A) \[{{f}_{2}}>f\]and\[{{f}_{1}}\] becomes negative
B) \[{{f}_{1}}\]and\[{{f}_{2}}\] both become negative
C) \[{{f}_{1}}={{f}_{2}}<f\]
D) \[{{f}_{1}}>f\]and\[{{f}_{2}}\]becomes negative
Correct Answer: D
Solution :
\[\frac{1}{f}=\left( \frac{3}{2}-1 \right)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\]\[\frac{1}{f}=\frac{1}{2}\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] ?.(1) \[\frac{1}{{{f}_{1}}}=\left( \frac{\frac{3}{2}}{\frac{4}{3}}-1 \right)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\]\[\frac{1}{{{f}_{1}}}=\frac{1}{8}\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\]?.(2) \[\frac{1}{{{f}_{2}}}=\left( \frac{\frac{3}{2}}{\frac{5}{3}}-1 \right)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\]\[\frac{1}{{{f}_{2}}}=-\frac{1}{10}\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\]..(3) From (1)\[f=\frac{2}{\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)}\] From (2) \[{{f}_{1}}=\frac{8}{\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)}\] From (3)\[{{f}_{2}}=-\frac{10}{\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)}\]You need to login to perform this action.
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