A) 72
B) 60
C) 36
D) 20
Correct Answer: D
Solution :
For the first condition \[\frac{1}{{{f}_{1}}}=\frac{1}{v}-\frac{1}{u}\Rightarrow \frac{1}{10}=\frac{1}{v}+\frac{1}{30}\] \[\frac{1}{v}=\frac{1}{10}-\frac{1}{30}\] \[\Rightarrow \] \[v=\frac{30}{2}=15\,cm\] where \[{{f}_{1}}=10\,cm,\,u=-30\,cm\] For the second condition when concave lens is placed \[v=(15+45)cm=60\,cm\] \[\frac{1}{F}=\frac{1}{v}-\frac{1}{u}\] (where \[F=\]focal length of combination) \[\therefore \] \[\frac{1}{F}=\frac{1}{60}+\frac{1}{30}\] \[F=\frac{60}{3}cm=20cm\] The magnitude of focal length of concave lens \[\frac{1}{F}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}\Rightarrow \frac{1}{20}=\frac{1}{10}+\frac{1}{{{f}_{2}}}\Rightarrow {{f}_{2}}=-20\,cm\](Negative sign for concave lens)You need to login to perform this action.
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