A) \[\sqrt{2}\]
B) \[\frac{4}{3}\]
C) \[\sqrt{3}\]
D) \[\frac{3}{2}\]
Correct Answer: B
Solution :
\[\frac{1}{{{f}_{1}}}=\frac{1}{2}\times \frac{2}{18}=\frac{1}{18}\] \[\frac{1}{{{f}_{2}}}=\frac{({{\mu }_{1}}-1)}{-18}\] when\[{{\mu }_{1}}\]is filled between lens and mirror \[P=\frac{2}{18}-\frac{2}{18}({{\mu }_{1}}-1)=\frac{2-2{{\mu }_{1}}+2}{18}\] \[={{F}_{m}}=-\left( \frac{18}{2-{{\mu }_{1}}} \right)\] \[2=6-3{{\mu }_{1}}\] \[3{{\mu }_{1}}=4\] \[{{\mu }_{1}}=4/3\].
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