# Diffraction

## Young's Double Slit Experiment

This is important because it:

• Provides evidence of the interference of light.
• Can be used to find the wavelength of light.

Light is allowed to pass through a narrow slit and then through a pair of closely spaced slits to form an image on a screen. The light diffracts through the slits. The two sets of light waves emerging from the double slit interfere with each other, and the image formed on the screen consists of evenly spaced light and dark bands, or fringes.

Young correctly traced the origin of the fringes to the differences in the path lengths, of the two sets of waves. Some waves would arrive in phase to create the bright fringes, and others would arrive out of phase, leaving dark bands on the screen.

#### Finding the Wavelength of Light

Using Young's double slits to find the wavelength of light:

You should have an idea of the values you would expect in this formula. For instance, D should be in metres, whereas l should be around x10-7 m.

You will need to be able to show how to do similar experiments with a ripple tank or a microwave kit.

Question:

## Diffraction from a Single Slit

Waves diffract as they move through small gaps - that's the easy bit! But how can we calculate how much diffraction there will be, and what use is it to us anyway?

Calculating the angle of diffraction:

We can calculate the position of the dark fringes using the formula above, where:

θ = angle from the centre to the minimum

n = order of the minimum

b = width of the slit (m)

λ = wavelength of wave being diffracted (m)

What can we use this for?

Diffraction is not only an issue with light waves, it also affects all the electromagnetic waves, water waves and sound waves, etc. This means that we have to take diffraction into account when designing equipment such as loud speakers and satellite dishes.

For example:

The size of a speaker will affect the amount of diffraction of the sound waves leaving it, and will therefore change the width of the central maxima of the diffraction pattern produced!

Question:

## Diffraction from a Diffraction Grating

The pattern that you get with a large number of slits (a diffraction grating) is similar to the double slit pattern in that there are bright fringes on a dark background, but there are far fewer fringes and the gaps between them are much larger.

Double slit pattern:
(closely spaced bright fringes on a dark background)

Grating pattern:
(widely spaced bright fringes on a dark background)

Common Trick:

In a question on diffraction gratings, it is common that you are told that the grating has, for example, 600 lines per millimetre.

To use this information in the diffraction grating equation, we need to convert it into a slit spacing, d. To do this:

1. Convert it to lines per metre (for instance, 600x103 lines per metre)
2. Invert it. So 600 lines mm-1 becomes a slit spacing,

Diffraction can be used to analyse the light from stars, by separating the different wavelength waves that are emitted with a diffraction grating.

Different elements in stars will emit different frequencies from the electromagnetic spectrum, so we can work which elements are present in each star.

Question:

## Exam-style Questions

1. a) Explain the meaning of the term diffraction.

(2 Marks)

b) i) Describe how transverse water waves with a plane wavefront may be produced in a ripple tank.

(2 Marks)

ii) State how the wavelength of the waves could be shortened.

(1 Mark)

c) Fig.5.1 shows plane water waves in a ripple tank approaching a narrow gap, the size of which is approximately the same as the wavelength of the waves.

i) On Fig. 5.1, draw the pattern of the wavefronts emerging from the gap.

(2 Marks)

ii) Describe how the pattern of wavefronts emerging from the gap would change if the size of the gap were significantly increased.

(2 Marks)

iii) State why, under normal circumstances, light seems to travel in a straight line and does not appear to be diffracted.

(1 Mark)

(Marks available: 10)

2. A single slit is illuminated by a monochromatic (single colour) light source. A screne is placed 5.0 m away from this slit and two very narrow, parallel slits, 0.5 mm apart, are placed half way between the single slit and the screen. Interference fringes are visible on the screen and 10 fringe spaces measure 20 mm on the screen.

a) What is the wavelength of the light?

(3 Marks)

b) What happens if you use double slits with half the spacing between them?

(1 Mark)

c) What happens if you cover one of the double slits?

(2 Marks)

(Marks available: 6)

Answer outline and marking scheme for question:

1. a) the spreading out of the wavefront/waves

{do not allow 'the spreading out of light/sound' or 'bending of waves'}

when they pass through a gap (OR pass an obstacle)

(2 Marks)

b) i) a straight strip (OR bar OR ruler)

is vibrated vertically OR up and down (in the water)

(2 Marks)

ii) increase the frequency (of the waves/wave source)

OR use a shallower depth of water

(1 Mark)

c) i) semicircular wavefronts drawn

no change in : i.e. approx. same before & after gap

(2 Marks)

ii) less diffraction occurs

wavefronts only slightly curved at edges (OR shown in a diagram)

{full marks may be scored from a valid diagram}

(2 Marks)

iii) Wavelength of light much smaller than most gaps

(1 Mark)

(Marks available: 10)

2. a)

 λ = ax (1 Mark) D

where,

λ = wavelength

a = double slit spacing = 0.5 m m

x = fringe spacing = 20/10 = 2.0 mm

D = double slit to screen distance = 2.5 m

 So, λ = 0.5 x 10-3 x 2.0 x 10-3 = 4.0 x 10-7 m (1 Mark) 2.5 (1 Mark)

(3 Marks)

b) If a is halved and D and λ remain the same, the fringe spacing (x) must be doubled.

(1 Mark)

c) The fringes disappear (1 Mark) and you can see a single slit pattern. This is a bright central line with parallel, fainter lines visible on each side.

(2 Marks)

(Marks available: 6)

## Diffraction, Interference and Superposition

#### Diffraction

A wave will diffract (spread out) as it goes through a gap or past an obstacle.

