Question

An object is located 13.0 cm in front of a convex mirror, the image being 6.80 cm behind the mirror. A second object, twice as tall as the first one, is placed in front of the mirror, but at a different location. The image of this second object has the same height as the other image. How far in front of the mirror is the second object located?
cm

Answer 1

The object distance is p = 13.0 cm

The image distance is q = -6.8 cm   (behind the mirror)

Using the mirror equation

1/p + 1/q = 1/f

1/13 - 1/6.80 = 1/f

f = -14.26 cm [negative sign indicates that the mirror is convex]

The magnification M = -q/p = 6.80 /13 = 0.52

Let h is the height of the object then height of the image is 0.52h

The height of the second object is 2h

And the height of the image of the second object is 0.52h

The magnificaion M = image height/object height = 0.52h / 2h = 0.26

The image distance q = -Mp

q = -0.26p

The mirror equation is

1/p + 1/q = 1/f

1/p - 1/0.26p = 1/-14.26

p = 40.59 cm