A) \[\operatorname{f}'\,\,=\,\,f,\,\,f''=\,\,2f\]
B) \[\operatorname{f}'\,\,=\,\,2f,\,\,f''=\,\,f\]
C) \[\operatorname{f}'\,\,=\,\,f,\,\,f''=\,\,f\]
D) \[\operatorname{f}'\,\,=\,\,2f,\,\,f''=\,\,2f\]
Correct Answer: A
Solution :
Since the lens is equiconvex, the radius of curvature of each half is same, say R according to lens maker?s formula \[\frac{1}{f}=(\mu -1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] (if lens is placed in air) Here, \[{{\operatorname{R}}_{1}}=R,\,\,{{R}_{2}}=-\,R\] (by sign convention) \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{f}=(\mu -1)\,\frac{2}{R}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,(\mu -1)\frac{1}{R}=\frac{1}{2f}\] ... (i) If we cut the lens along XOX?, then two halves of the lens will be having the same radii of curvatures and so, \[\operatorname{Focal} length\,\,f' = \,f\] But when we cut it along YOY? then, we will have \[{{R}_{1}}=R\,\,but\,\,{{R}_{2}}=\infty \] \[\therefore \,\,\,\,\,\frac{1}{f''}=(\mu -1)\left( \frac{1}{R}-\frac{1}{\infty } \right)=(\mu -1)\frac{1}{R}=\frac{1}{2f}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\operatorname{f}'' = 2f\]
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