# Two source interference

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We examine interference from two coherent sources.

## Waves on a pond:

Think of when you drop a pebble into a pond, you will see circular waves eminate from the point where you dropped the pebble.

When you drop two pebbles side by side you will see a much more complicatedpattern:

Likewise with electromagnetic waves, you can get interesting interference phenomenawhen waves eminate from two point sources.

## Two point sources

Lets take a particular example of two point sources separated by a distance d. The waves emitted by point source are spherical and thus can be written $\stackrel{⃗}{E}=\frac{{\stackrel{⃗}{E}}_{0}}{r}{\mathrm{cos}}\left(kr-\omega t\right)$ To make the problem easier we will make the $k$ 's the same for the two sources. Also lets set the ${E}_{0}$ 's to be the same as well.

The the only difference in the waves will be the $r$ 's, that is ${\stackrel{⃗}{E}}_{1}=\frac{{\stackrel{⃗}{E}}_{0}}{{r}_{1}}{\mathrm{cos}}\left(k{r}_{1}-\omega t\right)$ ${\stackrel{⃗}{E}}_{2}=\frac{{\stackrel{⃗}{E}}_{0}}{{r}_{2}}{\mathrm{cos}}\left(k{r}_{2}-\omega t\right)$ Now there is a slightly subtle point here that is important to understand. In the denominator it is sufficient to say that ${r}_{1}\approx {r}_{2}$ and just call it $r$ . We assume that we are far enough away that the differences between ${r}_{1}$ and ${r}_{2}$ are too small to matter. However this is not true in the argument of the harmonic function. There, very small differences between ${r}_{1}$ and ${r}_{2}$ can have a big effect. So lets define ${r}_{1}={r}_{2}=R$ $\begin{array}{c}I\propto {⟨{\left(\frac{{E}_{0}}{R}{\mathrm{cos}}\left(k{r}_{1}-\omega t\right)+\frac{{E}_{0}}{R}{\mathrm{cos}}\left(k{r}_{2}-\omega t\right)\right)}^{2}⟩}_{T}\\ ={⟨\frac{{E}_{0}^{2}}{{R}^{2}}{{\mathrm{cos}}}^{2}\left(k{r}_{1}-\omega t\right)⟩}_{T}+{⟨\frac{{E}_{0}^{2}}{{R}^{2}}{{\mathrm{cos}}}^{2}\left(k{r}_{2}-\omega t\right)⟩}_{T}\\ \text{ }+{⟨2\frac{{E}_{0}^{2}}{{R}^{2}}{\mathrm{cos}}\left(k{r}_{1}-\omega t\right){\mathrm{cos}}\left(k{r}_{2}-\omega t\right)⟩}_{T}\\ =\frac{1}{2}\frac{{E}_{0}^{2}}{{R}^{2}}+\frac{1}{2}\frac{{E}_{0}^{2}}{{R}^{2}}+\\ \text{ }+{⟨2\frac{{E}_{0}^{2}}{{R}^{2}}{\mathrm{cos}}\left(k{r}_{1}-\omega t\right){\mathrm{cos}}\left(k{r}_{2}-\omega t\right)⟩}_{T}\end{array}$ Now to evaluate the final term we use ${\mathrm{cos}}\left(\theta -\phi \right)={\mathrm{cos}}\theta {\mathrm{cos}}\phi +{\mathrm{sin}}\theta {\mathrm{sin}}\phi$ and write  So we have $\begin{array}{c}I\propto \frac{1}{2}\frac{{E}_{0}^{2}}{{R}^{2}}+\frac{1}{2}\frac{{E}_{0}^{2}}{{R}^{2}}+2\frac{{E}_{0}^{2}}{{R}^{2}}\frac{1}{2}{\mathrm{cos}}k\Delta r\\ =\frac{1}{{R}^{2}}\left({E}_{0}^{2}+{E}_{0}^{2}{\mathrm{cos}}k\Delta r\right)\\ =\frac{1}{{R}^{2}}{E}_{0}^{2}\left(1+{\mathrm{cos}}k\Delta r\right)\end{array}$

