Waves are characterized by their ability to constructively and destructively *interfere*. Light waves which interfere with themselves after interaction with a small aperture or target are said to *diffract*. Light creates interference patterns when passing through holes (“slits”) in an obstruction such as paper or the surface of a CD, or when passing through a thin film such as soap.

\begin{align*} m \lambda = d \sin{\theta} \end{align*}

Double slit interference maxima. \begin{align*}m\end{align*} is the order of the interference maximum in question, \begin{align*}d\end{align*} is the distance between slits. and \begin{align*}\theta\end{align*} is the angular position of the maximum.

\begin{align*} m \lambda = d \sin{\theta} \end{align*}

Single slit interference maxima. \begin{align*}m\end{align*} and \begin{align*}\theta\end{align*} are defined as above and \begin{align*}d\end{align*} is the width of the slit.

\begin{align*} m \lambda = d \sin{\theta} \end{align*}

Diffraction grating interference maxima. \begin{align*}m\end{align*} and \begin{align*}\theta\end{align*} are defined as above and \begin{align*}d\end{align*} is the distance between the lines on the grating.

\begin{align*} m \lambda = 2nd\end{align*}

Thin film interference: \begin{align*}n\end{align*} is the index of refraction of the film, \begin{align*}d\end{align*} is the thickness of the film, and \begin{align*}m\end{align*} is an integer. In the film interference, there is a \begin{align*}\lambda/2\end{align*} delay (phase change) if the light is reflected from an object with an index of refraction greater than that of the incident material.

\begin{align*} \frac{1}{f} = \frac{1}{d_0} +\frac{1}{d_i} \end{align*}

#### Example 1

A typical experimental setup for an interference experiment will look something like this:

\begin{align*} \text{Recall: for constructive interference we require a path difference of} \; m \lambda = d \sin{\theta} \end{align*}

\begin{align*} \text{d = spacing of slits}\end{align*}

\begin{align*} \text{m = number of maximum (i.e. m = 0 is the central maximum, m = 1 is the 1st maximum and thus the first dot to the right and left of the center, etc.)}\end{align*}

\begin{align*} \text{L = distance form diffraction grating to screen} \end{align*}

\begin{align*} \Delta y = \text{distance from nth spot out from the center (if m = 1 then it is the 1st spot)} \end{align*}

Remember, **maximum** = place where waves constructively interfere, and **minimum** = place where waves destructively interfere.

Because the screen distance *L* is much larger than the slit distance *d*, one can see that

\begin{align*} \frac{\Delta y}{L} = \text{tan} \theta \approx \text{sin} \theta \end{align*}.

Thus, the condition for a first maximum becomes

\begin{align*} \lambda = d \sin{\theta} = d \frac{\Delta y}{L}\end{align*}

One can now easily calculate where the first maximum should appear if given the wavelength of the laser light, the distance to the screen and the distance between slits.

First Maximum: \begin{align*} \Delta y = \frac {L \lambda}{d} \end{align*}

#### Example 2

White light (which is comprised of all wavelengths and thus all colors) separates into a rainbow pattern as shown below.

Each wavelength of light has a unique interference pattern given by the equation above. Thus all the wavelengths (i.e. colors) have a unique \begin{align*} \Delta y \end{align*} based on the equation given at the end of Example 1. This is how white light separates out into its individual wavelengths producing a rainbow after going through a diffraction grating.

### Interactive Simulation

### Review

- In your laboratory, light from a \begin{align*}650 \;\mathrm{nm}\end{align*} laser shines on two thin slits. The slits are separated by \begin{align*}0.011 \;\mathrm{mm}\end{align*}. A flat screen is located \begin{align*}1.5 \;\mathrm{m}\end{align*} behind the slits.
- Find the angle made by rays traveling to the third maximum off the optic axis.
- How far from the center of the screen is the third maximum located?
- How would your answers change if the experiment was conducted underwater?

- Again, in your laboratory, \begin{align*}540 \;\mathrm{nm}\end{align*} light falls on a pinhole \begin{align*}0.0038 \;\mathrm{mm}\end{align*} in diameter. Diffraction maxima are observed on a screen \begin{align*}5.0 \;\mathrm{m}\end{align*} away.
- Calculate the distance from the central maximum to the first interference maximum.
- Qualitatively explain how your answer to (a) would change if you:
- move the screen closer to the pinhole
- increase the wavelength of light
- reduce the diameter of the pinhole

- Students are doing an experiment with a Helium-neon laser, which emits \begin{align*}632.5 \;\mathrm{nm}\end{align*} light. They use a diffraction grating with \begin{align*}8000\end{align*} lines/cm. They place the laser \begin{align*}1 \;\mathrm{m}\end{align*} from a screen and the diffraction grating, initially, \begin{align*}95 \;\mathrm{cm}\end{align*} from the screen. They observe the first and then the second order diffraction peaks. Afterwards, they move the diffraction grating closer to the screen.
- Fill in the
**Table**(below) with the*expected*data based on your understanding of physics. Hint: find the general solution through algebra*before*plugging in any numbers. - Plot a graph of the first order distance as a function of the distance between the grating and the screen.
- How would you need to manipulate this data in order to create a
*linear*plot? - In a real experiment what could cause the data to deviate from the expected values? Explain.
- What safety considerations are important for this experiment?
- Explain how you could use a diffraction grating to calculate the unknown wavelength of another laser.

Distance of diffraction grating to screen \begin{align*}(cm)\end{align*} Distance from central maximum to first order peak \begin{align*}(cm)\end{align*} \begin{align*}95\end{align*} \begin{align*}75\end{align*} \begin{align*}55\end{align*} \begin{align*}35\end{align*} \begin{align*}15\end{align*} - Fill in the
- A crystal of silicon has atoms spaced \begin{align*}54.2 \;\mathrm{nm}\end{align*} apart. It is analyzed as if it were a diffraction grating using an \begin{align*}x-\end{align*}ray of wavelength \begin{align*}12 \;\mathrm{nm}\end{align*}. Calculate the angular separation between the first and second order peaks from the central maximum.
- Laser light shines on an oil film \begin{align*}(n = 1.43)\end{align*} sitting on water. At a point where the film is \begin{align*}96 \;\mathrm{nm}\end{align*} thick, a \begin{align*}1^{st}\end{align*} order dark fringe is observed. What is the wavelength of the laser?
- You want to design an experiment in which you use the properties of thin film interference to investigate the variations in thickness of a film of water on glass.
- List all the necessary lab equipment you will need.
- Carefully explain the procedure of the experiment and draw a diagram.
- List the equations you will use and do a sample calculation using realistic numbers.
- Explain what would be the most significant errors in the experiment and what effect they would have on the data.

### Review (Answers)

- a. \begin{align*}10.2^\circ\end{align*} b. \begin{align*}27\;\mathrm{cm}\end{align*} c. \begin{align*}20\;\mathrm{cm}\end{align*}
- a. \begin{align*}0.72\;\mathrm{m}\end{align*}
- a. Expected data: 48.07, 37.95, 27.83, 17.71, 7.59 b. Δy = 0.506L c. The plot is linear d. mis-measurement of the fringes and their spacing e. When handling lasers, always wear eye protection
- \begin{align*}13.5^\circ\end{align*}
- \begin{align*}549 \;\mathrm{nm}\end{align*}
- Answers will vary