JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Geometrical Optics

The image returns to the object only when light strikes the mirror normally and retraces its path, which happens when the object sits at the focal point of the lens system. With no liquid, that focal length is $f=18$ cm, and the lens-maker's relation for the equiconvex lens, $\frac{1}{f}=(1.5-1)\frac{2}{R}$, gives $R=18$ cm. When a liquid of index ${\mu }_1$ fills the gap between lens and mirror, it acts as a plano-concave lens in contact with the glass lens, and the combination now focuses from $27$ cm. Writing $\frac{1}{27}=\frac{1}{f_{\text{glass}}}+\frac{1}{f_{\text{liquid}}}$ with $\frac{1}{f_{\text{glass}}}=\frac{1}{18}$ and $\frac{1}{f_{\text{liquid}}}=({\mu }_1-1)\left(\frac{1}{-18}\right)$, we get $\frac{1}{27}=\frac{1}{18}-\frac{{\mu }_1-1}{18}$. Multiplying through by $18$, $\frac{18}{27}=1-({\mu }_1-1)$, so $\frac{2}{3}=2-{\mu }_1$ and ${\mu }_1=\frac{4}{3}$.