Example 13.1: Concave mirrors

Question: An object of height $h=4$cm is placed a distance $p=15$cm in front of a concave mirror of focal length $f=20$cm. What is the height, location, and nature of the image? Suppose that the object is moved to a new position a distance $p=25$cm in front of the mirror. What now is the height, location, and nature of the image?
 
Answer: According to Eq. (358), the image distance $q$ is given by

$$q = \frac{1}{1/f - 1/p} = \frac{1}{(1/20-1/15)} = - 60\,{\rm cm}.$$

Thus, the image is virtual (since $q<0$), and is located $60$cm behind the mirror. According to Eq. (352), the magnification $M$ of the image is given by

$$M = -\frac{q}{p} = -\frac{(-60)}{(15)} = 4.$$

Thus, the image is upright (since $M>0$), and magnified by a factor of $4$. It follows that the height $h'$ of the image is given by

$$h' = M\,h = (4)\,(4) = 16\,{\rm cm}.$$

If the object is moved such that $p=\,25$cm then the new image distance is given by

$$q = \frac{1}{1/f - 1/p} = \frac{1}{(1/20-1/25)} = 100\,{\rm cm}.$$

Thus, the new image is real (since $q>0$), and is located 100cm in front of the mirror. The new magnification is given by

$$M = -\frac{q}{p} = -\frac{(100)}{(15)} = -6.67.$$

Thus, the image is inverted (since $M<0$), and magnified by a factor of $6.67$. It follows that the new height of the image is

$$h' = M\,h = -(6.67)\,(4) = -26.67\,{\rm cm}.$$

Note that the height is negative because the image is inverted.