Physics 6 Tutorial 7 - Interference
Contents |
This is quite a long and challenging tutorial. Work through it slowly and carefully. If you don't understand it, ask for help from your tutor.
Before you attempt this tutorial, you should review Waves Tutorial 7 and Waves Tutorial 8. Remember also that coherent waves have:
the same frequency,
(nearly) the same amplitude,
and a constant phase relationship.
They do not have to be in phase, as long as the phase difference remains the same. Light waves of the same frequency are called monochromatic.
Note that the AQA Physics codes used in Waves Tutorial 7 are different to the SQA codes used here:
Wavelength is the same (l);
Distance to the screen is the same (D);
Slit separation is d here (instead of s);
Fringe separation is Dx (instead of w);
Number of wavelengths or orders is m (instead of n).
We know that constructive interference results when two waves interact and their displacements are in the same direction. If the displacements are in the opposite directions, the result is destructive interference.
In this diagram there are 8 blue waves from Source 1 and 12 red waves from Source 2. The waves are coherent. (They are different colours to show the different trains of waves.) The path difference is 4 wavelengths, and the two waves are in phase when they interact:
In other words, the crests coincide with crests, and the troughs coincide with troughs. Therefore they interfere constructively so the resultant has a large amplitude. The constructive interference is the result of the path difference being a whole number of wavelengths. Or we can say that the path difference is an even number of half-wavelengths.
Now in this case, the path difference is 4.5 wavelengths. Since the waves are p radians out of phase, there is destructive interference. In other words the crests coincide with troughs.
The resultant has an amplitude of zero because the waves have the same amplitude, but the directions are opposite. We say that the path difference is an odd number of half wavelengths.
What would happen to the resultant if the path difference was not quite an odd number of half wavelengths? |
We have seen how path difference is the difference between two distances travelled by two light rays from their sources. Consider these diagrams.
It doesn't take a genius to see that the path length between S1 to P is the same as S2 to P. The path difference is zero. So we can say:
S_{2}P - S_{1}P = 0
Now look at this:
We know that the difference between S_{2}P and S_{1}P is a whole number of wavelengths, m, because we have a bright region or bright fringe due to constructive interference. So:
S_{2}P - S_{1}P = ml
The rays meet at the first bright fringe, because there is a path difference of 1 wavelength.
In the diagram below, the path difference is (m + ½) wavelengths, because we have a dark fringe due to destructive interference:
So we can write:
S_{2}P - S_{1}P = (m + ½)l
In all these cases, the optical path difference is the same as the geometrical path difference. Now consider this:
In this case, the geometrical path difference is zero, as we would expect. However the optical path difference is (m + ½) wavelengths, because we have destructive interference to give a dark spot. The waves are out of phase by p radians.
The reason for this is that light waves travel more slowly in optically denser materials. In a material with refractive index of 1.5, the speed of light is 2.0 × 10^{8} m s^{-1}, as compared with 3.0 × 10^{8} m s^{-1} for air. The key point to remember is that the frequency remains the same. The colour of the light, governed by frequency, does not alter. Therefore to comply with the wave equation, the wavelength must be reduced by the same proportion. Therefore:
Therefore the optical path length can be worked out:
optical path length = geometrical path length × refractive index
So the optical path difference, G, in this case is:
G = (S_{2}P × n) - S_{1}P = (m + ½)l
The strange looking symbol, G, that looks like a gallows is "Gamma", a Greek capital letter 'G'. It is used in several sources.
We can now extend this to write a more general expression. Let's call S_{2}P distance d_{2}, and S_{1}P distance d_{1}:
G = (d_{2}n_{glass}) - (d_{1}n_{air}) = (m + ½) l
If the two distances are the same, we can write for a minimum:
G = d(n_{glass} - n_{air}) = (m + ½) l
If the result is constructive interference (leading to a maximum) the relationship is changed to:
G = d(n_{glass} - n_{air}) = m l
Worked example A coherent source of microwaves of wavelength 6.00 mm is split between two slits. Each microwave beam is 20 cm long. The top beam passes through air, while the bottom ray passes through a block of material of refractive index 1.51. A maximum reading is observed when the two rays meet.
