A) equal to 40 cm
B) equal to 20 cm
C) equal to 10 cm
D) None of these
Correct Answer: A
Solution :
[a] \[\frac{1}{{{f}_{\omega }}}=\left( ^{\omega }{{\mu }_{g}}-1 \right)\left( \frac{1}{{{r}_{1}}}-\frac{1}{{{r}_{2}}} \right)\] [\[{{f}_{\omega }}\] is focal length of lens in water] \[\frac{1}{f}={{(}^{a}}{{\mu }_{g}}-1)\left( \frac{1}{{{r}_{1}}}-\frac{1}{{{r}_{2}}} \right)\] [f is focal length of lens in air] Dividing, \[\frac{f}{{{f}_{\omega }}}=\frac{{{(}^{\omega }}{{\mu }_{g}}-1)}{{{(}^{a}}{{\mu }_{g}}-1)};\]\[^{\omega }{{\mu }_{g}}=\frac{^{a}{{\mu }_{g}}}{^{a}{{\mu }_{w}}}\] \[=\frac{3/2}{4/3}=\frac{3}{2}\times \frac{3}{4}=\frac{9}{8}\] \[\frac{f}{{{f}_{w}}}=\frac{9/8-1}{\frac{3}{2}-1}=\frac{1/8}{1/2}=\frac{1}{4}\] \[{{f}_{w}}=4\times f=4\times 10=40\,\,cm.\]
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