm{~cm}


Option: 3

f_{0}=7.2 \mathrm{~cm}$ and $f_{e}=5 \mathrm{~cm}


Option: 4

f_{0}=30 \mathrm{~cm}$ and $f_{e}=6 \mathrm{~cm}


Answers (1)

best_answer

 We know that, \mathrm{f}_{0}+\mathrm{f}_{\mathrm{e}}=36 \quad \ldots(1)

Where \mathrm{\ell=} length of the tube \mathrm{=36 \mathrm{~cm}}
\mathrm{\frac{\mathrm{f}_{0}}{\mathrm{f}_{\mathrm{e}}}=\mathrm{m}=5}\quad \ldots(2)

Where \mathrm{\mathrm{m}=} magnification \mathrm{=5}

Solving (1) and (2) we obtain,

\mathrm{5 f_{e}+f_{e}=36}

\Rightarrow \quad f_{e}=6 \mathrm{~cm}$ and $f_{0}=5 f_{e}=30 \mathrm{~cm}

\therefore \left ( D \right )

Posted by

Ritika Harsh

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

UP NEET UG Counselling 2025: BDS fee structure out for non-minority private dental colleges
anam-khan

NEET admissions, exit tests, registration: NCAHP chief on allied and healthcare sciences’ future
anam-khan

KEA extends Karnataka NEET UG 2025 registration deadline to July 17 for MBBS, BDS, AYUSH admissions

Latest Question