A) 50 cm
B) 100 cm
C) 200 cm
D) 150 cm
Correct Answer: B
Solution :
Key Idea: For a plano-convex lens radius of curvature of plane surface is infinity. We know that focal length of lens is \[\frac{1}{f}=(n-1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\]where\[f\] is focal length, n the refractive index and\[{{R}_{1}},{{R}_{2}}\] the radii of curvatures. For a plano-convex lens \[{{R}_{1}}=60\,cm,\,{{R}_{2}}=\infty ,\] \[n=1.6.\] \[\therefore \] \[\frac{1}{f}=(1.6-1)\left( \frac{1}{60}-\frac{1}{\infty } \right)=\frac{0.6}{60}\] \[\Rightarrow \] \[f=100\,cm\]
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