Two Source Interference

### 10.3.1 Explain by means of the principle of superposition, the interference pattern produced by waves from two coherent point sources.

Coherent sources: means that the phase difference between the two sources does not change with time. i.e. if they start in phase they will always be in phase.

When two coherent point sources interfere they produce a interference pattern similar to that shown below: The dark lines are areas where the two waves interfere destructively, in between the dark points are areas of maximum amplitude where the waves interfere constructively. When the two waves have both traveled an integer value of wavelengths they interfere constructively (assuming the sources are in phase). If the two waves have both traveled an integer value plus one half of a wavelength they interfere destructively.

Conditions for interference:

Constructive:$\text{Path Difference} = n \lambda$

Destructive: $\text{Path Difference} = (n + \frac{1}{2}) \lambda$

### 10.3.3 Outline Young’s double slit experiment for light and draw intensity distribution of the observed fringe pattern.

In 1804 Thomas Young made the first convincing experiment to prove that light diffracted. A sketch of this apparatus is shown below: For a source he used sunlight, which passed through a filter so it was monochromatic (single colored), the light then passed through a lens designed to create parallel wavefronts (this was required to show diffraction). The light then passed through a single slit and then a double slit. If the light did not diffract the screen at would have been completely dark. Instead of being dark a diffraction pattern was found as shown below: As a side note: this experiment was very convincing proof that light was a wave, after all how can particles diffract and interfere? They do and it’s a bit weird.

So how do we predict where the light and dark fringes will appear? The geometry of the situation determines where the fringes will appear. In Young’s experiment the distance between the silts d is very small when compared to the distance between the silts and the screen at point P. Therefore we can treat the lines $\overline{S_1P}$ and $\overline{S_2P}$ as parallel. The line $\overline{S_1Z}$ is drawn such that $\overline{S_1P}$ and $\overline{ZP}$ are identical lengths. This means that the angle $\widehat{S_2S_1ZP}$ is very close to θ. So the path difference of the two sources is $\overline{S_2Z}$. If the path difference is an integer number of wavelengths the sources will constructively interfere and a bright fringe will appear. So the conditions for a bright fringe to appear is:

(1)
\begin{align} S_2 Z = n \lambda \end{align}

or re-writing the length $S_2 Z$ in terms of d and $\theta$:

(2)
\begin{align} d \sin \theta = n \lambda \end{align}

This last equation is given in the IB formula book.

Now lets look at the spacing between fringes. From the sketch below we can state:

(3)
\begin{align} \tan \theta = \frac{y_n}{D} \end{align}

Where $y_n$ is the distance from the center point to the nth bright fringe and D is the distance between the screen and the slits.

Because θ is small we can make the approximation $\sin \theta = \tan \theta$ , therefore we can substitute into our earlier equation (2) and get:

(4)
\begin{align} d \frac{y_n}{D} = n \lambda \end{align}

Solving for $y_n$:

(5)
\begin{align} y_n = \frac{n \lambda D}{d} \end{align}

Now if we look at:

(6)
\begin{align} \Delta y = y_{n+1} - y_n \end{align}
(7)
\begin{align} \Delta y = \frac{\lambda D}{d} \end{align}

Or in IB notation:

(8)
\begin{align} s = \frac{\lambda D}{d} \end{align}

This last formula is in the IB handbook and gives the distance between two successive fringes.