Young's double slit experiment

In a nutshell

Young's double slit experiment shows the interference pattern for two coherent point sources. The pattern of light and dark fringes are a result of constructive and destructive interference. From this experiment, Young proved an equation for working out the wavelength of light from measurements taken.

Equations

Description	Equation
Young's double slit equation	$\lambda = \dfrac{ax}{D}$
Diffraction grating equation	$d\sin{\theta} = n\lambda$

Variable definitions

Quantity Name	Symbol	Derived Unit	SI Unit
$wavelength$	$\lambda$	$m$	$m$
$slit \space spacing$	$a$	$m$	$m$
$fringe \space spacing$	$x$	$m$	$m$
$distance$	$D$	$m$	$m$
$distance\space between\space slits$	$d$	$m$	$m$
$angle\space to \space zeroth \space order$	$\theta$	$\degree$	$rad$
$order$	$n$

Young's double slit experiment

Young's double slit experiment involves light from two coherent point sources interfering and producing a pattern that can be observed on a screen at a distance from the sources.

Physics; Waves; KS5 Year 12; Young's double-slit experiment

1.	Monochromatic laser source
2.	Double slit
3.	Diffraction occurs at each slit
4.	Screen
A.	Path difference of $\frac{\lambda}{2}$ $\to$ Destructive interference $\to$ Dark fringe
B.	Path difference of $0$ $\to$ Constructive interference $\to$ Light fringe
C.	Path difference of $\lambda$ $\to$ Constructive interference $\to$ Light fringe

The coherent point sources are created by using a double slit. Monochromatic (single-wavelength) laser light is directed at the double slit and diffraction takes place at each slit. The diffracted light from the slits acts as two point sources producing coherent light. It is important for the light to be coherent as only coherent light can form stable interference patterns.

The slit width of the double slits have to be around the same size as the wavelength of light generated by the laser. This is because the most diffraction occurs when the size of the gap (the slit) is the same size as the wavelength.

On the screen, a pattern of light and dark fringes will be produced. The light fringes are produced by constructive interference of the two waves. The dark fringes are produced by destructive interference.

Curiosity: This experiment can be performed with microwaves using a microwave generator and a probe! A pattern of strong and weak signals will be observed by moving the probe.

Young's double slit equation

Young used the results of his experiment to prove the equation for wavelength that he had previous derived.

The proof of this equation uses small angle approximations. These are able to be used as long as the distance between the slit and the screen $D$ , is much larger than the spacing between the slits $a$ . ( $D>>a$ )

Physics; Waves; KS5 Year 12; Young's double-slit experiment

The lines $S_1X$ and $S_2X$ represent light rays from two slits. Using trigonometry, the path difference between these rays $S_1P$ can be found to be given by

$path \space difference = a\sin{\theta_1}$

Points X and Y are at points of constructive interference and are adjacent bright fringes. This means the path difference of the two rays is equal to $n\lambda$ , where $n$ is an integer. As Y is the central bright fringe and X is adjacent to it, $n$ is equal to 1. The path difference is $\lambda$ .

Small angle approximations can also be used to derive a formula with $\tan{\theta_1}$ as the subject.

$\sin{\theta_1} \approx \tan{\theta_1} \approx \dfrac{\lambda}{a}$

Trigonometry can also be used to get an equation for $\theta_2$ in terms of the fringe spacing $x$ and the distance to the screen $D$ .

$\tan{\theta_2} = \dfrac{x}{D}$

Note: Fringe spacing refers to the distance between two adjacent maxima or two adjacent minima.

The angle $\theta_1$ is approximately equal to the angle $\theta_2$ so $\tan{\theta_1} \approx \tan{\theta_2}$

$\dfrac{\lambda}{a} \approx \dfrac{x}{D} \newline \ \newline \lambda = \dfrac{ax}{D}$

This is Young's double slit equation that he proved by using the measurements of his experiment.

The equation shows that in order for the fringes to be noticeable, there need to be a high ratio of distance from slits to screen $D$ to slit spacing $a$ to account for the small wavelength of light.

Tip: This derivation is not needed for an exam but it is useful to know where the final equation comes from!

Example

An interference pattern can be observed on a screen placed $5.0 \,m$ from two coherent laser light sources. The frequency of the sources is $430 \,THz$ . The slit spacing is measured as $0.3\,mm$ . Calculate the spacing between the light fringes observed on the screen.

First, write out the values given and check they are in the correct form:

$D = 5.0\,m\newline f = 430 \,THz = 430 \times10^{12} \,Hz\newline a = 0.3\,mm = 0.3\times10^{-3}\,m$ 

Next, write down the equations needed and rearrange if necessary:

$c = f\times \lambda\:\: \rightarrow \:\:\lambda = \dfrac{c}{f}\newline\ \newline \lambda = \dfrac{ax}{D}\:\:\rightarrow\:\: \dfrac{c}{f} = \dfrac{ax}{D} \rightarrow\:\: x = \dfrac{c}{f}\dfrac{D}{a}$

Then, substitute the values into the equation:

$x = \frac{3\times10^8}{430\times10^{12}} \frac{5.0}{0.3\times10^{-3}} = 1.1627... \times10^{-2} \,m$ 

Make sure to include units and round the lowest number of significant figures of the values given by the question:

$fringe \space spacing, x = 1.2\times10^{-2}\,m$ 

$\underline{The \space fringe \space spacing \space is \space 1.2\times10^{-2}\,m}$

Diffraction gratings

Young's double slit experiment can be modified to use many slits instead of two. This is done by replacing the double slit with a diffraction grating with slits that are a distance $d$ apart.

