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 a certain concave, spherical mirror has a focal length of 10.0 cm. (a) Locate the image and find the magnification for an object distance of 25.0 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 cm and (c) 5.00 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 cm. Calculate the magnification and describe the image.
Use the mirror equation to find the image distance:
{\frac{1}{p}}+{\frac{1}{q}}={\frac{1}{f}}
Substitute and solve for q. According to Table 23.1, p and f are positive.
{\frac{1}{25.0\mathrm{~cm}}}+{\frac{1}{q}}={\frac{1}{10.0\mathrm{~cm}}}
q = 16.7 cm
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}
M =\frac{h^{\prime}}{h}= -\frac{q}{p} [23.2]
The image is smaller than the object because |M| < 1, and it is inverted because M is negative. (See Fig. 23.13a.)
(b) Locate the image distance when the object distance is 10.0 cm. Calculate the magnification and describe the image.
The object is at the focal point. Substitute p = 10.0 cm and f = 10.0 cm into the mirror equation:
{\frac{1}{10.0\,\mathrm{cm}}}+{\frac{1}{q}}={\frac{1}{10.0\,\mathrm{cm}}}
{\frac{1}{q}}=0\quad\rightarrow{~~~{}{q=\infty}}
Because M = -q/p, the magnification is also infinite.
(c) Locate the image distance when the object distance
is 5.00 cm. Calculate the magnification and describe the image.
Once again, substitute into the mirror equation:
{\frac{1}{5.00~c\mathrm{m}}}+{\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 cm
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}}^{\prime} | + | – | ||
| Magnification | M | + | – | ||
