A) 3.0
B) 2.0
C) 2.5
D) 1.5
Correct Answer: D
Solution :
We know, \[\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\] If the object is placed at a distance of 20 cm from the lens, its image will be formed at the same place, \[{{f}_{lens}}=2\times {{f}_{mirror}}=2\times 30=60\,\,cm\] Given, \[v=-{{R}_{lens}};\,\,u=-20\,\,cm\] \[\therefore \] \[\frac{1}{60}=\frac{1}{{{R}_{lens}}}+\frac{1}{20}\] or \[\frac{1}{{{R}_{lens}}}=\frac{1}{20}-\frac{1}{60}=\frac{1}{30}\] Now, \[\frac{1}{f}=\frac{9(\mu -1)}{R}\] \[\Rightarrow \] \[\frac{1}{60}=\frac{(\mu -1)}{30}\] \[\therefore \] \[\mu =1.5\]and radius of curvature of lens\[~=30\,\,cm\].
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