A) 0.1 m
B) 0.75 m
C) 0.4 m
D) 0.2 m
Correct Answer: D
Solution :
After silvering the plane surface, piano convex lens behaves as a concave mirror of focal length \[\frac{1}{F}=\frac{2}{{{f}_{lens}}}\] But \[F=0.2\,m\] \[\therefore \] \[{{f}_{lens}}=2F=2\times 0.2=0.4\,m\] Now from lens makers formula \[\frac{1}{{{f}_{lens}}}=(\mu -1)\,\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] \[\therefore \] \[\frac{1}{0.4}=(1.5-1)\times \frac{1}{{{R}_{1}}}\] \[[\because \,\,{{R}_{2}}=\infty ]\] \[\Rightarrow \] \[{{R}_{1}}=0.5\times 0.4\] \[=0.2\,\,m\]
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