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\[P\propto \frac{1}{focal\,\,length}\] Hence, power will become half. New power\[=\frac{4}{2}=2D\]