A) \[48\,\,cm\]
B) \[36\,\,cm\]
C) \[24\,\,cm\]
D) \[12\,\,cm\]
Correct Answer: A
Solution :
Focal length in air is given by, \[\frac{1}{{{f}_{a}}}={{(}_{a}}{{\mu }_{g}}-1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] The focal length of lens immersed in water is given by \[\frac{1}{{{f}_{1}}}={{(}_{l}}{{n}_{g}}-1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] where,\[{{R}_{1}},\,\,{{R}_{2}}\]are radii of curvatures of the two surfaces of lens and ;n» is refractive index of glass with respect to liquid. Also, \[_{l}{{n}_{g}}=\frac{_{a}{{n}_{g}}}{_{a}{{n}_{l}}}\] Given,\[_{a}{{n}_{g}}=1.5,\,\,{{f}_{a}}=12\,\,cm,\,{{\,}_{a}}{{n}_{l}}=\frac{4}{3}\] \[\therefore \] \[\frac{{{f}_{l}}}{{{f}_{a}}}=\frac{{{(}_{a}}{{n}_{g}}-1)}{{{(}_{l}}{{n}_{g}}-1)}\] \[\frac{{{f}_{l}}}{12}=\frac{(1.5-1)}{\left( \frac{1.5}{4/3}-1 \right)}=\frac{0.5\times 4}{0.5}\] \[\Rightarrow \] \[{{f}_{1}}=4\times 12=48\,\,cm\]
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