Question

A plano-convex lens is made of a material of refractive index $n$. When a small object is placed $30 \,cm$ away in front of the curved surface of the lens, and image of the double the size of the object is produced. Due to reflection from the convex surface of the lens, another faint image is observed at a distance of $10\, cm$ away from the lens. Which of the following statement (s) is (are) true?

• A. The refractive index of the lens is 2.5
• B. The radius of curvature of the convex surface is 45 cm
• C. The faint image is erect and real
• D. The focal length of the lens is 20 cm

Answer 1

The Correct Option is D

for Refraction through lens $4 u =-30\, cm $ $ m =-2 $ $ m =\frac{ f }{ f + u } $ $-2=\frac{ f }{ f -30}$ $-2 f +60= f $ $ f =20\, cm$ for Reflection through convex surface $ u =-30, v =10 $ $\frac{1}{10}+\frac{1}{-30}=\frac{1}{ f _{0}} $ $ f _{0}=15 $ $ R =2 f _{0}=30$ Now for plano-convex lens $\frac{1}{f}=\frac{( n -1)}{ R }$ $\frac{1}{20}=\frac{( n -1)}{30}$ $ \Rightarrow n =2.5$