Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Can someone help me with this, In a compound microscope the focal length of objective lens is 1.2 cm and focal length of eye piece is 3.0 cm. When object is kept at 1.25 cm in front of objective, final image is formed at infinity. Magnifying power of

 In a compound microscope the focal length of objective lens is 1.2 cm and focal length of eye piece is 3.0 cm. When object is kept at 1.25 cm in front of objective, final image is formed at infinity.  Magnifying power of the compound microscope should be :

  • Option 1)

    200

  • Option 2)

    100

  • Option 3)

    400

  • Option 4)

    150

 

Answers (1)

best_answer

As we have learned

Compound Microscope -

m= -\frac{v_{o}}{u_{o}}\cdot \frac{D}{u_{e}}
 

- wherein

v_{o}\, and \, u_{o} is distance from objective.

u_{e} Distance from eyepiece.

Maximum magnification m= -\frac{v_{o}}{u_{0}}\left ( 1+\frac{D}{f_{e}} \right )

 

 f_e = 3cm \\ f_0 = 1.2 cm \\ u_0 = 1.25 cm , v_e = \infty \\ \Rightarrow \frac{1}{V_0 }= \frac{1}{f_0}+ \frac{1}{u_0}= \frac{1}{1.2}-\frac{1}{1.25}

 

Magnification = \frac{v_0}{u_0}\cdot \frac{D}{f_e}= \frac{30}{1.25}\times \frac{25}{3}= 200 

 

 

 

 


Option 1)

200

Option 2)

100

Option 3)

400

Option 4)

150

Posted by

Avinash

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 Result 2024 Live: JEE April results at jeemain.nta.ac.in likely today; cutoff, rank predictor, update
anam-khan

JEE Main 2024 session 2 final answer key out at jeemain.nta.ac.in; steps to check probable scores
anam-khan

JEE Main Result 2024 Live: Download session 2 final answer key at jeemain.nta.ac.in; rank predictor, cutoff

Latest Question