Given: Height of the needle, \\\\(h_1 = 4.5 cm\\\\).Object distance, \\\\(u = −12 cm\\\\).The focal length of the convex mirror, \\\\(f=15cm\\\\).Image distance, \\\\( v\\\\) The value of \\\\(v\\\\) can be obtained using the mirror formula.\\\\(\\dfrac{1}{v}+\\dfrac{1}{u}=\\dfrac{1}{f}\\\\)\\\\(\\dfrac{1}{v}+\\dfrac{1}{-12}=\\dfrac{1}{15}\\\\)\\\\(\\dfrac{1}{v} = \\dfrac{1}{12}+\\dfrac{1}{15}\\\\)\\\\(\\dfrac{1}{v} = \\dfrac{9}{60}\\\\)\\\\(\ herefore v \\approx 6.7cm\\\\)Hence, the image of the needle is 6.7 cm away from the mirror. Also, it is on the other side of the mirror.The image size is given by the magnification formula.\\\\(m=\\dfrac{h^{'}}{h}=-\\dfrac{v}{u}\\\\)\\\\(h'=\\dfrac{6.7\ imes4.5}{12}\\\\)\\\\(\\Rightarrow h'=+2.5cm\\\\)So, \\\\(m=\\dfrac{2.5}{4.5}\\\\)\\\\(m=0.56\\\\)The height of the image is \\\\(2.5 cm\\\\). The positive sign indicates that the image is erect, virtual, and diminished. If the needle is moved farther from the mirror, the size of the image will reduce gradually.