JEE Main 2014 — Ray Optics Question with Solution

If a lens of refractive index $\mu$ is immersed in a medium of refractive index ${\mu }_1$, then its focal length in medium is given by $$\frac{1}{{\mathrm{f}}_{\mathrm{m}}}=\left({}_{\mathrm{m}}{\mu }_l-1\left(\frac{1}{{\mathrm{R}}_1}\right)-\frac{1}{{\mathrm{R}}_2}\right)$$ If ${\mathrm{f}}_{\mathrm{a}}$ is the focal length of lens in air, then $$\begin{array}{rl}& \frac{1}{{\mathrm{f}}_{\mathrm{a}}}=({}_{\mathrm{a}}{\mu }_l-1\left(\frac{1}{{\mathrm{R}}_1}-\frac{1}{{\mathrm{R}}_2}\right)\\ & \Rightarrow \frac{{\mathrm{f}}_{\mathrm{m}}}{{\mathrm{f}}_{\mathrm{a}}}=\frac{\left({}_{\mathrm{a}}{\mu }_l-1\right)}{\left({}_{\mathrm{m}}{\mu }_l-1\right)}\end{array}$$ If ${\mu }_1>\mu$, then ${\mathrm{f}}_{\mathrm{m}}$ and ${\mathrm{f}}_{\mathrm{a}}$ have opposite signs and the nature of lens changes i.e. a convex lens diverges the light rays and concave lens converges the light rays. Thus given option (a) is correct.