A) 48 cm
B) 36 cm
C) 24cm
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 \[_{l}{{n}_{g}}\]is refractive index of glass with respect to liquid. Also, \[_{l}{{n}_{g}}=\frac{_{a}{{n}_{g}}}{_{a}{{n}_{l}}}\] Given, \[_{a}{{n}_{g}}=1.5,{{f}_{a}}=12cm,{{\,}_{a}}{{n}_{l}}=\frac{4}{3}\] \[\therefore \] \[\frac{{{f}_{l}}}{{{f}_{a}}}=\frac{({{\,}_{a}}{{n}_{g}}-1)}{({{\,}_{l}}{{n}_{g}}-1)}\] \[\frac{{{f}_{1}}}{12}=\frac{(1.5-1)}{\left( \frac{1.5}{4/3}-1 \right)}=\frac{0.5\times 4}{0.5}\] \[\Rightarrow \] \[{{f}_{1}}=4\times 12=48\,cm\]
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