A) \[{{f}_{1}}=18\,\,cm,\,\,{{f}_{2}}=10\,\,cm\]
B) \[{{f}_{1}}=20\,\,cm,\,\,{{f}_{2}}=28\,\,cm\]
C) \[{{f}_{1}}=20\,\,cm,\,\,{{f}_{2}}=18\,\,cm\]
D) \[{{f}_{1}}=24\,\,cm,\,\,{{f}_{2}}=18\,\,cm\]
Correct Answer: C
Solution :
Since, the doublet is corrected for spherical aberration, it satisfies the following condition. \[{{f}_{1}}-{{f}_{2}}=d=2\,\,cm\] \[\Rightarrow \] \[{{f}_{1}}={{f}_{2}}+2\,\,cm\] Let the equivalent focal length\[=F\] \[\therefore \] \[F=\frac{{{f}_{1}}{{f}_{2}}}{{{f}_{1}}+{{f}_{2}}-d}=10\,\,cm\] Solving it\[,\] \[{{f}_{1}}=20\,\,cm,\,\,{{f}_{2}}=18\,\,cm\].
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