question_answer

A symmetric double convex lens is cut in two equal parts by a plane perpendicular to the principal axis. If the power of the original lens is 4D, the power of a cut lens will be

A)  2D                                         

B)  3D                         

C)  4D                         

D)         5D

Correct Answer: A

Solution :

Biconvex lens is cut perpendicularly to the principal axis, it will become a plano-convex lens.             Focal length of biconvex lens \[\frac{1}{f}=(n-1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\]     \[\frac{1}{f}=(n-1)\frac{2}{R}\]                              \[(\because \,{{R}_{1}}=R,\,{{R}_{2}}=-R)\] \[\Rightarrow \]               \[f=\frac{R}{2\,(n-1)}\]                                  ?(i) For plano-convex lens \[\frac{1}{{{f}_{1}}}=(n-1)\,\left( \frac{1}{R}-\frac{1}{\infty } \right)\] \[{{f}_{1}}=\frac{R}{(n-1)}\]                                        ?(ii) Comparing Eqs. (i) and (ii), we see that focal length becomes double. As power of lens \[P\propto \frac{1}{\text{focal}\,\text{length}}\] Hence, power will become half. New power\[=\frac{4}{2}=2\,D\]