# Interactive textbook: Diffraction and interference of light

### From qualitative to quantitative So far we have dealt with diffraction and interference on a qualitative level; by now you should understand the basics and make some predictions of increases or decreases but not calculate values. You'll know that reducing the width of a single slit increases the width of the central diffraction maxima but can't calculate how wide it is. You'll know that double slits produce interference fringes but can't calculate how far apart they are.

To derive equations we need to start from the model used to represent the effect: Huygens' construction.

#### Using vectors to represent waves

Each point on a transverse wave moves up and down with simple harmonic motion. The equation for its vertical displacement at time $$t$$ is:

$$y=A\sin\omega t$$

• $$A$$ is amplitude ($$\text{m}$$)
• $$ω =2πf$$, the angular frequency ($$\text{s}^{-1}$$)

This is the same equation as the vertical component of the displacement from the centre of a point moving in a circle of radius $$A$$. We can use this rotating vector model to represent the displacement of a point on a wave. This simplifies the process of adding waves that meet at a point.

When two waves meet the resultant displacement is the vector sum of the two displacements. If the waves are in phase the displacements add. The combined amplitude is the sum of the individual amplitudes:

Notice how the amplitude of the combined wave is the vector sum of the green and purple vectors. Now we will add two out of phase waves:

This time the vectors are in opposite directions so cancel, resulting in zero amplitude. When we consider two waves an angle $$ϕ$$ out of phase that we can see the benefit of this approach:

This time it would be quite difficult to add the waves but adding the vectors is easy. The angle between the vectors is the same as the phase angle. This graph shows two sine waves. The phase angle is:

0.5 rads = 0.5 x 360/2π degrees

### Single slit diffraction

When light passes through a narrow slit it diffracts giving a characteristic intensity distribution. According to Huygens’ construction a wavefront can be considered to be made up of a large number of small wavelet sources. The resulting wave at any point in front of the slit is found by adding up all the wavelets as shown below. It is rather tricky to deal with all these waves together. Let's consider one point at a time. The amplitude at P can be found by adding vectors that represent each wavelet. The phase difference between each wavelet depends on the path difference, but this is complicated by the geometry. We will simplify matters further by considering a distant point. Here the waves can be considered parallel. Considering parallel waves we can see that the waves straight ahead will add together. Each small vector represents a wavelet. At a small angle to this position the waves will be slightly out of phase. To see the full picture we can model the situation using GeoGebra.

The curling vectors represent the addition of the wavelets at different angles. The resultant amplitude is the line joining the ends. The graph shows the square of the amplitude at these angles, proportional to the intensity of the light. We are particularly interested in the point where the wavelets first cancel out. This is the position of the first minimum, which defines the angular width of the central maximum. We can see that the phase difference between the first and last wavelets at this angle is $$2π$$, meaning the path difference is $$λ$$. We can represent the rays of the first and last wavelets - the rays from each side of the slit (of width $$b$$) - on a simple diagram. • The angle $$θ$$ is the angle to the first minimum.
• The rays are almost parallel and meet at a distant screen.
• The blue line bisects the rays at $$90°$$ showing that the lower one will travel a distance $$\lambda$$ further than the upper one.

$$\sin\theta={\lambda\over b}$$

If the angle is small and in radians, applying the small angle approximation:

$$\theta={\lambda\over b}$$

Decreasing the slit width $$b$$ increases the width of the central maximum but less light passes through the slit so the intensity is reduced.

The diagram represents the waves from each wavelet. The rays are parallel because:

The rays that form the principal maxima are not really parallel, this is an approximation to make the geometry easier.

At the first maxima the phase angle between the first and last wavelet is: The diagram represents the waves from the first and last wavelets forming the first minima. The angle θ = ### Resolution

Resolution deals with the ability to see detail. A low resolution photograph of the night sky shows only one star at A: A high resolution photograph reveals that there are two: The resolution of a photograph depends upon the number of pixels and diffraction of light as it passes through the lens aperture (opening). Diffraction spreads the light from the stars causing overlap, making them impossible to distinguish.

A simple camera consists of a lens and a light-sensitive screen (CCD). Light from a distant star is parallel but the lens focuses this light on the screen. Light from a second star arrives at a different angle and is focussed on a different point on the screen. The angle between these rays is the angle subtended at the camera.

