A bit of history and applications of the double slit

Early 19th century. Everyone is convinced that light consists of a stream of particles, because that is what the famous and unquestionable Isaac Newton thought. And this, although Newton, who allegedly had an apple falling on his head, could not explain how diffraction occurs or how the Newtonian rings named after him are formed, which you can for example observe on the surface of a soap bubble. He was firmly convinced that light propagates in rays consisting of many spherical light particles.

An ophthalmologist and physicist named Thomas Young was tired of Newtons assertions - he wanted to show that light is a wave. So to prove this theory, Young had to come up with an experiment that would make visible properties of light that no ordinary particle could ever have. And one of those properties is diffraction. Light must be able to propagate around a corner, just like a sound wave, if the corner is small enough.

A particle could never do that! It would hit an obstacle and bounce back. A big enough wave, on the other hand, would simply pass the obstacle and continue to propagate unhindered. A light wave should behave like a sound wave that is bent around a corner and is audible to a person standing around the corner. But light should also show interference like for example water waves. So if two light waves interfere constructivly with eacth other, then the resulting wave should be brighter. It would be easy to think up such an experiment with water... But not with light - otherwise Newton would certainly have already tried it.

The double-slit experiment was born! Probably one of the most important experiments in physics, with which you can directly show that light has a wave character, as if it were like sound or water.

Setup of a double-slit experiment

The double-slit experiment setup is simple! All you need is:

A monochromatic light source

A detector screen

Two narrow slits close together (thats why its called the double-slit experiment).

Ready is the setup! It would be advantageous to use a light source which produces monochromatic (i.e. unicolored), parallel light. (The waves are then said to be coherent). A laser or a monochromatic lamp very far away from the double slit has these two properties. In this way you get much sharper result on the detector screen!

What is observed in the double slit experiment?

Now what do you observe on the detector screen when you switch on the light source? The light moves to the double slit and passes through the two slits. Then it lands on the screen and creates alternating bright and dark light stripes. It produces the same pattern as a water wave passing through two slits. The whole appearance on the screen is called interference pattern.

The band in the middle is a bright light stripe of zeroth order. All the following maxima are 1st order maxima, 2nd order maxima and so on. And dark stripes are called 1st minima, 2nd minima and so on. What you see here on the screen is a typical behaviour of waves! If you would bombard the double slit with any particles, you would never get such a pattern! You would simply see two stripes on the screen where the particles have landed. Exactly what you would normally expect from particles; one stripe because of one slit and the other stripe because of the other slit.

This is why the classical double-slit experiment is so fascinating: it shows a property of light that Newton and others did not believe. It seems that light does not behave like a particle, but rather like a wave.

How is the interference pattern at the double slit explained?

Wave interference

In order to understand how this light pattern comes about, you need to know what happens when waves hit each other. They interfere! The waves can interfere constructively or destructively.

Constructive interference occurs exactly when the two identical waves have a path difference \( \Delta s\) that corresponds to a multiple of the wavelength:

Formula anchor$$ \begin{align} \Delta s ~=~ m \, \lambda \end{align} $$

\(m\) is an integer: \( m ~=~ ...-2,1,0,1,2... \). If it does not matter if e.g. the lower or upper 2nd maximum is meant, then you consider only positive integers \( m ~=~ 0,1,2... \).

For destructive interference we shift the wave by half the wavelength and add a multiple of the wavelength. So the path difference is now:

Formula: Alternative condition for destructive interference

Formula anchor$$ \begin{align} \Delta s ~=~ \left( \frac{1}{2} ~+~ m \right) \, \lambda \end{align} $$

With \( m ~=~ ...-2,1,0,1, 2... \). Since the interference stripes occur symmetrically on both sides of the main maximum, it is sufficient to look at only one side of the screen. For this purpose, we use only positive integers for \( m ~=~ 0, 1, 2... \). Also we want to reserve \( m = 0 \) for the zeroth order maximum. So \(m\) thus goes from one to infinity: \( m ~=~ 1, 2... \)

