A) 15 cm
B) 20 cm
C) 30 cm
D) 10 cm
Correct Answer: C
Solution :
\[{{\left( r-\frac{3}{10} \right)}^{2}}+{{3}^{2}}={{r}^{2}}\] \[{{r}^{2}}-\frac{6r}{10}+\frac{9}{10}+9={{r}^{2}}\] \[-\frac{6r}{10}+\frac{909}{100}=0\] \[r=15.15\text{ }m\] \[\frac{1}{f}=(\mu -1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] \[{{R}_{plane}}=\infty \] \[{{R}_{convex}}=15\text{ }cm\] (approx.) \[\mu =\frac{3\times {{10}^{8}}}{2\times {{10}^{8}}}=\frac{3}{2}\] \[\frac{1}{f}=\frac{1}{2}\times \frac{1}{15}\] \[f=30\text{ }cm\]You need to login to perform this action.
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