A plano-convex lens of refractive index 1.5 has a curved face of radius of curvature 30 cm; its plane face is silvered. A point object lies on the…

Light refracts at the curved face, reflects at the plane silvered face and refracts again at the curved face, so the system acts as a mirror of power $2P_1$ where $P_1=(n-n_m)/R$ and $n_m$ is the outer medium. In air $2P_1=2(0.5)/0.30=3.33\ \mathrm{D}$, i.e. an equivalent concave mirror of focal length $30\ \mathrm{cm}$; so (A), which claims $15\ \mathrm{cm}$, is wrong. With $u=-60\ \mathrm{cm}$, $1/v=-1/30+1/60$ gives $v=-60\ \mathrm{cm}$ and magnification $-1$: a real, inverted image of the same size at $60\ \mathrm{cm}$, so (C) is correct. In the liquid $P_1=(1.5-4/3)/0.3=0.556\ \mathrm{D}$ and $f=n_m/(2P_1)=1.2\ \mathrm{m}=120\ \mathrm{cm}$, not $90\ \mathrm{cm}$; (B) is wrong. The object at $60\ \mathrm{cm}$ is then inside the focal length, so $1/v=-1/120+1/60$ gives $v=+120\ \mathrm{cm}$: a virtual, magnified image, so (D) is wrong. Answer: C only.