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 \[F=40\,cm\]