A thin plano-convex glass lens (refractive index 1.5) has its curved surface of radius of curvature 30 cm facing a small point object on the axis, 20…

Light enters through the curved surface, reflects from the silvered plane surface and leaves through the curved surface again. The lens, mirror, lens combination acts as a single mirror of power $P=2P_L$ (two passes through the lens, the plane mirror adding no power). The focal length of the lens is given by $\dfrac1{f_L}=(\mu-1)\dfrac1R=\dfrac{0.5}{30}$, so $f_L=60\ \mathrm{cm}$ and the equivalent mirror has focal length $F=f_L/2=30\ \mathrm{cm}$, concave. With the object at $u=-20\ \mathrm{cm}$ and $f=-30\ \mathrm{cm}$ the mirror equation gives $\dfrac1v=\dfrac1f-\dfrac1u=-\dfrac1{30}+\dfrac1{20}=\dfrac1{60}$. So $v=+60\ \mathrm{cm}$, a virtual image $60\ \mathrm{cm}$ behind the silvered surface. Since the object lies inside the focal length, the image is virtual and magnified, as expected. The distance from the lens is $60\ \mathrm{cm}$.