Answer:

Correct Answer: C

Solution:

Given $ {\mu _g}=1.5 $ $ {\mu _{oil}},=1.7 $ $ R=20cm $

From Lens Maker’s formula for the piano convex lens $ \frac{1}{f}=(\mu -1)[ \frac{1}{R _1}-\frac{1}{R _2} ] $

Here, $ R _1=R $ and for plane surface $ R _2=\infty $

$ \therefore $ $ \frac{1}{f _{lens}}=(1.5-1)( \frac{1}{R}-0 ) $

$ \Rightarrow $ $ \frac{1}{f _{lens}},=\frac{0.5}{R} $

When the intervening medium is filled with oil, then focal length of the concave lens formed by the oil

$ \frac{1}{{f _{concave}}}=(17-1)-( -\frac{1}{R}-\frac{1}{R} ) $ $ =-0.7\times \frac{2}{R}=\frac{-14}{R} $

Here, we have two concave surfaces So.

$ \frac{1}{f _{eq}}=2\times \frac{1}{f}+\frac{1}{f} $ $ =2\times \frac{0.5}{R}+( \frac{-14}{R} )=\frac{1}{R}-\frac{14}{R}=-\frac{0.4}{R} $

$ \therefore $ $ f _{eq}=-\frac{R}{0.4}=-\frac{20}{0.4}=-50cm $