#### Diffraction at the aperture

When light passes through the lens aperture it diffracts in the same way as when light passes through a narrow slit except that the diffraction is in two dimensions. The result is a bright circle in the centre with rings around the outside. The equation for the position of the first minimum is:

$$\sin\theta=1.22{\lambda\over b}$$

• $$b$$ is the diameter of the aperture ($$\text{m}$$)

Applying the small angle approximation:

$$\theta=1.22{\lambda\over b}$$ When the light from the two stars arrives at the screen it is spread out. If the two stars are too close the two circles of light will overlap making them look like one. This can be shown in the animation below showing how the image on the screen would change as the angle subtended by the stars changes.

#### Rayleigh criterion

The Rayleigh criterion gives the condition for two point sources to be resolved:

Two points are resolved if the principal maximum of one diffraction pattern coincides with the first minimum of the other. So if the image of the second dot is such that its principal maximum is coincident with the first minimum of the first it will just be resolved. Let's consider the point of just being resolved. W can write two equations for the angle $$θ$$. From the diffraction equation we know:

$$\theta=1.22{\lambda\over b}$$

This is the same as the angle subtended at the aperture, $$\tan\theta\approx\theta={d\over D}$$: $${d\over D}=1.22{\lambda\over b}$$

The distance between two stars that can be resolved at distance $$D$$ is:

$${d}=1.22{\lambda D\over b}$$

To increase resolution you need to increase the aperture size. That's why high resolution telescopes use large diameter mirrors. Resolution also depends on wavelength:

• Radio telescopes have dishes with very large diameters
• To increase the resolution of microscopes (of small diameters), short wavelength light can be used

Two points of light are NOT resolved by the eye.

They might be resolved if:

Shorter wavelength means the diffraction pattern is narrower.

The first diagram shows the rays of light incident on a telescope from two distant stars, the second shows the diffraction of light by the aperture.

The stars are:

The angle subtended by the stars is bigger than the angle between the central maximum and the first minimum.

### Two slit interference (Young's slits)

When laser light is shone on two parallel slits it spreads out due to diffraction at each slit. If the slits are close to each other the light will overlap on a distant screen. The light from each slit has travelled a different distance so there will be a phase difference between the two sources, resulting in interference fringes. If non laser light is used then a narrow slit must be placed between the double slits and the source. If this isn't done light from different parts of the source will pass through each slit so the light from each beam will not have a constant phase difference (not coherent) or similar amplitude.

The following image shows the screen. The dots are called interference fringes. To help understand how the fringes are formed we can consider two identical waves of wavelength $$2\text{ cm}$$ starting from two points separated by a distance $$6\text{ cm}$$. At this point both waves travel the same distance so are in phase. The waves interfere constructively forming a bright dot.

Moving to a position above the central line we can see that the lower wave travels further. This introduces a phase difference. At this particular point the path difference is $$1\text{ cm}$$. This is $$λ\over2$$ which means the waves are $$π$$ out of phase so the interfere destructively creating a dark area. Further still: The path difference is $$2\text{ cm}$$. This is equal to $$λ$$, which means the waves have a phase difference of $$2π$$. The waves are in phase again creating another bright dot.

We can simplify the diagram above by just drawing the rays that form the first maximum in intensity (where the waves add). It is important to understand this diagram:

1. The angles are small so we can use the approximation $$\tan θ = \sin θ = θ$$.
2. The line BC is longer than AC by $$λ$$. This length is shown using the grey perpendicular construction lines.
3. The two angles labelled $$θ$$ are equal.

We can write two equations for the angle $$θ$$:

$$\theta={s\over D}$$ and $$\theta={\lambda\over d}$$

$${s\over D}={\lambda\over d}$$

$$\Rightarrow s={\lambda D\over d}$$

#### Diffraction envelope

The reason that each beam spreads out is due to diffraction, so each beam forms a single slit diffraction pattern. The slits are very close so the two patterns coincide. Looking at the image below you can clearly see the effect of diffraction causing the change in intensity of the fringes. Changing the slit width changes the diffraction pattern. Note in the simulation the intensity is relative; reducing slit width will reduce the intensity of the peaks.

Changing slit separation changes the fringe spacing. If C is the position of the 3rd maximum and the wavelength = 500 nm, the path difference =

3 x 500 nm = 1500 nm In this example the slit spacing is:

The angle between the principal diffraction max and the 1st minimum is much bigger than the angle from the central max to the first interference max.

