Question 23.3: Images Formed by a Convex Mirror GOAL Calculate properties o...
Images Formed by a Convex Mirror
GOAL Calculate properties of a convex mirror.
PROBLEM An object 3.00 cm high is placed 20.0 cm from a convex mirror with a focal length of magnitude 8.00 cm. Find (a) the position of the image, (b) the magnification of the mirror, and (c) the height of the image.
STRATEGY This problem again requires only substitution into the mirror and magnification equations. Multiplying the object height by the magnification gives the image height.
(a) Find the position of the image.
Because the mirror is convex, its focal length is negative.
Substitute into the mirror equation:
{\frac{1}{p}}+{\frac{1}{q}}={\frac{1}{f}}
{\frac{1}{20.0\,\mathrm{cm}}}+{\frac{1}{q}}={\frac{1}{-8.00\,\mathrm{cm}}}
Solve for q:
q = -5.71 cm
(b) Find the magnification of the mirror.
Substitute into Equation 23.2:
M={}-\frac{q}{p}={}-\left(\frac{-5.71\mathrm{~cm}}{20.0\mathrm{~cm}}\right)=~0.286
M=\frac{h^{\prime}}{h}=-\frac{q}{p} [23.2]
(c) Find the height of the image.
Multiply the object height by the magnification:
h^{\prime}=h M=(3.00\mathrm{~cm})(0.286)=~0.858
REMARKS The negative value of q indicates the image is virtual, or behind the mirror, as in Figure 23.13c. The image is upright because M is positive.
