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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