An object is placed at a distance of 30 cm and it is 3cm high placed perpendicular to the principal axis of a concave lens of focal length 15 cm. The image is formed at a distance of 10 cm. What would be the size of the image formed?

Sign Conversion of concave lens:

Here, Image distance, $\mathrm{v}=-10 \mathrm{cm}$, focal length, $\mathrm{f}=-15 \mathrm{cm}$, object distance, $\mathrm{u}=-30 \mathrm{cm}$, object size, ${\mathrm{h}}_{\mathrm{o}}= 3 \mathrm{cm}$

We know,
 $\mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}}$, where $\mathrm{m}=$ the magnification of the image
$\Rightarrow \mathrm{Image} \mathrm{size}=\frac{\mathrm{v}}{\mathrm{u}}\times \mathrm{object} \mathrm{size}$
$\Rightarrow {\mathrm{h}}_{\mathrm{i}}=-\frac{10}{-30}\mathrm{ }\times 3$
$\Rightarrow \mathrm{Image} \mathrm{size}=1 \mathrm{cm}$

Therefore, the correct option is 1 cm.