• # question_answer 43) 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

Biconvex lens is cut perpendicularly to the principal axis, it will become a plano-convex lens.                      Focal length of biconvex lens            $\frac{1}{f}=(n-1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)$ $\frac{1}{f}=(n-1)\frac{2}{R}$ $(\because \,{{R}_{1}}=R,{{R}_{2}}=-R)$ $\Rightarrow$ $f=\frac{R}{2(n-1)}$ 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 $P\propto \frac{\text{1}}{\text{focal}\,\text{length}}$ Hence, power will become half. New power $=\frac{4}{2}=2\,D$