Question

An equiconvex lens is cut into two halves a long (i) XOX and (ii) YOY as shown in the figure let $f.f^{\prime },f^{\prime \prime }$ be the focal lengths of the complete lens of each half in case (i) and of each half in case (ii) respectively
Choose the correct statement from the following :

• A. $f^{\prime }=f,f"=f$
• B. $f^{\prime }=2f,f"=2f$
• C. $f^{\prime }=f,f"=2f$
• D. $f^{\prime }=2f,f"=f$

Answer 1

The correct option is C $f^{\prime }=f,f"=2f$

Initially the focal length of equicontex lens is

$\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$ ...(i)

$\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{-R})=\frac{2(\mu -1)}{R}$

Case I

When lens is cut along XOX, then each half id again equiconvex with

$R_1=+R,R_2=-R$

Thus, $\frac{1}{f}=(\mu -1)[\frac{1}{R}-\frac{1}{-\left(R\right)}]=(\mu -1)\frac{2}{R}=\frac{1}{f^{\prime }}$

Case II

When lens is cut along XOY, then each half becomes plano-convex with

$R_1=R,R_2=\infty$

Thus $\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$

$=\left(\frac{\mu -1}{R}\right)=\frac{1}{2f}$

Hence $f^{\prime }=f,f"=2f$