Assumption 18.3.1. Huygen’s Principle.
Every point along a wave front can be considered a source of a waves emanating from that point in the direction of light propagation.
Section 18.3 Two-slit Interference
Start by playing with the
light interference simulation for a few minutes. Click on the “Slits” tab and select the laser. On the right hand side, increase the number of slits to two.Write down some observations. What happens when you move the slit width to its minimum and maximum values? What happens as you increase the slit separation? What happens when you increase and decrease the frequency?
Subsection Video Lesson: Two-slit Interference
Subsection Deriving the Interference Pattern
A powerful wave property allows complicated light waves to be treated as superpositions of simpler light waves. In particular, any wave front of light can be treated as a collection of point sources of spherical waves, an idea known as Huygen’s Principle.
As a simple example, when a coherent beam of light is incident on two very narrow slits, as shown in the figure below, the light that emerges from the slits can be approximated as two spatially-separated point sources of coherent, in-phase spherical light waves.
Applying the general rules for Maximum Constructive Interference and Complete Destructive Interference from the Interference and Path Length Difference Sections allows you to determine the spacing of the interference pattern by looking at the path length difference of the rays exiting the slits. The figure below gives a magnified look at the paths traveled by the two rays.
If the screen is far away from the sources, the two rays in the figure above may be treated as parallel, allowing you to calculate the path-length difference using trigonometry.
Assumption 18.3.5. Two-Source Path Length Difference.
For a screen that is far away from two sources of light, the path length difference can be approximated as
\begin{equation*}
\Delta D = d \sin \theta
\end{equation*}
where \(d\) is the distance between slits and \(\theta\) is the angle between the center line and the path to a spot on the screen.
Historical Note 18.3.6. The Double-Slit Experiment.
The Double-Slit Experiment was originally conducted by Thomas Young in 1801 to show that light does indeed behave like a wave. Just like water waves interacting with each other, as light travels through two slits, the crests and troughs of the light waves will interact with each other and produce an interference pattern on the screen behind the slits.
Representation 18.3.7. Intensity Graph.
An intensity graph shows the intensity of light in some region, often on a screen. The example below is for light on a distant screen that has passed through two extremely narrow slits.
Exercises Activities
1. Constructive Interference.
Combine the geometric path-length difference found above and the general rule for Maximum Constructive Interference to find an equation relating \(d\text{,}\) \(\lambda\text{,}\) and \(\theta\) for the bright spots on the screen.
Answer.
\begin{equation*}
d \sin \theta = m\lambda
\end{equation*}
2. Destructive Interference.
Combine the geometric path-length difference found above and the general rule for Complete Destructive Interference to find an equation relating \(d\text{,}\) \(\lambda\text{,}\) and \(\theta\) for the dark spots on the screen.
Answer.
\begin{equation*}
d \sin \theta = \left( m + \frac{1}{2} \right)\lambda
\end{equation*}
3. The Small Angle Approximation.
In the video, you learned about using the Small-angle Approximation for interference. Use the small-angle approximation to write an equation relating \(d\text{,}\) \(y\text{,}\) \(L\text{,}\) and \(\lambda\) for points of maximum constructive interference.
Answer.
\begin{equation*}
\sin\theta \approx \theta \approx \tan\theta = \frac{y}{L}
\end{equation*}
Plugging this into the maximum constructive interference expression above gives:
\begin{equation*}
d \frac{y}{L} = m\lambda
\end{equation*}
4. Measure the Wavelength.
You are conducting the double slit experiment and you have a light source with an unknown wavelength. The slits are separated by a distance of \(0.5 \mathrm{~mm}\) and your screen is \(3.4 \mathrm{~m}\text{.}\) You measure the third bright fringe to be at a position of \(4.7 \mathrm{~mm}\) to the right of the center line on the screen. What is the wavelength of light?
