A convex mirror always form virtual image, which is erect and small in size of an object kept in front of it. Focal length of convex mirror f = + 15 cm Object distance u = – 12 cm Using mirror formula $\frac{1}{v} + \frac{1}{u}$ = $\frac{1}{f}$ $\frac{1}{v} + \frac{1}{-12}$ = $\frac{1}{15}$ or $\frac{1}{v}$ = $\frac{1}{15} + \frac{1}{12}$ or $\frac{1}{v}$ = $\frac{4+5}{60}$ v = + $\frac{60}{9}$ cm = + 6.66 cm Therefore, image is virtual, formed at 6.67 cm at the back of the mirror. Magnification m = - $\frac{v}{u}$ m = - $\frac{\frac{60}{9}}{-12}$ ⇒ m = $\frac{5}{9}$ = + 0.55 Size of image $h_2$ = m × $h_1$ = $\frac{5}{9} \times h_1$ = $\frac{5}{9}$ × 4.5 ⇒ $h_2$ = 2.5 cm It shows the image is erect, small in size and virtual. When the needle is moved farther from mirror the image also move towards focus and decreasing in size. As u approaches ∞, v approaches focus but never beyond f.