PHY C15: Single-Slit Interference

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.

Image result for single slit interference gif

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:

Image result for single slit interference graph

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.

For example,

  • 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:

  • Order
  • Wavelength
  • 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:

Via trigonometry,

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
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.

⇐ Previous in Physics: Diffraction
⇒ Next in Physics: Diffraction Grating

2 thoughts on “PHY C15: Single-Slit Interference

  1. Pingback: PHY C15: Diffraction Grating – ProDuckThieves

  2. Pingback: PHY C15: Diffraction – ProDuckThieves

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