If a lens of refractive index μ is immersed in a medium of refractive index μ1, then its focal length in medium is given by fm1=(mμl−1(R11)−R21) If fa is the focal length of lens in air, then fa1=(aμl−1(R11−R21)⇒fafm=(mμl−1)(aμl−1) If μ1>μ, then fm and fa 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.