A) \[-160cm\]
B) \[-100cm\]
C) \[+10\text{ }cm\]
D) \[+100cm\]
E) \[+\text{ }160\text{ }cm\]
Correct Answer: A
Solution :
From lens formula \[\frac{1}{f}={{(}_{a}}{{\mu }_{g}}-1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] \[=(1.5-1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] ?. (1) Also, \[_{l}{{\mu }_{g}}=\frac{{{\mu }_{g}}}{{{\mu }_{l}}}=\frac{1.5}{1.6}\] \[\therefore \] \[\frac{1}{f}={{(}_{l}}{{\mu }_{g}}-1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] \[\frac{1}{f}=\left( \frac{15}{16}-1 \right)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] ??(2) \[\Rightarrow \] \[\frac{f}{f}=\frac{\left( \frac{15}{16}-1 \right)}{(1.5-1)}=-\frac{1}{16\times .5}\] \[\Rightarrow \] \[f=-160\,cm\]
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