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\[=\frac{4}{2}=2\,D\]You need to login to perform this action.
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