Note: The wavelength remains the same before and after the gap.

Remember this: The nearer the slit size is to the wavelength, the more the wave will diffract.

1. The smaller the gap the greater the diffraction.
2. The longer the wavelength the greater the diffraction.

You should be able to describe experiments such as the ripple tank or microwave kit that will show diffraction.

Interesting Point:

It is not only waves that can be diffracted. In 1923, Davisson and Germer showed that electrons and all sub-atomic particles could also be diffracted. This is supported by De Broglie's theory of wave/particle duality, which shows that electrons have a mass but can behave just like waves with no mass, such as light waves.

#### Single Slit Diffraction Pattern

If a wave goes through a slit a diffraction pattern can be detected on the other side, with regions where the wave is intense and regions where the intensity falls to zero. graph of intensity against distance from the centre of the pattern can be drawn:

The reason for this pattern is explained in 'interference and superposition'

Did you know that X-rays can be diffracted by the gaps between atoms in crystal structures, so by analysing the diffraction patterns we can work out the shape of the crystal structure. This only works because the gaps between the atoms are roughly the same size as the wavelength of X-rays!

Coherence:

Coherent waves are waves with a constant phase difference. (Note: They don't have to be in phase for this to be true.) They will have the same frequency and wavelength (they are normally produced from one source).

#### Interference and Superposition

When two waves meet they will interfere and superpose. After they have passed they return to their original forms. This is true if they are coherent or not.

At the point they meet, the two waves will combine to give a resultant wave whose amplitude (or intensity) may be greater or less than the original two waves.

The resultant displacement can be found by adding the two displacements together:

This is called the Principle of Superposition.

If two waves of the same type and the same frequency combine so that the crest of one coincides with the trough of the other, they will completely cancel each other out. This is called destructive interference. Alternatively, the two waves could combine when their crests coincide; then there would be constructive interference and the resultant amplitude would be equal to the sum of the separate amplitudes:

Superposition will occur whether waves are coherent or not. (However, if the waves are coherent, they will interfere to produce a fixed pattern.)

The same rules apply... the resultant displacement at any point is always the sum of the separate displacements of the wave at that point:

#### Path Difference

You will need to be able to work out whether there will be constructive or destructive interference at a point. We do this by comparing how far the two waves have travelled to reach the point. The difference in the distances will tell us if the waves are in phase or not.

In the diagram above the waves are in phase when they leave each slit.

A bright fringe occurs at P if S2P - S1P = nλ

In other words, for the waves to be in phase and constructively interfere, the difference in the distances travelled by the waves must be a whole number of wavelengths.

A dark fringe occurs if S2P - S1P = nλ + l/2λ

In other words, for the waves to be out of phase and destructively interfere, the difference in the distances travelled by the waves must be exactly half a wavelength, or 1 1/2 , or 2 1/2 , etc.

If you don't have a nice 1/2 or whole number of wavelengths you can find the phase difference using:

Note 1: The wave that has travelled the furthest will have lost more energy, so will have a slightly smaller amplitude than the other. This means that the waves might not completely cancel each other out.

Note 2: If a wave is reflected off a surface it will have a 180 degree phase change which must be taken into consideration!

#### Fringes

So why do we get fringes?

The same principle shown in this diagram produces fringes even when it is only diffraction through a single slit. This is because a wave front can be thought of as being made up of lots of wave sources in a row that will then interfere with each other as you move away from the wave front!

Question:

## S-Cool Revision Summary

#### Diffraction

A wave will diffract (spread out) as it goes through a gap or past an obstacle.

Note: The wavelength remains the same before and after the gap.

Remember this: The nearer the slit size is to the wavelength, the more the wave will diffract.

1. The smaller the gap the greater the diffraction.
2. The longer the wavelength the greater the diffraction.

You should be able to describe experiments such as the ripple tank or microwave kit that will show diffraction.

#### Single Slit Diffraction Pattern

If a wave goes through a slit a diffraction pattern can be detected on the other side, with regions where the wave is intense and regions where the intensity falls to zero. A graph of intensity against distance from the centre of the pattern can be drawn:

Coherence:

Coherent waves are waves with a constant phase difference. (Note: They don't have to be in phase for this to be true.) They will have the same frequency and wavelength (they are normally produced from one source).

Young's Double Slits:

The pattern formed is of close, bright "fringes" of light.

A bright fringe occurs at P if S2P - S1P = nλ

A dark fringe occurs at P if S2P - S1P = nλ + ½ λ

#### Interference and Superposition

When two waves meet they will interfere and superpose. After they have passed they return to their original forms. This is true if they are coherent or not.

#### Path Difference

You will need to be able to work out whether there will be constructive or destructive interference at a point. We do this by comparing how far the two waves have travelled to reach the point. The difference in the distances will tell us if the waves are in phase or not.

#### Finding the Wavelength of Light

Using Young's double slits to find the wavelength of light:

λ = wavelength

a = distance between slits

x = fringe spacing

D = distance from slits to screen

#### Diffraction from a Diffraction Grating

Using a diffraction grating to find the wavelength of light:

nλ = dsinθ

d = slit spacing

θ = angle from centre

n = order of maximum

The pattern that you get with a large number of slits (a diffraction grating) is similar to the double slit pattern in that there are bright fringes on a dark background, but there are far fewer fringes and the gaps between them are much larger.

Double slit pattern:

(closely spaced bright fringes on a dark background)

Grating pattern:

(widely spaced bright fringes on a dark background)