Chapter 44: Problem 4

Question: A Luminous point is moving at speed v_O toward a spherical mirror, along its axis. Show that the speed of its image is moving at

$$v_I~=~-v_O \left({R\over 2O-R}\right)$$

Assume that the mirror is concave, with $R=15~{\rm cm}$ and $v_O=5.0~{\rm cm/s}$. Find the speed of the image if the object is far outside the focal point ( $O=75~{\rm cm}$), if it is near the focal point ( $O=7.7~{\rm cm}$) and close to the mirror ( $O=0.15~{\rm cm}$).

Solution: For an oject located at O, we know that the image is located at

$${1\over O}+{1\over I}~=~{2\over R}$$

which can be written as

$$I~=~{RO\over 2O-R}$$

Taking a single time derivative gives

$${dI\over dt}~=~-{R^2\over (2O-R)^2} {dO\over dt}$$

which is what we are required to prove.

1.

$O=75~{\rm cm}$ gives $v_I~=~-0.062~{\rm cm/s}$

2.

$O=7.7~{\rm cm}$ gives $v_I~=~-7.0\times 10^3~{\rm cm/s}$

3.

$O=0.15~{\rm cm}$ gives $v_I~=~-5.21~{\rm cm/s}$