A) 0.45 m
B) 0.15 m
C) 0.30 m
D) 0.6 m
Correct Answer: D
Solution :
Given, \[{{f}_{a}}=0.15m,{{\mu }_{g}}=\frac{3}{2},{{\mu }_{w}}=\frac{4}{3}\] According to Lens makers formula \[\frac{1}{f}=(\mu -1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\]where\[\mu =\frac{{{\mu }_{L}}}{{{\mu }_{M}}}\] \[\frac{1}{{{f}_{a}}}=\left( \frac{{{\mu }_{g}}}{{{\mu }_{a}}}-1 \right)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] \[=\left( \frac{(3/2)}{1}-1 \right)C\]where\[C=\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] or \[\frac{1}{{{f}_{a}}}=\frac{C}{2}\] ?(i) Also,\[\frac{1}{{{f}_{w}}}=\left( \frac{{{\mu }_{g}}}{{{\mu }_{w}}}-1 \right)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)=\left( \frac{(3/2)}{(4/3)}-1 \right)C\] or \[\frac{1}{{{f}_{w}}}=\frac{C}{8}\] ?(ii) From Eqs. (i) and (ii), we get \[\frac{{{1}_{w}}}{{{f}_{a}}}=\frac{C}{2}\times \frac{8}{C}=4\] or \[{{f}_{w}}=4{{f}_{a}}\] \[=4\times 0.15=0.6m\]
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