However, there is a little problem if we insert one for \(m\). The path difference \(\Delta s\) of the first-order minimum would then be three half of the wavelength: \( \Delta s = \frac{3}{2}
\, \lambda \). But this should be one half of the wavelength: \( \Delta s = \frac{1}{2}
\, \lambda \). To correct this, we rewrite the condition 2. That goes with the following trick: You add +1 in the parenthesis of 2 and subtract -1 immediately after. Added up you have added a zero ( +1 - 1 = 0), which does not change anything at the condition 2:

Add a zero to the interference condition

Formula anchor$$ \begin{align} \Delta s ~=~ \left( 1/2 ~+~ 1 ~-~ 1 ~+~ m \right) \, \lambda \end{align} $$

Then you sum up \( \frac{1}{2} - 1 \) in 3. This gives \( - \frac{1}{2} \). And you add the remaining \( + 1 \) to \( m \): \( m + 1 \). So you add to each possible \( m \): +1 so that you get:

Ordinal number increases by 1

Formula anchor$$ \begin{align} m ~=~ 0+1,1+1, 2+1... ~=~ 1,2,3... \end{align} $$

The ordinal \( m \) for minima does not start from 0, but from 1. The adjusted condition is now as follows:

Formula anchor$$ \begin{align} \Delta s ~=~ \left( m ~-~ \frac{1}{2} \right) \, \lambda \end{align} $$

With \( m ~=~ 1,2... \).

What happens to light at the double slit?

Let's now send such plane waves to the double slit. The red lines represent the maxima of the waves seen from above. As soon as the light waves arrive at the double slit, we use the Huygens principle. This states that from every point on the wave front a spherical wave is created. If the slit is very thin, then it appears like a point source of light that emits a spherical wave. This can propagate in the whole space behind the slit. Hopefully nobody has switched off the light, so the spherical waves go constantly through the slit.

As you can see, the movement of light waves is not restricted by the slits as expected - unlike particles that would mainly move straight through the slits. But how does the pattern of light and dark stripes on the screen come about?

How is the interference pattern created on the screen?

You know that the red lines represent 'wave crests'. The 'wave troughs' are located exactly between two red lines. These were marked with dashed red lines so that you can follow them better.

When the waves from the two slits meet, they interfere. Mathematically, interference means that the amplitudes of the waves add up. Amplitude of one wave plus the amplitude of the other wave gives the amplitude of the resulting wave. So we know - where two solid lines intersect, a wave maximum meets another wave maximum and constructive interference takes place. The amplitude of the resulting wave is larger. And where two dashed lines meet, the minima of the waves amplify to form a resulting wave minimum twice as large - this is also constructive interference but the amplitude is just negative.

And do you know where the waves cancel each other out completely? Exactly where a solid line intersects a dashed line! These are points where a wave maximum meets a wave minimum.

And everything in between is partial interference - there is partly cancellation and partly amplification. That's why you don't see any sharp lines on the screen, but rather: the light stripes merge evenly into dark stripes.

If you follow the points of constructive interference up to the screen, you will reach the maxima. The zeroth order maximum occurs if the path difference is: \( \Delta s = m\, \lambda = 0 \), which is \( m=1 \). The 1st order maximum occurs at a path difference of exactly one wavelength: \( \Delta s = 1\, \lambda \), which is \( m=1 \). And so on! Now you understand where this numbering of the maxima comes from; namely from the integer \(m\), which indicates how large the path difference of the interfering waves is.

And what about the dark stripes? Follow the points of destructive interference up to the screen and you will see that they lead to the minima. These minima are described by the integer \(m\), which runs from 1 to infinity. For example: The 1st minimum with the path difference of half lambda, you get when \( m = 1\). And so on. Just use the Eq. 5.

For example, to find the wavelength of light \( \lambda \), using the distance \( a \) of the double slit to the screen, as well as the distance \( x \) of the main maximum to another maximum, and the distance \( g \) of the two slits, you can derive the following relation:

Formula anchor$$ \begin{align} \frac{m \, \lambda}{g} ~\approx~ \frac{x}{a} \end{align} $$

Analogously, instead of the distance between the main maximum and a secondary maximum, you can consider the distance between the main maximum and a secondary minimum. Then the above relation changes to:

+ Perfect for high school and undergraduate physics students + Contains over 500 illustrated formulas on just 140 pages + Contains tables with examples and measured constants + Easy for everyone because without vectors and integrals