The concave spherical mirror shown in Fig. $36-9$ has radius of curvature $4 \mathrm{~m}$. An object $O O^{\prime}, 5 \mathrm{~cm}$ high, is placed $3 \mathrm{~m}$ in front of the mirror. By ( $a$ ) construction and (b) computation, determine the position and height of the image $I I^{\prime}$.
In Fig. $36-9, C$ is the center of curvature, $4 \mathrm{~m}$ from the mirror, and $F$ is the principal focus, $2 \mathrm{~m}$ from the mirror.
(a) Two of the following three convenient rays from $O$ will locate the image.
1) The ray OA, parallel to the principal axis. This ray, like all parallel rays, is reflected through the principal focus $F$ in the direction AFA'.
(2) The ray $O B$, drawn as if it passed through the center of curvature $C$. This ray is normal to the mirror and is reflected back on itself in the direction $B C B^{\prime}$ '
(3) The ray $O F D$ which passes through the principal focus $F$ and, like all rays passing through $F$, is reflected parallel to the principal axis in the direction $D D^{\prime}$.
The intersection $I$ of any two of these reflected rays is the image of
O. Thus II' represents the position and size of the image of $O O^{\prime}$ ' The image is real, inverted, magnified, and at a greater distance from the mirror than the object. (Note: If the object were at $I I^{\prime}$, the image would be at $I I^{\prime}$ and would be real, inverted, and smaller.)
(b) Using the mirror equation in which $R=-4 \mathrm{~m}$,
$$\frac{1}{s_{o}}+\frac{1}{s_{i}}=-\frac{2}{R} \quad \text { or } \quad \frac{1}{3}+\frac{1}{s_{i}}=-\frac{2}{-4} \quad \text { or } \quad s_{i}=6 \mathrm{~m}$$
The image is real (since $s_{i}$ is positive) and located $6 \mathrm{~m}$ from the mirror. Also, since the image is inverted, both the magnification and $y_{i}$ are negative:
$$
M_{T}=-\frac{s_{i}}{s_{o}}=-\frac{6 \mathrm{~m}}{3 \mathrm{~m}}=-2 \quad \text { and so } \quad y_{i}=(-2)(5 \mathrm{~cm})=-0.10 \mathrm{~m}
$$