Question

a student wants to scale down michelsons light - speed experiment to a size that will fit in one room. a six - sided mirror is available, and the stationary mirror can be mounted 16 m from the rotating mirror.
part a
if the arrangement is otherwise as shown in fig. 22 - 10 in the textbook, at what minimum rate must the mirror rotate? express your answer using two significant figures.
$\omega=\square\\ rev/s$

Explanation:

Step1: Calculate the time for light to travel to the stationary mirror and back

The distance \(d = 16\space m\), and the speed of light \(c=3\times 10^{8}\space m/s\). The time \(t\) for light to travel to the stationary mirror and back is \(t=\frac{2d}{c}\). Substituting the values, we get \(t=\frac{2\times16}{3\times 10^{8}}=\frac{32}{3\times 10^{8}}\space s\).

Step2: Determine the angular displacement of the mirror

For a six - sided mirror, to get the light to reflect properly (so that the next face of the mirror is in place when the light returns), the mirror must rotate by \(\frac{1}{6}\) of a full - rotation (\(\Delta\theta=\frac{2\pi}{6}=\frac{\pi}{3}\space rad\)) in the time \(t\) that the light is traveling.

Step3: Use the formula for angular speed \(\omega=\frac{\Delta\theta}{t}\)

Substitute \(\Delta\theta = \frac{\pi}{3}\space rad\) and \(t=\frac{32}{3\times 10^{8}}\space s\) into \(\omega=\frac{\Delta\theta}{t}\). Then \(\omega=\frac{\frac{\pi}{3}}{\frac{32}{3\times 10^{8}}}\space rad/s\). Simplifying, \(\omega=\frac{\pi\times10^{8}}{32}\space rad/s\).
To convert from \(rad/s\) to \(rev/s\), use the conversion factor \(1\space rev = 2\pi\space rad\). So \(\omega=\frac{\frac{\pi\times 10^{8}}{32}}{2\pi}\space rev/s=\frac{10^{8}}{64}\space rev/s\approx1.6\times 10^{6}\space rev/s\)

Answer:

\(\omega = 1.6\times 10^{6}\space rev/s\)