Question

A particle moves towards a concave mirror of focal length 30 c m along its acis and with a constant speed of 4 cm/s what is the speed of its Image when the particle is at 90 cm from the mirror

Answer 1

Answer on Question #55455, Physics / Optics

A particle moves towards a concave mirror of focal length $30~\mathrm{cm}$ along its axis and with a constant speed of $4~\mathrm{cm/s}$. What is the speed of its image when the particle is at $90~\mathrm{cm}$ from the mirror?

Solution:

The equation for image formation by rays near the optic axis (paraxial rays) of a mirror has the same form as the thin lens equation:

$$\frac {1}{o} + \frac {1}{i} = \frac {1}{f}$$

where $o =$ object distance, $i =$ image distance, $f =$ focal length.

Differentiation with respect to time

$$\frac {d}{d t} \left(\frac {1}{o}\right) + \frac {d}{d t} \left(\frac {1}{i}\right) = \frac {d}{d t} \left(\frac {1}{f}\right)$$

f=const

Thus,

$$- \frac {1}{o ^ {2}} \frac {d o}{d t} - \frac {1}{i ^ {2}} \frac {d i}{d t} = 0$$ $$\frac {d i}{d t} = - \frac {i ^ {2}}{o ^ {2}} \frac {d o}{d t}$$

Here:

$$\frac {d o}{d t} = 4~\mathrm{cm/s}$$ $$o = 90~\mathrm{cm}$$

From mirror equation

$$\frac {1}{i} = \frac {1}{30} - \frac {1}{90} = \frac {2}{90}$$ $$i = \frac {90}{2} = 45~\mathrm{cm}$$

Hence,

$$\frac {d i}{d t} = - \frac {45 ^ {2}}{90 ^ {2}} \cdot 4 = - 1~\mathrm{cm/s}$$

Answer: So image is moving at 1 cm/s away from mirror

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