Physics - Geometrical Optics Question with Solution | TestHub

Using $\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$, the centre of curvature lies on the incidence side, so $R<0$.

If ${\mu }_2>{\mu }_1$ the surface power is negative: a real object always forms a virtual image (A correct), but a converging (virtual) object close to the surface can still give a real image (B wrong).

If ${\mu }_1>{\mu }_2$ the power is positive: a virtual object always forms a real image (C correct), while a real object can form a real image (D wrong).