question_answer

Focal length of an equiconvex lens is\[20cm\]. If we cut it once perpendicular to principal exis in and then along principal axis. Then, focal length of each part will be:

A)  \[20cm\]                     

B)  \[10cm\]

C)  \[40cm\]            

D)  \[5cm\]

Correct Answer: C

Solution :

    For fig.   (i) \[\frac{1}{20}=(\mu -1)\left( \frac{1}{R}-\frac{1}{R} \right)\]                  ?..(i) For fig. (ii) \[\frac{1}{F}=(\mu -1)\,\left( \frac{1}{R}-\frac{1}{\infty } \right)\]                           ?..(ii) From the equations (i) and (ii) we get                         \[F=40\,cm\]