My name is Alexander FufaeV and here I write about:

# Double-Slit Experiment with Light

## Important Formula

What do the formula symbols mean?

## Path difference

Unit
The path difference tells by how many wavelengths $$\lambda$$ the light wave through the upper slit, differs from the light wave through the lower slit. In the case of constructive interference, the difference is a multiple of the wavelength: $\Delta s = m \, \lambda$ With destructive interference, the path difference is: $\Delta s = \left( m-\frac{1}{2} \right) \, \lambda$ Here $$m$$ is an integer 1, 2, 3, ... It numbers the light and dark stripes, respectively. For the first maximum / minimum is $$m = 1$$. For the fifth maximum / minimum is $$m = 5$$.

## Slit distance

Unit
Distance between the two slits (measured from the center of the slit). The further apart the two slits are, the closer the interference fringes are to each other.

## Fringe separation

Unit
Distance of the main maximum (which is exactly in the center of the screen) to a bright or dark interference fringe. If the condition for constructive interference is used for the path difference $$\Delta s$$, then this is the distance to a bright fringe (maximum). If the destructive interference condition is used, this is the distance to a dark fringe (minimum).

## Screen distance

Unit
The distance of the double slit to the detector screen. Here the approximation is used that the screen is far away from the double slit, compared to the fringe distance, that is: $$a$$ must be much larger than $$x$$ for this formula to be used.

## 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:

$$m$$ 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:

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:

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:

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

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:

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:

## Exercises with Solutions

Use this formula eBook if you have problems with physics problems.

### Exercise #1: Generating at Least 15 Interference Fringes with the Double Slit

You have a double slit setup with a screen that is only $$15 \, \text{cm}$$ wide. The double slit and the screen are attached to each other at a distance of $$3 \, \text{m}$$, and the slit spacing is $$0.15 \, \text{mm}$$. On the screen, you want to create an interference pattern, and you want to see at least $$15$$ bright stripes there!

What wavelength $$\lambda$$ do you need to use for this?

#### Solution to Exercise #1

From the condition for interference maxima: $$\Delta s ~=~ m \, \lambda$$ and the sketch of the double slit (with the approximation that the screen is far away from the double slit): $$\frac{x}{a} ~=~ \frac{\Delta s}{g}$$ you get the wavelength: $$\lambda ~=~ \frac{ x \, g }{ a \, m }$$

Substituting the given values yields: $\lambda ~=~ \frac{ 0.15\cdot10^{-3}\text{m} ~\cdot~ 0.15\text{m} }{ 3\text{m} ~\cdot~ 15 } ~=~ 5\cdot10^{-7} \, \text{m}$

So, you need to use light with at least $$500 \, \text{nm}$$ wavelength (red light) to generate at least 15 stripes on a $$15 \, \text{cm}$$ wide screen (at a distance of $$3 \, \text{m}$$ from the double slit).

### Exercise #2: Wavelength of Monochromatic Parallel Light at the Double Slit

Monochromatic light strikes the double slit in parallel, with both slits spaced $$g ~=~ 0.8 \, \mathrm{mm}$$ apart. You observe interference stripes on the screen, which is $$a ~=~ 5 \, \mathrm{m}$$ away from the double slit. Two bright fringes are $$x ~=~ 3.5 \, \mathrm{mm}$$ apart.

What wavelength $$\lambda$$ does the light have?

#### Solution to Exercise #2

Since the monochromatic light strikes the double slit in parallel, the light waves arrive at both slits in phase. Because the waves are in phase, they satisfy the condition for constructive interference as follows: $$\Delta s ~=~ m \, \lambda$$

Here, $$\Delta s$$ is the path difference. We have taken the condition for constructive and not destructive interference because the task specifies the distance between two bright rather than dark fringes.