### Multiple slit interference

A diffraction grating is a piece of glass with thousands of very fine lines printed to form a series of parallel slits. Each slit diffracts the light causing it to spread out. The light from each slit overlaps so the waves interfere creating areas of light and dark. The effects of having so many narrow and close slits are:

• the diffraction maximum is very wide
• the fringes are far apart
• the fringes are very narrow Increasing the number of slits per millimetre makes each slit narrower and further apart. The equation for the maxima is derived in a similar way to the equation for double slits except this time we will consider parallel light. This makes the geometry easier.

Consider a laser incident on an a number of equally-spaced very narrow slits. The light will be diffracted in all directions but let's isolate the direction of the first maximum. In this image you can see that each wave travels a different distance. If the waves are to add up then the path difference between each of the waves must be a multiple of $$λ$$. If we consider two adjacent slits the path difference for the first maximum must be $$λ$$. $$\sin\theta={\lambda \over d}$$

Maxima occur when...

$$\theta=\sin^{-1}({m\lambda\over d})$$

...where $$m$$ is a whole number.

Increasing the number of slits has two effects:

1. Intensity of the light increases
2. Sharpness of the lines increases

The latter needs a bit of explanation. First let's consider two slits.

To model the superposition of light passing through two slits we can apply the vector model used in diffraction. We can represent the light from two slits by two vectors. To model the way light superposes at different angles we can change the angle between the vectors as the angle is changed.

The blue dots plot the resultant of the two rotating vectors at different angles. The result is a series of equally spaced bright and dark fringes which is the result expected for double slits.

Lets now try this with three slits.

Now we get small subsidiary maxima between the principal maxima.

For four slits:

Now there are 2 subsidiary maxima and the principal maxima are brighter and narrower. With 1000 lines there will be 998 very small subsidiary maxima and the principal maxima are very bright and narrow.

The image represents laser light passed through multiple slits. How many slits are there?

There are 7 subsidiary maxima so 9 slits.

A 5mm wide laser beam is passed through a 2cm wide diffraction grating with100 lines per mm.

How many subsidiary maxima will there be?

the laser is 5mm wide so will pass through 500 lines.

The image represents light passing through 3 slits. This is close to:

The resultant is not zero and not as big as for a principal max where they add in a line.

### Resolvance

If light consisting of 2 different wavelengths is passed through a grating each wavelength forms a set of lines at different angles so a spectrum is produced. #### Spectrometry A spectrometer is used to measure the wavelength of the lines in a spectrum. Light is passed through a diffraction grating and the telescope (on the right) is rotated so that the line can be viewed. The angle is measured and the wavelength calculated.

If the two wavelengths are very close the lines will be close. If they are too close the lines will overlap and won't be resolved. Whether they overlap or not depends on how broad the lines are and that depends on the number of lines on the grating. The extent of overlap also depends on the wavelength. When you have a large number of slits the rotating vector model gives the same pattern as the single slit diffraction.

We know that the width of the central maximum depends on the wavelength:

Two lines will be resolved if:

$$\Delta\lambda={\lambda\over mN}$$

• $$Δλ$$ is the difference in wavelength ($$\text{m}$$)
• $$λ$$ is the mean wavelength of the two wavelengths to be resolved ($$\text{m}$$)
• $$m$$ is the order of the lines
• $$N$$ is the number of slits

Note that $$N$$ is the number of slits not the number per millimetre. The optics of the spectrometer spreads out the light so that it passes through all the slits on a grating. This might not be the case in exam questions; a laser will only pass through the slits within the width of its beam.

a 5mm wide beam of light passes through a 2cm wide diffraction grating with100 lines per mm. The light is made of two wavelengths, 500 nm and 501 nm.

Will the two 2nd order interference maxima be resolved?

$$\Delta\lambda={\lambda\over mN}$$

λ = 500 nm

m = 2

N = 500

Δλ = 0.5 nm

The lines will be resolved

### Thin film interference

When light reflects off a soap bubble some of the light reflects off the first surface and some passes into the soap. Some of this light is then reflected off the second surface. The two reflected waves are coherent so interfere.

