A) \[\frac{3}{2}\]
B) \[\frac{4}{3}\]
C) \[\frac{5}{3}\]
D) \[2\]
Correct Answer: C
Solution :
When the lens in air, we have \[{{P}_{a}}=\frac{1}{{{f}_{a}}}=\frac{{{\mu }_{g}}-{{\mu }_{a}}}{{{\mu }_{a}}}\left[ \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right]\] When the lens is in liquid, we have \[{{P}_{l}}=\frac{1}{{{f}_{l}}}=\frac{{{\mu }_{g}}-{{\mu }_{l}}}{{{\mu }_{l}}}\left[ \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right]\] Here,\[{{P}_{a}}=5,\,\,{{P}_{l}}=-1,\,\,{{\mu }_{a}}=1,\,\,{{\mu }_{g}}=1.5\] On solving, we get,\[{{\mu }_{l}}=\frac{5}{3}\]
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