Q. 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
Rajasthan PMTRajasthan PMT 2008Ray Optics and Optical Instruments
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 $
