Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

A biconcave lens is made of glass with a refractive index of 1.5 and has radii of curvature of 20 cm and 30 cm. If the 20 cm surface is silvered then the effective focal length of the mirror formed is:

Option: 1

\frac{-60}{11} cm

 


Option: 2

\frac{-60}{9}cm


Option: 3

\frac{-50}{11}cm


Option: 4

\frac{-50}{9}cm


Answers (1)

best_answer

For silvered lens -     \frac{1}{f_{e}}= \frac{1}{f_{m}} - \frac{2}{f_{l}}
 

 

Therefore, focal length of lens      \frac{1}{f_{e}}=\left ( 1.5-1 \right ) \left ( \frac{1}{30}-\frac{1}{-20} \right )  or    {f_{e}}=24cm

Focal length of the mirror,   {f_{m}}=\frac{R_{2}}{2}=\frac{-20}{2}=-10cm

Effective focal length of silverd lens is,   \frac{1}{f_{e}}=\frac{1}{f_{m}}-\frac{2}{f_{e}}  \Rightarrow f_{e}=\frac{-60}{11}cm

The silvered convex lens behaves as a concave mirror of focal length \frac{60}{11} cm.

Posted by

Gaurav

View full answer

Similar Questions