Chapter 24, Problem 6P

A concave spherical mirror has a radius of curvature of 10.0 cm. Calculate the location and size of the image formed of an 8.00-mm-tall object whose distance from the mirror is (a) 15.0 cm. (b) 10.0 cm. (c) 2.50 cm. and (d) 10.0 m.

6. Repeat the previous problem, except use a convex mirror with the same magnitude of focal length.

24.6. Set Up: To find the location of the image use $\frac{1}{s}+\frac{1}{s\prime }=\frac{2}{R}$. For a convex mirror R < 0. To find the height of $m=\frac{y\prime }{y}=\frac{s\prime }{s}$

Solve: (a) s = 15.0 cm. $\frac{1}{s\prime }=\frac{2}{R}-\frac{1}{s}=\frac{2s-R}{Rs}$, so s =    = $s\prime =\frac{Rs}{2s-R}=\frac{(-10.0\text{ cm)(15}\text{.0 cm)}}{30.0\text{ cm +10}\text{.0 cm}}$ = -3.75 cm.

$m=-\frac{s\prime }{s}=\frac{3.75\text{ cm}}{\text{15}\text{.0 cm}}$ = +0.250

y′ = my = +2.00 mm. The image is 3.75 cm behind the mirror. It is 2.00 mm tall and is upright.

(b) s = 10.0 cm. $s\prime =\frac{(-10.0\text{ cm)(10}\text{.0 cm)}}{20.0\text{ cm +10}\text{.0 cm}}$= −3.33 cm. $m=\frac{s\prime }{s}=\frac{-3.33\text{ cm}}{\text{15}\text{.0 cm}}=-3.75$ = +0.333.

y′ = my = +2.67 nun. The image is 3.33 cm behind the mirror. It is 2.67 mm tall and is upright.

(c) s = 2.50 cm. $s\prime =\frac{(-10.0\text{ cm)(25}\text{.0 cm)}}{5.00\text{ cm +10}\text{.0 cm}}$= −1.67 cm. $m=-\frac{s\prime }{s}=\frac{-1.67\text{ cm}}{\text{2}\text{.50 cm}}$ = +0.667.

y′ = my = +5.33 nun. The image is 1.67 cm behind the mirror. It is 5.33 mm tall and is upright.

(d) s = 1000 cm. $s=\frac{(-10.0\text{ cm)(1000 cm)}}{2000.0\text{ cm +10}\text{.0 cm}}$ = −4.98 cm.

$m=-\frac{s\prime }{s}=\frac{-4.98\text{ cm}}{\text{1000 cm}}$ = +4.98 × 10⁻³

y′ = my = +0.040 mm. The image is 4.98 cm behind the mirror. It is 0.040 mm tall and is upright.