Additionally, the distance from the slit to the screen $$a ~=~ 3 \, \text{m}$$ is much larger than the slit spacing $$g ~=~ 0.3 \, \text{mm}$$. This means you can use the following approximation: $$\tan(\phi) ~\approx~ \sin(\phi) ~=~ \frac{x}{a}$$

From the right triangle, where the opposite side is the path difference $$\Delta s$$, you gain the information: $$\sin(\phi) ~=~ \frac{\Delta s}{g}$$

Set this equation equal to the above approximation: $$\frac{x}{a} ~=~ \frac{\Delta s}{g}$$

Also, use the condition for constructive interference for the path difference: $$\frac{x}{a} ~=~ \frac{m \, \lambda}{g}$$

Now, you have derived a relationship that contains only quantities given in the exercise. Just rearrange the last equation to solve for the desired wavelength: $$\lambda ~=~ \frac{x \, g}{m \, a}$$

Assuming that the bright fringes always have the same spacing and nothing further is mentioned in the exercise, you can consider the distance from the 0th to the 1st maximum: $$m ~=~ 1$$. Substituting the given values yields: \begin{align}\lambda ~&=~ \frac{ 3.5 \cdot 10^{-3}\text{m} ~\cdot~ 0.8 \cdot 10^{-3}\text{m} }{ 1 ~\cdot~ 5\text{m} } \\ ~&=~ 5.6 \cdot 10^{-7} \, \text{m}\end{align}

This path difference is equivalent to a wavelength of $$560 \text{nm}$$, corresponding to the typical wavelength of green light. $$\lambda ~=~ \frac{ 3.5 \cdot 10^{-3}\text{m} ~\cdot~ 0.8 \cdot 10^{-3}\text{m} }{ 1 ~\cdot~ 5\text{m} } ~=~ 5.6 \cdot 10^{-7} \, \text{m}$$

### Exercise #3: Calculating the Spacing of the Double Slit via Minimum Distance

A double slit is illuminated with red light of wavelength $$\lambda ~=~ 650 \, \text{nm}$$ in parallel. At a distance $$a ~=~ 3 \, \text{m}$$ from the double slit, there is a screen on which a sharp interference pattern is visible. Since it is not easy to determine the distance from the main maximum to the 1st maximum, instead, the distance from the 5th minimum to the opposite 5th minimum is measured, which is $$\Delta x ~=~ 6 \, \text{cm}$$.

What is the spacing $$g$$ between the two slits through which the light passed?

#### Solution to Exercise #3

Since the red light in parallel strikes the double slit, the light waves arrive at both slits in phase. And, because the waves are in phase, the condition for destructive interference applies as follows: $\Delta s ~=~ \left( m ~-~ \frac{1}{2} \right) \, \lambda$

Here, $$\Delta s$$ is the path difference, and $$m ~=~ 1,2,3...$$ indicates the order of the minima. We have taken the condition for destructive and not constructive interference because the task specifies the distance between two minima. Minima are the points on the screen that are dark. The light waves have canceled out at these points.

Regarding the slit spacing: It is unknown. But what you can say about it just by looking at it is that it is very small... (I calculated it, it IS small *cough*). The distance from the slit to the screen $$a ~=~ 3 \, \text{m}$$ is much larger than the unknown slit spacing $$g$$. This means you can use the following approximation: $\tan(\phi) ~\approx~ \sin(\phi) ~=~ \frac{x}{a}$

The position $$x$$ on the screen (measured from the center) is only indirectly known. The distance between the 5th minima was measured. Since the interference pattern is symmetrical, the distance from the main maximum to the 5th minimum is just half of the measured value. This is also the sought position $$x$$ on the screen: $$x ~=~ \frac{\Delta x}{2}$$. Substitute this position into the above equation: $\sin(\phi) ~=~ \frac{\Delta x}{2a}$

From the right triangle, where the opposite side is the path difference $$\Delta s$$, you can read off: $\sin(\phi) ~=~ \frac{\Delta s}{g}$

Set these last two formulas equal: $\frac{\Delta x}{2a} ~=~ \frac{\Delta s}{g}$

Since you want to look at the minima, also set the condition for destructive interference: $\frac{x}{a} ~=~ \frac{ \left( m ~-~ \frac{1}{2} \right) \, \lambda }{g}$

Now you have derived a relationship that contains only quantities given in the task. Just rearrange it with respect to the desired slit spacing $$g$$: $g ~=~ \frac{ 2a \, \left( m ~-~ \frac{1}{2} \right) \, \lambda }{ \Delta x }$

Substituting the given values yields: $g ~=~ \frac{ 2 \cdot 3\text{m} ~\cdot~ \left( 5 ~-~ \frac{1}{2} \right) ~\cdot~ 650 \cdot 10^{-9}\text{m} }{ 0.06\text{m} } ~=~ 2.925 \cdot 10^{-4} \, \text{m}$

This corresponds to a slit spacing of approximately $$0.3 \text{mm}$$, which is hardly measurable with a ruler... but fortunately, it works with the double slit experiment!