A) 20 cm
B) 30 cm
C) 40cm
D) 10cm
Correct Answer: B
Solution :
From lens formula \[\frac{1}{f}=(\mu -1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] If focal length of lens in air is\[{{f}_{a}}\]and in liquid is \[{{f}_{1}},\]then \[\frac{1}{{{f}_{a}}}={{(}_{a}}{{\mu }_{g}}-1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] \[\frac{1}{{{f}_{e}}}={{(}_{l}}{{\mu }_{g}}-1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] \[\frac{{{f}_{l}}}{{{f}_{a}}}=\frac{{{(}_{a}}{{\mu }_{g}}-1)}{{{(}_{l}}{{\mu }_{g}}-1)}\] \[\frac{{{f}_{l}}}{20}=\left[ \frac{\frac{3}{2}-1}{\frac{4}{3}-1} \right]\] \[=\frac{3}{2}\] \[{{f}_{l}}=30\,cm\]
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