A point object ‘O’ is placed in front of a glass rod having spherical end of radius of curvature 30 cm. The image would be formed at

By using formula: $\frac{{\mathrm{\mu }}_2}{\mathrm{v}}-\frac{{\mathrm{\mu }}_1}{\mathrm{u}}=\frac{{\mathrm{\mu }}_2-{\mathrm{\mu }}_1}{\mathrm{R}}$

$\frac{1.5}{\mathrm{v}}-\frac{1}{-15}=\frac{0.5}{30}$

$\frac{1.5}{\mathrm{v}}=\frac{1}{60}-\frac{1}{15}=\frac{-3}{60}$

$V=-30\mathrm{cm}$

Negative sign shows that, image is obtained on the same side of object i.e. towards left.