Mathematics of Interference
15.2 Mathematics of Interference
By the end of this section, you will be able to:
- Determine the angles for bright and dark fringes for double slit interference
- Calculate the positions of bright fringes on a screen
What we will do now is make sense of this diagram of the double slit experiment:
Watch this video to see how we can determine what the diffraction pattern will look like when we shine light through two slits:
Let’s break down what we can get from this picture:
Notice there is a right triangle that has a base = R and a height equal to y. Because this is a right triangle, we can say
That distance, y, is measured from the center of the pattern to some point on the pattern. If the distance y is measured to a bright spot, we know that is due to constructive interference between light rays r1 and r2, so the path difference is equal to 1 wavelength, or 2 wavelengths, or 3 wavelengths, etc. We say, at this position y:
where m is an integer (m = 0, 1, 2, 3, …). The angle θ in this expression is the same angle as in , so we will want to relate these two equations.
We are going to use a small angle approximation, because this angle θ is really, really small. You can’t tell from my drawing, but R >> d so my drawing is definitely NOT to scale. If you imagine that R is something like 2 or 3 meters and d is something like 1 µm, then you will see how small θ really is. The approximation we will use is, for small angles:
We can say this because, if you look at a plot of sin θ and the tan θ, where θ is small (near 0°) the graphs overlap, meaning they are the same value:
So the equation for constructive interference at a point y on the screen becomes
If you know d (the size of the spacing between the slits), R (the distance from the slits to the screen), and λ (the wavelength of the light), then you can predict all the places on the screen where you will see a bright spot:
When you measure from the center of the pattern (where y = 0), you will see a bright spot at
and another at
and another at
and so on.
You can also use this relationship to determine other things. Like, if you don’t know the spacing between the slits but you know R, λ, and you can measure y, then you can solve for d. This is a way to measure microscopic things (like a slit spacing of ~ 1 µm) with a macroscopic measurement (y ~ cm).
Now let’s say you measure y from the center of the pattern to a dark spot. This is due to destructive interference between the rays that come from the two slits. We can use the condition for destructive interference (the path difference is equal to 1/2λ, or 3/2λ, or 5/2λ, etc.) and the same small angle approximation to get
When you measure from the center of the pattern (where y = 0), you will see a dark spot at
and another dark spot at
and another at
and so on.
|A plane monochromatic light wave is incident on a double slit. As the slit separation increases, what happens to the separation between the interference fringes on the screen?|
|Suppose Young’s double-slit experiment is performed in air using red light and then the apparatus is immersed in water. What happens to the interference pattern on the screen?|
|Coherent monochromatic light of wavelength 632.8 nm passes through a pair of thin parallel slits. The figure shows the central portion of the pattern of bright fringes viewed on a screen 1.40 m beyond the slits.|
What is the distance between the two slits?
The images below show the interference pattern formed by two wave sources (located at the red dots) emitting waves of the same wavelength in phase with one another. Looking at the images, what seems to affect the spacing of the interference maxima (light areas) and minima (dark areas)? Describe the differences and similarities between the two images.
You measure θ to be 0.0190° for the first order maximum. What is the wavelength of this light?
How far apart are the slits’ midpoints?