Answer :
We are given,
Size of the needle is h1=4.5 cm
Distance of object is u=-12 cm
Focal length of the convex mirror is f= 15 cm
Image distance is represented by v
$\frac{1}{u}$+ $\frac{1}{v}$= $\frac{1}{f}$
$\frac{1}{v}$= $\frac{1}{f}$- $\frac{1}{u}$
$\frac{1}{15}+\frac{1}{12}=\frac{4+5}{60}=\frac{9}{60}$
∴ $\frac{60}{9}$
= 6.7cm
The picture is obtained on the other side of the mirror, at a distance of 6.7cm between the needle image and the mirror.
Using magnification formula:
$m=\frac{h_{2}}{h_{1}}=-\frac{v}{u}$
$ h_{2}=\frac{-v}{u}\times h_{1}$
$=\frac{-6.7}{-12}\times 4.5$
= +2.5cm
$m=\frac{h_{2}}{h_{1}}=\frac{2.5}{4.5}$
= 0.56
The image has a height of 2.5cm. The picture is erect, imaginary, and decreased since it has a positive indication.
As the needle moves away from the mirror, the picture moves away from the mirror as well, resulting in a reduction in image size.