Clearly I will be a maximum when the cosine is = +1 $k\Delta r=2n\pi \text{ }n=0,1,2\dots$ $\frac{2\pi }{\lambda }\Delta r=2n\pi$ $\Delta r=n\lambda$ There will be a minimum when the cosine is = -1 $k\Delta r=n\pi \text{ }n=1,3,5\dots$ $\Delta r=\frac{n\lambda }{2}\text{ }n=1,3,5\dots$ So you get light and dark bands which are called interference fringes.To reiterate; we have two rays of light eminating from two point sources. Wehave looked at the combined wave at some point, a distance ${r}_{1}$ from the first source and a distance  from the second source. In that case we find that the intensity is proportional to $\frac{1}{{R}^{2}}{E}_{0}^{2}\left(1+{\mathrm{cos}}k\Delta r\right)\text{.}$ To make things easier we can redefine ${E}_{0}$ to be the amplitude of the waves at the point under consideration, that is $I={\epsilon }_{0}c{E}_{0}^{2}\left(1+{\mathrm{cos}}k\Delta r\right)\text{.}$ Or we can say ${I}_{0}=\epsilon c{E}_{0}^{2}/2$ and write $I=2{I}_{0}\left(1+{\mathrm{cos}}k\Delta r\right)\text{.}$

Say we place a screen a distance S away from the two sources:

In this case we see that $\Delta r=d{\mathrm{sin}}\theta$ So we have maxima at $\Delta r=n\lambda =d{\mathrm{sin}}\theta \text{.}$ The angle between two maxima is given by ${\mathrm{sin}}{\theta }_{n+1}-{\mathrm{sin}}{\theta }_{n}=\frac{\lambda }{d}$ or for small $\theta$ $\Delta \theta =\frac{\lambda }{d}$ Notice how when the sources are moved far apart the effect maxima become very closetogether so the screen appears to be uniformly illuminated. If a screen is placed a distance S away the maxima on the screen will occur such that $d{\mathrm{sin}}\theta =n\lambda$ but in the small angle limit ${\mathrm{sin}}\theta ={\mathrm{tan}}\theta =\frac{y}{S}$ which implies $y=\frac{n\lambda S}{d}$ likewise minima will occur at $y=\frac{n\lambda S}{2d}\text{ }n=1,3,5\dots$ using ${\mathrm{cos}}\theta =2{{\mathrm{cos}}}^{2}\frac{\theta }{2}-1$ we can rewrite $I=2{I}_{0}\left(1+{\mathrm{cos}}k\Delta r\right)$ as $I=4{I}_{0}{{\mathrm{cos}}}^{2}\frac{k\Delta r}{2}$

## Young's double slit

Young's double slit.is an excellent example of two source interference. The equations for this are what we worked out for two sources above. Interferenceis an excellent way to measure fine position changes. Small changes in $\Delta r$ make big observable changes in the interference fringes.

## Michelson interferometer

A particularly useful example of using interference is the Michelson interferometer. This can be used to measure the speed of light in a medium,measure the fine position of something, and was used to show that the speed of light is a constant in all directions.

When $\Delta r$ , the path length difference in the two arms is $\Delta r=n\lambda$ then the rays of light in the traveling down the center of the apparatus will interfere constructively. As you move off axis the light travels slightlydifferent lengths and so you get rings of interference patterns. If you have set up the apparatus so that $\Delta r=n\lambda$ and then move one of the mirrors a quarter wavelength then $\Delta r=n\lambda +\frac{1}{2}\lambda$ and you get destructive interference of the central rays. Thus you can easily position things to a fraction of a micron with such a set up.

What really matters is the change in the optical pathlength. For example you could introduce a medium in one of the paths that has a different index ofrefraction, or different velocity of light. This will change the optical pathlength and change the interference at the observer. Thus you can measurethe velocity of the light in the introduced medium.

Michelson and Morely used this technique to try to determine if the speed of light is different in different directions. They put the whole apparatus on arotating table and then looked for changes in the interference fringes as it rotated. They saw no changes. In fact they went so far as to wait to see whathappened as the earth rotated and orbited and saw no changes. They thus concluded that the speed of light was the same in all directions (which nobodyat the time believed, even though that is the conclusion you draw from Maxwell's equations.)

## Ring gyroscope

Another application of interference is a a gyroscope, ie. as device to measure rotations.

If the apparatus is rotating, then the pathlengths are different in different directions and so you can use the changes in the interference patterns tomeasure rotations. This is in fact how gyroscopes are implemented in modern aircraft.

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