A) \[+10D\]
B) \[+20D\]
C) \[-5D\]
D) \[-30D\]
Correct Answer: A
Solution :
We know, \[\frac{1}{f}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}\] where, \[f=\frac{1}{5}m=20\,\,cm\] \[\therefore \] \[\frac{1}{20}=\frac{1}{10}+\frac{1}{{{f}_{2}}}\] \[\Rightarrow \] \[{{f}_{2}}=20\,\,cm\] When lenses are separated, then \[\frac{1}{f}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}-\frac{d}{{{f}_{1}}{{f}_{2}}}\] \[=\frac{1}{10}-\frac{1}{20}=\frac{10}{10\times (-20)}\] \[\Rightarrow \] \[f=10\,\,cm=0.1\,\,m\] \[\therefore \] \[p=10\]diopter.
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