What happens when one side of a convex lens is made fully reflecting? Assume that I have got a bowl half-filled with mercury. In this bowl, I place a convex lens which make one side of the lens fully reflecting. What will happen to the lens, will it behave like a concave mirror and what will be the change in the sign of the focal length??
 A: It will behave as a concave mirror with a convex lens in front of it, or just as a convex mirror if the other side is reflecting (with a lens behind it that doesn't really do anything because it's blocked). You would have to do ray tracing to see more precise behavior for the former case. 
A: As others have pointed out, it will behave as a lens and a mirror.  I wanted to point out a particular version of this with interesting applications: the cat's eye retroreflector:

The cat's eye retroreflector is a transparent sphere that is silvered on one side, much like the eye of a cat with its reflective tapetum lucidum at the back.  This structure has the curious property that it always reflects light directly back at the source.  This makes it "light up" if you hit it with a flash light or headlight.  This of course, leads to its use on roads

Not all lenses mirrored on one side will have this precise effect, but I think it's pretty awesome!
A: Yes. Convex mirrors are typically made by depositing a metal coating on the curved surface of a convex piece of glass. But the focal lengths are incomparable.
A: As other answers have pointed out, if one surface of the lens is made reflective by for instance coating it with a thin reflective film, for light incident from the coated side it will behave as a convex mirror, and for light incident from the other side it will behave as a concave mirror. Assuming the lens is sufficiently thin, we can also calculate how "concave" this mirror is, i.e. what the effective focal length is.
Let $R_1$ be the radius of curvature of the non-coated surface and $R_2$ that of the coated surface, both assumed positive if the surface is convex. For an object placed a distance $x$ away from the non-coated side of the system, by the lensmaker's equation we have for the image through the lens at $x'$
$$ \frac{1}{x} + \frac{1}{x'} = (n - 1)\left(\frac{1}{R_1} + \frac{1}{R_2}\right). $$
For the image through the mirror located at $y'$, we have
$$ -\frac{1}{x'} + \frac{1}{y'} = \frac{2}{R_2}. $$
For the final image back though the lens located at $y$, we have
$$ -\frac{1}{y'} + \frac{1}{y} = (n - 1)\left(\frac{1}{R_1} + \frac{1}{R_2}\right). $$
Adding up all three equations,
$$ \frac{1}{f} = \frac{2}{R} = \frac{1}{x} + \frac{1}{y} = 2\frac{n-1}{R_1}+\frac{2}{R_2} $$
where $f$ is the effective focal length and $R$ is the radius of curvature of the equivalent concave mirror. Rearranging, we have
$$ f = \frac{1}{2} \frac{R_1 R_2}{(n-1)R_2 + R_1} $$
$$ R = 2f = \frac{R_1 R_2}{(n-1)R_2 + R_1}. $$
Edit: My above statement that the lens behaves as a concave mirror for light incident from the non-coated side assumes that the lens is biconvex. For a concavo-convex lens, depending on the radii of curvature and the refractive index of the lens, it can also behave as a planar or a convex mirror. If for instance $R_1 < 0$, $R_2 > 0$ and $R_2 < -R_1 < (n-1)R_2$, the lens by itself is still converging but the lens-mirror system behaves as a convex mirror.