The interference pattern produced monochromatic light passed through a diffraction grating is very sharp. This is because light diffracts at each slit which produces many beams that reinforce the pattern.

Physics; Waves; KS5 Year 12; Young's double-slit experiment

1.	Monochromatic laser source
2.	Double slit
3.	Screen
4.	Diffraction pattern (order lines)

The maxima produced by interference are labelled as numbered order lines. Each order line has an order number $n$ that corresponds to it.

The zero order line ( $n=0$ ) is the line of maximum brightness that is produced at the centre. The lines either side of this are called the first order line ( $n=1$ ). This pattern continues for all the maxima.

Diffraction grating equation

You can use an equation to calculate variables associated with a diffraction grating.

Physics; Waves; KS5 Year 12; Young's double-slit experiment

At each slit light diffracts and interferes with the diffracted waves from other slits.

For the first order maximum, light from one slit aligns with light from the previous slit which has travelled one wavelength further. This means the path difference between the two adjacent slits is equal to one wavelength, $\lambda$ .

Using trigonometry, the equation for the first maximum can be derived

$\sin{\theta} = \dfrac{\lambda}{d} \newline \ \newline d\sin{\theta} = \lambda$

The path difference between lines aligning at the other maxima follows the pattern of being $n\lambda$ where $n$ is the order of the line. This means the equation can be written generally as

$d\sin{\theta} = n\lambda$

The equation shows that the larger the wavelength of light used, the more spread out the pattern will be. It also shows that the larger the distance between the slits, the less spread out the pattern will be.

It is impossible for $\sin{\theta}$ to produce a value greater than $1$ . This means there will be a limit to how many orders of maxima are produced.

Tip: If the calculation produces a 'math error' and all the values used are correct, then this probably means the order does not exist!

Finding slit spacing

The slit spacing may not be given explicitly in a question and instead the question will give the number of slits per a unit of length. To convert from this to slit spacing, use the formula:

$d = \dfrac{length}{number\space of \space slits \space in \space length}$

Note: This is not an assessed equation, however the formula is useful to remember for converting slits per length into slit spacing.

Example

A diffraction grating has $325$ slits per $cm$ . Calculate the slit spacing of the grating.

First, write out the values given and check they are in the correct form:

$length = 1\,cm = 1\times 10^{-2} \,m\newline number \space of \space slits \space in \space length = 325$ 

Then, write down the equation needed:

$d = \dfrac{length}{number\space of \space slits \space in \space length}$ 

Next, substitute the values into the equation:

$d = \dfrac{1\times10^{-2}}{325} = 3.076...\times 10^{-5}$ 

Make sure to include units and round to the lowest number of significant figures of the values given in the question:

$d = 3.08 \times 10^{-5}\,m$ 

$\underline{The \space slit \space spacing \space of \space the \space grating \space is\space\, 3.08 \times 10^{-5}\,m}$

Example

Green laser light of wavelength $540\,nm$ is directed at a diffraction grating. The diffraction grating has $630$ slits per $mm$ . A diffraction pattern is produced at the screen. Calculate the angle between the zero order line and the second order line.

First, write out the values given and check they are in the correct form:

$\lambda = 540\,nm = 540 \times 10 ^{-9}\,m\newline d = \frac{1\,mm}{630}= \frac{1}{630\,000} \,mm\newline n = 2$ 

Next, write down the equation needed and rearrange if necessary:

$\sin{\theta} = \dfrac{n\lambda}{d}\newline \theta = \sin^{-1}({\dfrac{n\lambda}{d}})$ 

Then, substitute the values into the equation:

$\theta = \sin^{-1}({\frac{2\times 540 \times 10^{-9}}{\frac{1}{630\,000}}}) = 42.87...$ 

Make sure to include the correct units and round to the lowest number of significant figures of the values given in the question

$\theta = 43 \degree$ 

$\underline{The \space angle\space between\space the\space zero\space order\space line\space and\space the\space second\space order\space line\space is\space 43\degree}$ .

Uses of diffraction gratings

White light is made up of different wavelengths of light. As different wavelengths of light diffracts by different amount, the pattern produced is a spectrum for each order (except the zeroth order which stays white).

The spectra produced by diffraction gratings can help inform astronomers and chemists what elements the light came from.

It's also useful for working out spacing in crystals, as the crystal acts as a diffraction grating for light. This is done using X-rays as the wavelength of light is similar to the spacing between the atoms. The process is called X-ray crystallography.