Question

An equiconvex lens of glass of focal , length $0.1m$ is cut along a plane perpendicular to principle axis in to two equal parts .The ratio of focal length of new lenses formed is

• A. $1:1$
• B. $1:2$
• C. $2:1$
• D. $2:\frac{1}{2}$

Answer 1

The correct option is A $1:1$

Focal length of lens
$\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2}$
After cutting lens along a plane perpendicular to principal axis ,we get two plano-convex lens. For then $R_1=R,R_2=\infty$
$\therefore$ New focal length ,
$\frac{1}{f_1}=(\mu -1)(\frac{1}{R_1}-\frac{1}{\infty })$
$\frac{1}{f_1}=\frac{(\mu -1)}{R}$.........................(i)
For second part
$\frac{1}{f_2}=(\mu -1)(\frac{1}{\infty }+\frac{1}{R})$
$\frac{1}{f_2}=\frac{(\mu -1)}{R}$........................(ii)
From Eqs.(i) and (ii)
$f_1:f_2=1:1$