$A$ $4.5 \, cm$ needle is placed $12 \, cm$ away from a convex mirror of focal length $15 \, cm$. Give the location of the image and the magnification. Describe what happens to the image as the needle is moved farther from the mirror.

(A) Given: Object height $h_{1} = 4.5 \, cm$,Object distance $u = -12 \, cm$,Focal length $f = +15 \, cm$ (for a convex mirror).
Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
Substituting the values: $\frac{1}{15} = \frac{1}{v} + \frac{1}{-12}$.
Rearranging: $\frac{1}{v} = \frac{1}{15} + \frac{1}{12} = \frac{4 + 5}{60} = \frac{9}{60} = \frac{3}{20}$.
Thus,$v = \frac{20}{3} \approx +6.67 \, cm$.
The image is formed $6.67 \, cm$ behind the mirror.
Magnification $m = -\frac{v}{u} = -\frac{6.67}{-12} \approx +0.56$.
As the needle is moved farther from the mirror,the object distance $u$ increases. Consequently,the image moves closer to the focus $F$ and its size decreases.