A) 0.45 m
B) 0.15 m
C) 0.30 m
D) 0.6 m
Correct Answer: D
Solution :
Given, \[{{f}_{a}}=0.15\,\,m,\,{{\mu }_{g}}=\frac{3}{2},{{\mu }_{w}}s=\frac{4}{3}\] According to Lens maker's 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 \[\frac{{{f}_{w}}}{{{f}_{a}}}=\frac{C}{2}\times \frac{8}{C}=4\] or \[{{f}_{w}}=4\,{{f}_{a}}\] \[=4\times 0.5=0.6\,m\]
You need to login to perform this action.
You will be redirected in
3 sec
