Question 23.2: Images Formed by a Concave Mirror Goal Calculate properties ...
Images Formed by a Concave Mirror
Goal Calculate properties of a concave mirror.
Problem Assume that a certain concave spherical mirror has a focal length of 10.0 \mathrm{~cm}. (a) Locate the image and find the magnification for an object distance of 25.0 \mathrm{~cm}. Determine whether the image is real or virtual, inverted or upright, and larger or smaller. Do the same for object distances of (b) 10.0 \mathrm{~cm} and (\mathbf{c}) 5.00 \mathrm{~cm}.
Strategy For each part, substitute into the mirror and magnification equations. Part (b) involves a limiting process, because the answers are infinite.
(a) Find the image position for an object distance of 25.0 \mathrm{~cm}. Calculate the magnification and describe the image.
Use the mirror equation to find the image distance:
\text { (1) } \frac{1}{p}+\frac{1}{q}=\frac{1}{f}
Substitute and solve for q. According to Table 23.1, p and f are positive.
\begin{aligned} \frac{1}{25.0 \mathrm{~cm}}+\frac{1}{q} & =\frac{1}{10.0 \mathrm{~cm}} \\ q & =16.7 \mathrm{~cm} \end{aligned}
Because q is positive, the image is in front of the mirror and is real. The magnification is given by substituting into Equation 23.2:
M=-\frac{q}{p}=-\frac{16.7 \mathrm{~cm}}{25.0 \mathrm{~cm}}=-0.668
The image is smaller than the object because |M|<1, and inverted because M is negative. (See Fig. 23.13a.)
(b) Locate the image distance when the object distance is 10.0 \mathrm{~cm}. Calculate the magnification and describe the image.
The object is at the focal point. Substitute p=10.0 \mathrm{~cm} and f=10.0 \mathrm{~cm} into the mirror equation:
\begin{aligned} \frac{1}{10.0 \mathrm{~cm}}+\frac{1}{q} & =\frac{1}{10.0 \mathrm{~cm}} \\ \frac{1}{q} & =0 \rightarrow q=\infty \end{aligned}
Since M=-q / p, the magnification is infinite, also.
(c) Locate the image distance when the object distance is 5.00 \mathrm{~cm}. Calculate the magnification and describe the image.
Once again, substitute into the mirror equation:
\begin{aligned} \frac{1}{5.00 \mathrm{~cm}}+\frac{1}{q} & =\frac{1}{10.0 \mathrm{~cm}} \\ \frac{1}{q} & =\frac{1}{10.0 \mathrm{~cm}}-\frac{1}{5.00 \mathrm{~cm}}=-\frac{1}{10.0 \mathrm{~cm}} \\ q & =-10.0 \mathrm{~cm} \end{aligned}
The image is virtual (behind the mirror) because q is negative. Use Equation 23.2 to calculate the magnification:
M=-\frac{q}{p}=-\left(\frac{-10.0 \mathrm{~cm}}{5.00 \mathrm{~cm}}\right)=2.00
The image is larger (magnified by a factor of 2) because |M|>1, and upright because M is positive. (See Fig. 23.13b.)
Remarks Note the characteristics of an image formed by a concave, spherical mirror. When the object is outside the focal point, the image is inverted and real; at the focal point, the image is formed at infinity; inside the focal point, the image is upright and virtual.
TABLE 23.1
Sign Conventions for Mirrors
| Quantity | Symbol | In Front | In Back | Upright Image | Inverted Image |
| Object location | p | + | — | ||
| Image location | q | + | — | ||
| Focal Length | f | + | — | ||
| Image height | h’ | + | — | ||
| Magnification | M | + | — |
