## Section46.10Fundamentals of Optics Bootcamp

### Subsection46.10.8Miscellaneous

Refer to the Fizeau's experiment shown in Figure 46.3.2. Prove that the speed of light will be equal to $2nNd\text{,}$ where $N$ is the total number of teeth in the wheel, $n$ is the rotation rate of the wheel in number of revolutions per second when the returned light passes through the next opening in the wheel, and $d$ is the distance from the apparatus to the mirror.

Hint

Equated the time for back and forth of ray to the time to rotate to bring next opening of the wheel to line up.

See solution.

Solution

Using the symbol $v$ for the speed of light, we find that the time for light to return to the wheel will be

\begin{equation*} \Delta t = \dfrac{2d}{v}. \end{equation*}

When we look at the motion of the wheel, during this time the wheel needs to rotate so that its the time between the subsequent openings in the wheel in the path of the light. In 1 sec the wheel will rotate through $n\times N$ teeth. Therefore, the time between teeth will be

\begin{equation*} \Delta t' = \dfrac{1}{nN}. \end{equation*}

Equating $\Delta t$ with $\Delta t'$ gives

\begin{equation*} \dfrac{2d}{v} = \dfrac{1}{nN},\ \ \Longrightarrow\ \ v = 2 d nN. \end{equation*}

Refer to the Michelson's apparatus for measuring the speed of light shown in Figure 46.3.4. Prove that the speed of light will be equal to $16nd\text{,}$ where $n$ is the rotation rate of the octagon mirror in the number of revolutions per second, and $d$ is the distance from rotating mirror to the fixed mirror on the Lookout Mountain. Ignore the distance between the curved mirror and the flat mirror on the Lookout Mountain.

Hint

Equate time for light to return to time for wheel to rotate by $\dfrac{1}{8}$ of a turn.

See solution.

Solution

Using the symbol $v$ for the speed of light, we find that the time for light to return to the rotating wheel will be

\begin{equation*} \Delta t = \dfrac{2d}{v}. \end{equation*}

When we look at the motion of the rotating wheel the time to rotate $\dfrac{1}{8}$ of the full rotation will be

\begin{equation*} \Delta t' = \dfrac{1}{8}\times \dfrac{1}{n}. \end{equation*}

Equating $\Delta t$ with $\Delta t'$ gives

\begin{equation*} \dfrac{2d}{v} = \dfrac{1}{8n},\ \ \Longrightarrow\ \ v = 16nd. \end{equation*}

A small light bulb is at the bottom of a tank that has a layer of water and a layer of oil as shown in Figure 46.10.21. Find the path of the ray shown as it comes out in the air. Use a protractor to measure the angle of incidence. If you do not have a protractor, use angle of incidence equal to $27^{\circ}\text{.}$

Hint

We use a protractor to find the angle of incidence at the water/oil interface.

$\theta_{\textrm{oil}} = 30^{\circ}\text{,}$ $\theta_{\textrm{air}} = 37^{\circ}\text{.}$

Solution

The incidence angle turns out to be approximately $27^{\circ}\text{.}$ Using Snell's law at this interface we find that the angle of refraction for the given ray will be $30.2^{\circ}\text{.}$ Since the interfaces are parallel, the angle of incidence at the oil/air interface is also $30.2^{\circ}\text{.}$ Now, an application of Snell's law at the oil/air interface gives the angle of refraction in air to be $37^{\circ}\text{.}$

A ray of light consisting of a red and a blue light is incident on a glass circular plate parallel to a diameter. If the refractive indices of red and blue lights for the given glass are 1.5 and 1.55 respectively, find the first two places where rays come out of the glass plate and the directions of the red and blue lights.

Hint

Apply Snells's law. The normal at each point of the circle is the radial line passing through center of the circle.