An equiconvex lens is cut into two halves along `(i) XOX^(')` and `(ii) YOY^(')` as shown in the figure. Let `f,f^(')f^('')` 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

Options

A. `f' = f,f'' = f`
B. `f' = 2f,f'' = 2f`
C. `f' = f, f'' = 2f`
D. `f' = 2f,f'' = f`

Answer

C

Explanation

Initial, the foca length length of equiconvex lens is
`(1)/(f)=(mu-1)((1)/(R_(1))-(1)/(R_(2)))`
`(1)/(f)=(mu-1)((1)/(R)-(1)/(-R))=(2(mu-1))/(R)`
Case I : When lens is cut along `XOX'` then each half is again equiconvex with
`R_(1)=+R,R_(2)=-R`
Thus, `(1)/(f)=(mu-1)[(1)/(R)-(1)/((-R))]`
`= (mu-1)[(1)/(R)+(1)/(R)]`
`=(mu-1)(2)/(R)=(1)/(f) rArr f'=f`
Case II : When lens is cut along `YOY'` then each half becomes plano-convex with
`(R_(1)=R_(1)R_(2)=infty`
Thus `(1)/(f'')=(mu-1)((1)/(R_(1))-(1)/(R_(2)))`
`= (mu-1)((1)/(R)-(1)/(infty))=((mu-1))/(R)=(1)/(2f)`
Hence `f'=f,f'' =2f`.