a
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

\(\frac{1}{{{f_m}}} = {(_m}{\mu _l} - 1)\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right)\)

If \(f_{a}\) is the focal length of lens in air, then

\(\frac{1}{{{f_a}}} = {(_a}{\mu _l} - 1)\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right)\)

\( \Rightarrow \frac{{{f_m}}}{{{f_a}}} = \,\frac{{{(_a}{\mu _l} - 1)}}{{{(_m}{\mu _l} - 1)}}\)

If \(\mu_{l}>\mu,\) then \(f_{m}\) and \(f_{a}\) have opposite signs and the nature of lenschanges i.e. a convex lens diverges the light rays and concave lens converges the light rays. Thus given option \((a)\) is correct.