Young's double slit equation is well known result that shows the light has a wave property.
- https://www.khanacademy.org/science/physics/light-waves/interference-of-light-waves/v/youngs-double-slit-part-2
- https://en.wikipedia.org/wiki/Young%27s_interference_experiment
A typical illustration of Young's interference experiment is shown in Figure 1. This figure confuses novice learners since at the point C, two parallel rays intersect. But how could possibly parallel rays intersect? In Euclidean geometry, parallel two lines will never intersect.
 |
| Figure 1: Typical explanation figure: Two parallel rays and path length difference \[\Delta x\] |
 |
| Figure 2: Precise figure: Two rays intersect at point C to cause interference. |
Figure 2 shows more precise figure of Young's interference experiment. At the point C, two rays meet and cause an interference depends on the path difference from the holes H1 and H2. When you zoom in to the two holes part, you see the precise path length difference \(\Delta x_r\) is slightly differ from the \(\Delta x\). The real path length difference \(\Delta x_r\) is
\[ d \sin (\theta - \alpha) = \Delta x_r,\]
where the distance between two slits is \(d\).
 |
Figure 3: Precise figure: Two rays meet at point C.
|
Many explanations noted: The distance \(f\) is enough large compare to the \(d\).
This implies \(\alpha\) is quite small. But I usually doesn't see this hint, ``\(\alpha\) is quite small''. I would like to see this hint.
If I see this hint about \(\alpha\), then I could say, ok, \(\alpha\) is small, then \(\cos \alpha \approx 1\), \(\sin \alpha \approx 0\). But I would like to say what is the order since this is also relative to \(\theta\). And \(\theta\) is usually relatively small, too. Why we don't say \(\sin \alpha \approx 0\), \(\cos \theta \approx 0\)? If someone said so, then distance is always 0 and this figure doesn't make any sense. When I was a high school student, I had this problem.
One of the reason here is when \(f\) is larger, actually \(\theta\) doesn't change and \(\alpha\) becomes smaller, so, I could figure out that
\[
\sin (\theta - \alpha) \approx \sin \theta.
\]
Maybe we should realize this by myself, but it is nice if the textbook tells that more directly instead of, we assume \(f\) is enough large.
One more observation is Maclaurin expansion of \(\sin\) and \(\cos\). If you learn calculus, this is simple, but for novice learner it is also nice to mention that.
When we expand the \(\sin (\theta - \alpha)\),
\[
\sin (\theta - \alpha) = \sin \theta \cos \alpha - \cos \theta \sin \alpha.
\]
Now, consider the Maclaurin expansion to approximate the \(\sin x\) and \(\cos x\), just the first two teams are:
\[
\sin x \approx x - \frac{x^3}{3!}\\
\cos x \approx 1 - \frac{x^2}{2!}
\]
When \(x\) is small, actually only the first term of \(\sin\) and \(\cos\) matters if they are multiplied. Since especially the second term of $\cos$ is much smaller than the \(\sin\)'s first term. \(\alpha\) is so small, we can easily assume it is less than 1 degree, which is 0.017 rad. Even \(\theta\) is 1 degree, 0.017 rad, the \(\cos\) part became 0.00029. It is almost 100 times smaller than the \(\sin\) term. And \(\alpha\) can be much smaller.
Therefore, we can use Figure 1 for how to calculate the interference pattern.
It is useful, but I had a problem of the figure, since it seems two parallel rays meet at the point C. And parallel rays should not intersect.
After I learned calculus and how to approximate the \(\sin\) and \(\cos\), then this became clear, but I think it is nice to write this small article for the learners like high school student me.
Here is my experiment results.
 |
Figure 4: Set up (left one hole, middle and right two holes).
|
 |
Figure 5: One hole case (no interference)
|
 |
| Figure 6: two holes (interference) |
Posts
