Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • What is the position and nature of image formed by lens combination shown in figure ? (f1, f2 are focal lengths ) (1) 40 cm from poin

What is the position and nature of the image formed by lens combination shown in the figure? (f1, f2 are focal lengths )

                                                 

Answers (1)

Thin lens formula -

\frac{1}{v}-\frac{1}{u}= \frac{1}{f}

- wherein

u \, and \, v are object and image distance from the lens.

For the first lens

u_{1}=-20,\, \, \, f_{1}=5cm

so, \frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f}

\frac{1}{v}-\frac{1}{-\left ( 20 \right )}=\frac{1}{5}

v_{1}=\frac{20}{3}

So  u_{2}  for 2nd  lens is given by

So  u_{2}=\frac{20}{3}-2=\frac{14}{3}

f_{2}=-5cm

So  \frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}

\frac{1}{v_{2}}-\frac{1}{\left ( 1_{\frac{4}{3}} \right )}=\frac{1}{-5}

V_{2}=70cm   and the image is real.

Posted by

lovekush

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 session 2 result date announced; Steps to check NTA session 2 scores
anam-khan

JEE Main 2024 session 2 final answer key soon at jeemain.nta.ac.in; result on April 25
anam-khan

JEE Main 2024 Session 2: Last day to challenge answer key today

Latest Question