A point object O is placed at 15 cm in front of a glass rod $(\mu =1.5)$ having spherical end of radius of curvature $30 cm.$ The image would be formed at

By using formula $\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}\Rightarrow \frac{1.5}{v}-\frac{1}{(-15)}=\frac{(1.5-1)}{+30}\Rightarrow v=-30\mathrm{cm}$ 

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