Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A compound microscope has an eye piece of focal length and an objective of focal length

A compound microscope has an eye piece of focal length \mathrm{10\ cm} and an objective of focal length \mathrm{4\ cm}. Calculate the magnification, if an object is kept at a distance of \mathrm{5\ cm} from the objective, so that the final image is formed at the least distance of distinct vision \mathrm{25\ cm}:

Option: 1

12


Option: 2

11


Option: 3

13


Option: 4

14


Answers (1)

best_answer

\mathrm{\text { Here, } \mathrm{f}_0=4 \mathrm{~cm}, \mathrm{f}_{\mathrm{e}}=10 \mathrm{~cm}, \mathrm{u}_0=-5 \mathrm{~cm}}
For objective
\mathrm{\begin{aligned} & \mathrm{\frac{1}{\mathrm{v}_0}-\frac{1}{-5}=\frac{1}{4}} \\ & \frac{1}{\mathrm{v}_0}=\frac{1}{4}-\frac{1}{5}=\frac{1}{20} \quad \text { or } \quad \mathrm{v}_0=20 \mathrm{~cm} \end{aligned}}
Magnification when the final image is formed at the least distance of distinct vision \mathrm{D(=25\ cm)} is
\mathrm{\mathrm{M}=\frac{\mathrm{v}_0}{\left|\mathrm{u}_0\right|}\left(1+\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{c}}}\right)=\frac{20}{5}\left(1+\frac{25}{10}\right)=14}

Posted by

chirag

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2024: NTA debars 39 candidates for 3 years on account of using unfair means
anam-khan

JEE Main 2024 session 2 paper 2 answer key objection window closes today; fees
anam-khan

JEE Main 2024 Topper: Farmer’s son Gajare Nilkrishna says ‘keep questioning until you understand’

Latest Question