The following data are utilized in an experiment to measure the...

Time taken by light to travel to the mirror and back is $t = \dfrac{2D}{c}$.

For the image to be completely eclipsed (not observed) at the minimum angular speed, the wheel must rotate such that the light passing through a gap is blocked by the adjacent tooth upon its return.

The angle between the center of a gap and the center of the adjacent tooth is \theta = \dfrac{360^{}^{{}^{\circ}}}{2N}.

Given $N = 180$, we get \theta = \dfrac{360^{}^{{}^{\circ}}}{2 \times 180} = 1^{}^{{}^{\circ}}.

The time taken by the wheel to rotate by this angle is $t = \dfrac{\theta}{\omega}$.

Equating the two expressions for time:

$\dfrac{2D}{c} = \dfrac{\theta}{\omega}$

$\omega = \dfrac{\theta c}{2D}$

Substituting the given values ($D = 12 \times 10^3 \text{ m}$, $c = 3 \times 10^8 \text{ m/s}$, \theta = 1^{}^{{}^{\circ}}):

\omega = \dfrac{1^{}^{{}^{\circ}} \times 3 \times 10^8}{2 \times 12 \times 10^3}

\omega = \dfrac{3 \times 10^5}{24} = 12500^{}^{{}^{\circ}}\text{/s}