Question

part a
what is the minimum angular speed at which michelsons eight - sided mirror would have had to rotate to reflect light into an observers eye by succeeding mirror faces (1/8 of a revolution, see the fugure)?(figure 1)
express your answer to two significant figures and include the appropriate units.

Explanation:

Step1: Calculate the time for light to travel

The distance \(d = 35\times10^{3}\text{ m}\), and the speed of light \(c=3\times 10^{8}\text{ m/s}\). The time for light to travel to the stationary mirror and back is \(t=\frac{2d}{c}\).

$$t=\frac{2\times35\times 10^{3}}{3\times 10^{8}}=\frac{70\times10^{3}}{3\times 10^{8}}=\frac{7}{3}\times10^{-4}\text{ s}$$

Step2: Relate angular displacement and time

The mirror rotates \(\frac{1}{8}\) of a revolution. The angular displacement \(\theta=\frac{2\pi}{8}=\frac{\pi}{4}\text{ rad}\). Using the formula \(\omega=\frac{\theta}{t}\).

$$\omega=\frac{\frac{\pi}{4}}{\frac{7}{3}\times10^{-4}}=\frac{3\pi}{28}\times10^{4}$$

$$=\frac{3\times3.14}{28}\times10^{4}\approx 3.4\times10^{3}\text{ rad/s}$$

Answer:

\(\omega = 3.4\times 10^{3}\text{ rad/s}\)