In a compound microscope, maximum magnification is obtained when the final image is formed at the least distance of distinct vision

use m=v/v, if=1/v + 1/u

all power of all lens

On substituting the given quantities in the lens maker?s equation, 1/f = (n2/n1 ? 1)(1/R1 ? 1/R2), we have 1/f = [(1.5/1.75)-1] [1/(-R) ? 1/R)] = (-0.25/1.75) (-2/R) = 0.5/1.75R, from which f = 3.5R. You should ensure that all your doubts on applying the sign convention are cleared before proceeding to the next question. In the lens maker?s equation, we applied negative sign to the radius of curvature (R) of the first surface since the concave lens presents a concave surface to the incident ray (and hence the centre of curvature is on the same side of the pole as the object is). The sign of the radius of curvature of the second surface is positive since it presents a convex surface to the ray proceeding towards it after the refraction at the first surface. We did not bother about the sign of the focal length of the lens, since it is the unknown quantity in this problem. The sign turns out to be positive, indicating that the concave lens behaves as a converging lens on immersing it in a denser liquid.