Pls answer this question...

Dear Student,
Given,
Height of the object, h_o = 6 cm
Focal length of the convex lens, f = 15 cm
Object distance, u = 10 cm
Let the image distance be v.
Then using the mirror equation:
$$\begin{gathered} \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \\ \frac{1}{15}=\frac{1}{v}-\frac{1}{10} \\ \frac{1}{v}=\frac{1}{15}+\frac{1}{10}=\frac{2+3}{30}=\frac{5}{30}=\frac{1}{6} \\ v=6 cm (on the back side of the convex mirror) \\ Magnification, M=-\frac{v}{u}=-\frac{6}{-10}=+0.6 \\ M=\frac{Image height}{Object height}=+0.6 \\ Image height=+0.6\times 6=3.6 cm (diminished and erect) \end{gathered}$$

So, the image formed will be virtual, erect and diminished.

Regards