If the waves are in phase they interfere constructively. The phase difference depends on the path difference which, if the angle of incidence is close to $$90°$$, is approximately twice the thickness of the bubble. The condition for constructive interference is only met for waves that are in phase. The colour of the reflected light is determined by those wavelengths. When a soap film is held vertically the soap drains downwards so the thickness at the bottom is greater than the top. Coloured bands form where the conditions for constructive interference is met for different wavelengths. Let's look at this more closely: If the angle of incidence is small the path difference is $$2t$$.

If the wavelength in the soap is $$λ_\text{s}$$, the phase difference caused by the difference in path is $$2\pi{2t\over \lambda_\text{s}}$$.

When light reflects off the soap it undergoes a $$-π$$ phase change but not when it reflects from soap to air. This gives an extra $$π$$ phase difference: $$2\pi{t\over \lambda_\text{s}}-\pi$$

The two waves will add constructively if the phase difference is $$2πm$$ (where $$m$$ is an integer):

$$2\pi{t\over \lambda_\text{s}}-\pi=2\pi m$$

$$2t=(m+{1\over2})\lambda_\text{s}$$

The refractive index from air to soap is defined as:

$$_\text{a}n_\text{s}={\sin i_\text{a}\over\sin i_\text{s}}={c_\text{a}\over c_\text{s}}={f\lambda_\text{a}\over f\lambda_\text{s}}$$ $$\lambda_\text{s}={\lambda_\text{a}\over_\text{a}n_\text{s}}$$

If you look up the refractive index of a material you will find the refractive index from a vacuum to the material. However the refractive index of air is about the same as a vacuum so we can assume that $$_\text{a}n_\text{s} = n_\text{s}$$, the absolute refractive index of soap.

$$2t=(m+{1\over2}){\lambda_\text{a}\over_\text{a}n_\text{s}}$$

The following images from a GeoGebra simulation shows how the wave is reflected to give constructive interference. The black wave passes through the soap film, notice how the wavelength decreases. The green wave is the wave reflected off the first surface; notice the $$π$$ change in phase. The peak that just passed through is a reflected trough.

Let's now look at the wave reflected off the second surface. The reflected wave is red. This time there is no phase change; the peak just passed through is reflected as a peak. You can see that the two reflected waves are in phase.

The film thickness is $$0.75λ_\text{s}$$ so:

$$2t=(1+{1\over2})\lambda_\text{s}=1.5\lambda_\text{s}$$

For destructive interference we can consider a film of thickness $$λ_\text{s}$$. First the wave reflected of the first surface... ...and the second: This time they are out of phase.

If the film is very thin... ...there is no path difference. There is only the phase difference due to reflection at the first surface so the reflected waves interfere destructively.

#### Coloured bands

The coloured bands in the soap bubble are caused when the condition for constructive interference is satisfied for a particular wavelength, they are like contours of thickness. The thickness of the film is therefore the same all along the blue line in the photo below. • The top, black band is where the film is so thin that all wavelengths interfere destructively.
• The white band is where the film is a bit thicker, no wave lengths are completely in phase or out of phase. All are reflected to some extent.
• The coloured line before the blue are due to the short wavelengths interfering destructively leaving the other colours (red, orange, green and yellow) to combine to give the colour.
• The first blue band is where blue light interferes constructively making blue much brighter than the other partially reflected colours.

A thickness of $$525\text{ nm}$$ will give constructive interference for red light of wavelength $$700\text{ nm}$$...

$$2t=(m+{1\over2})\lambda_\text{s}$$

$$2\times 525=(m+{1\over2})\times 700$$

$$m=1$$

...and also blue light of wavelength $$420\text{ nm}$$:

$$2t=(m+{1\over2})\lambda_\text{s}$$

$$2\times 525=(m+{1\over2})\times 420$$

$$m=2$$

A magenta band will form at this thickness.

#### Non-reflective coatings ​​​​​​​Thin film interference is used to reduce the reflection off lenses. The lens is coated with a thin layer of a material with higher refractive index than glass. Reflection off the coating-glass boundary will not undergo a phase shift where reflection off the air-coating boundary does. The coating is thin so that all wavelengths interfere destructively. The reduction in reflection increases the intensity of the transmitted light.

The diagram represents light waves reflected off a thin film. Which wave has undergone a phase change of π?

The green wave reflects off the more dense medium. A trough is reflected but a peak was incident.

Total Score:

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