According to the question;
Object distance = u;
Image distance (v) = -10cm;
Focal length = -15cm;
By lens formula;
\(\frac1v-\frac1u=\frac1f\)
⇒ \(\frac1{-10}-\frac1u=\frac1{10}\)
⇒ \(\frac1{u}=\frac1{15}-\frac1{10}\)
⇒ \(\frac1{u}=\frac{2-3}{30}-\frac{-1}{30}\)
⇒ \(\frac1{u}=\frac{-1}{30}\)
⇒ u = -30cm.
Thus, object should be placed 30cm in front of lens.
Now;
Height of object h1= 2cm;
Magnification = \(\frac{h_2}{h_1}=\frac{v}u\)
Putting values of v and u
Magnification = \(\frac{h_2}{2}=\frac{-10}{-30}\)
⇒ \(\frac{h_2}{5}=\frac{1}{3};\)
⇒ h2 = \(\frac53\) = 1.67
Height of image is 1.67 cm.
Thus, the image is virtual, diminished, and erect and one-third of the size of object