SOLUTION Let's consider the image formed by a shiny spherical Christmas SET UP (Figure 2) shows our diagram. (To limit its size, we drew it not to scale; the angles are exaggerated, and Santa would actually be much taller and farther ornament. Santa checks himself for soot by looking at his away.) The surface of the ornament closest to Santa acts as a convex mirror with radius R = -(7.20 cm) /2 = -3.60 cm and focal length reflection in a silvered Christmas tree ornament 0.750 m away ( f = R/2 = -1.80 cm. The object distance is s = 0.750 m = 75.0 cm. Figure 1). The diameter of the ornament is 7.20 cm. Let's estimate Santa's height as 1.6 m. Where and how tall is the SOLVE From 1/s + 1/s' = 1/f, image of Santa formed by the ornament? Is it upright or inverted? = f = -1.80 cm 75.0 cm = -1.76 cm The lateral magnification m is given by the following: m y' -1.76 cm = 2.35 x 10-2 75.0 cm Because m is positive, the image is upright, and it is only about 0.0235 as tall as Santa himself. Thus, the image height y' is y' = my = (0.0235) (1.6 m) = 3.8 x 10-2 m = 3.8 cm REFLECT The object is on the same side of the mirror as the incoming light, so the object distance s is positive. Because s' is negative, the image is behind the igure mirror-that is, in (Figure 2) it is on the side opposite to that of the outgoing light-and it is virtual. The image is about halfway between the front surface of the