An object of size $3.0 \, cm$ is placed $14 \, cm$ in front of a concave lens of focal length $21 \, cm$. Describe the image produced by the lens. What happens if the object is moved further away from the lens?

(N/A) Size of the object,$h_1 = 3.0 \, cm$.
Object distance,$u = -14 \, cm$.
Focal length of the concave lens,$f = -21 \, cm$.
Using the lens formula,$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
Substituting the values: $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{-21} + \frac{1}{-14} = \frac{-2 - 3}{42} = \frac{-5}{42}$.
Thus,$v = -8.4 \, cm$.
The negative sign indicates that the image is virtual and erect,formed $8.4 \, cm$ in front of the lens.
Magnification $m = \frac{v}{u} = \frac{-8.4}{-14} = 0.6$.
Image height $h_2 = m \times h_1 = 0.6 \times 3.0 = 1.8 \, cm$.
If the object is moved further away from the lens,the virtual image moves towards the focus of the lens,and the size of the image decreases.