Now for something less intuitive:
- Single-slit interference
- Equation: nλ = a sin θ
Let’s jump in!
What pattern does diffraction through a slit result in on a screen?
You might think that it causes an even spread of intensity (since no interference seems to occur), BUT an INTERFERENCE PATTERN actually shows up.
There are light bands called MAXIMA & dark bands called MINIMA.
The intensities at the screen can be shown by this graph:
Why does interference occur in single-slit diffraction?
This is to do with the WAVELETS.
The wavelets produced between edges of the slit can be said to interfere.
Let’s look at the maxima & minima:
There is a bright maxima at the centre of the screen, followed by a repeating pattern of maxima & minima going to the sides.
Since the middle is a bright maxima, we will consider the dark MINIMA to count the number of repetitions in the pattern.
The position of a minimum away from the middle is known as its ORDER.
- The first minimum has an order of 1, counted from the centre.
- The next minimum further away has an order of 2.
Generally, we say that a minimum has an order of n, where n = 1,2,3,…
The minima come from the DESTRUCTIVE interference between wavelets leaving the diffraction slit.
There is a relationship between the:
- Slit size
- Angle of the point on the screen from the centre
Once again, this is all due to PATH DIFFERENCE.
What formula can we use to relate the angle & the order of the minima?
Here’s a video with the full proof:
Here’s a brief summary:
Consider this diagram:
In real life (for light diffraction), the difference in angle between the points to the wall is very small, & can be considered to be negligible.
Thus, we can approximate that the angles are both the same θ.
We can pair every point on the top of the edge with a point exactly a/2 below it:
For destructive interference to occur at any point, the path difference of 2 waves to that point must be a whole ODD number of λ/2.
However, EVEN numbers of n are also accepted, if we pair up points a/4 apart:
This simplifies to our formula!
How do we apply this?
With this formula, we can:
- Find the width of a slit, given the other information
- Find the angle between the centre line & the first minimum, given the other information
- Calculate the wavelength of light, given the other information
- Calculate the order of a certain minimum, given the other information
- Find the total number of minima possible for a given setup
The first 4 are simple enough – just a matter of substituting the correct values.
We’ll take a deeper look at the last one:
How do you find the total number of minima possible for a given setup?
- a slit of width “a”
- light of wavelength λ
What is the total number of minima formed?
Identify the correct formula & list out the information
|Since it’s a single slit,|
Substitute the given information.
Obtain an expression for n in terms of sin θ
|STEP 3: |
Finding the range of n
|Remember a few things: |
We consider the angle from the centre to the very edge:
0 < θ < 90
0 < sin θ < 1
Thus, we can find the range of n:
0 < n < a/λ
Round down where appropriate
|n needs to be a WHOLE number.|
So, given a maximum value larger than n, you can conclude that n is the largest whole number smaller than a/λ.
Consider the question carefully! What are they looking for?
|If you need to find the TOTAL number of minima, this value of n must be multiplied by 2, since there are 2 minima for every order.|
If asked to find the MAXIMUM ORDER of minima, just leave the answer as n.