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, - A planolens of refractive indexand focal length is kept in contact with another planoconcave le - Optics - JEE Main

A plano lens of refractive index \mu _{1} and focal length f _{1} is kept in contact with another plano concave lens of refractive index \mu _{2} and focal length f _{2}. If the radius of curvature of their spherical faces is R each and f _{1} =2f_{2}, then \mu _{1} and \mu _{2} are related as: 

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

Answers (1)

best_answer

 

Lensmaker's Formula -

\frac{1}{f}= \left ( \frac{\mu _{2}}{\mu _{1}}-1 \right )\left ( \frac{1}{R_{1}}- \frac{1}{R_{2}}\right )
 

- wherein

\mu _{1}= refractive index of medium of object

\mu _{2}= refractive index of lens

R_{1}\, and \, R_{2} are radius of curvature of two surface

for lens 1

we have \frac{1}{f_{1}}=\left ( \mu _{1}-1 \right )\left ( \frac{1}{\infty }-\frac{1}{-R} \right )------(1)\\\\

for lens 2

\frac{1}{f_{2}}=\left ( \mu _{2}-1 \right )\left ( \frac{1}{\infty }-\frac{1}{-R} \right )------(2)\\\\

 

and f_{1}=2f_{2}------(3)

from (1),(2) and (3)

we get

\frac{1}{f_{1}}=\frac{1}{2f_{2}}\\\\\left ( \mu _{1}-1 \right )\left ( \frac{-1}{-R} \right )=\left ( \mu _{2}-1 \right )\frac{-1}{-R}\times \frac{1}{2}\\\\\\\Rightarrow \frac{\mu _{1}-1}{R}=\frac{\mu _{2}-1}{2R}\\\\\Rightarrow 2\mu _{1}-2=\mu _{2}-1\\\\\\2\mu _{1}-\mu _{2}=1

 

 


Option 1)

Option 2)

Option 3)

Option 4)

Posted by

admin

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 2026 session 2 application correction begins on jeemain.nta.nic.in; edit details by tomorrow
anam-khan

JEE Main 2026 Registration LIVE: Session 2 exam from April 2 to 9; shift timings, admit card, exam pattern
anam-khan

JEE Main 2026 session 2 registration ends today for BTech, BArch; apply at jeemain.nta.nic.in

Latest Question