What is the optical path difference in metres and wavelengths? |
Answer G = d(n_{material} - n_{air})
G = 0.20 m × (1.51 - 1.00) = 0.20 m × 0.51 = 0.102 m
m × 6.00 × 10^{-3} m = 0.102 m
m = 0.102 m ÷ (6.00 × 10^{-3} m) = 17 wavelengths. |
What would be the path difference for the example above, if the detector reads zero? |
In the last example we had the whole of one beam being transmitted by an optically denser material. Of course we don't have to have that. We could set up a laser with two slits and sneak in a thin piece of glass into the optical path - like this:
This is easier said than done.
Laser light of wavelength 620 nm is shone onto a screen as shown. A thin glass slide 0.1 mm thick is placed into the bottom ray. The glass has a refractive index of 1.51. (a) Calculate the optical path difference. (b) Discuss how the pattern on the screen might change, if at all. |
Interference has to use a Single Light Source
We can easily get a sound interference pattern from two loudspeakers. However, getting coherent light from two light sources (including lasers) is impossible. This is because photon generation is random, not continuous. Light is produced by the excitation of individual groups of atoms in bursts lasting less than nanoseconds (<1 × 10^{-9} s). There is no constancy in the phase relationships, even from a small region of the light source. Although we need not go into the explanation for this, it has been found that the coherence length for two rays of light rarely exceeds 1 mm. The phase relationships between the many millions of photons produced every second will be entirely random.
If we place a light bulb behind the two slits through which light passes, we will not achieve coherence. So we won't get an interference pattern. With a laser, we do, because the photons have a constant phase relationship.
If we place a single narrow slit in front or the light bulb, before the light passes to the two slits, the ray becomes more coherent, so an interference pattern can be seen.
Interference patterns are still difficult to see with a set-up like this. How anything meaningful could be observed with a candle...
Interference patterns in light are made using a single ray of light that is split into two. We can produce interference patterns by two methods:
1. Interference due to division of amplitude;
2. Interference due to division of wavefront.
Interference due to Division of Amplitude
A single light beam is split into two rays. One ray is transmitted and one is reflected at a boundary between two materials of different refractive indices. Consider a thin film of oil floating on water, as shown below.
The layer of oil has a thickness, t. The incident ray strikes the surface of the oil at an angle of q_{1}. It is refracted into the oil at an angle of q_{2}. When it has passed through the oil, it is refracted into the water at an angle of q_{3}.
The incident ray strikes the oil surface at an angle of incidence of 31 degrees. (a) Show that the angle of refraction into the oil is about 20 degrees. (b) Calculate the angle at which the ray is refracted into the water. Give your answer to an appropriate number of significant figures. |
Most of the light energy is transmitted into the water. However a small proportion of the energy is reflected at each boundary to make a weak reflected ray. In the diagram we can see the way these rays are reflected. Let's suppose 5 % of the energy is reflected at each boundary. The Incident ray has an energy of 100 units.
The first reflected ray, Ray 1, has an energy of 5 units.
The first refracted ray has an energy of 95 units.
At the boundary between the oil and the water, the second reflected ray has an energy of 0.05 × 95 units = 4.75 units.
At the boundary between the oil and the air, the second refracted ray (Ray 2) has an energy of 0.95 × 4.75 = 4.51 units
What would the energy of Ray 3 in units? What is the implication of this? |
The way Rays 1 and 2 interact depends on the thickness of the oil. If the optical path length of Ray 2 is one half wavelength different to the optical path length of Ray 1, we will get destructive interference. This explains why we get dark regions on a film of oil on water. We also get a spectrum of colours, because red light is refracted less than blue light.
Image by John - Wikimedia Commons
A similar effect is seen with soap bubbles, or coatings on glass. We will now look at the optical path difference in a thin film.
An important point before we start:
If the light goes from an optically less dense material (lower refractive index) to an optically more dense material (higher refractive index), there is a phase change of p radians when the light is reflected.
If the light goes from an optically more dense material (higher refractive index) to an optically less dense material (lower refractive index), there is no phase change when the light is reflected.
Optical Path Difference in a Thin Film
We will consider an incident ray of wavelength l that strikes an air-oil boundary at an angle of incidence of q_{1}. Most of the ray is transmitted, being refracted through an angle of refraction of q_{2}. However a weak reflected ray, Ray 1, is observed. The angle of reflection is, of course, q_{1}. It is important to note that the phase of the reflected ray is changed by p radians (180^{o}).
At the oil-water boundary, the ray is mostly refracted into the water. We are not interested in this, but we are interested in the ray that is reflected. Since the ray strikes a boundary where the water is optically less dense than the oil, there is zero phase change. The ray is reflected at an angle of q_{2}. It then is refracted through the oil-air boundary. Some is reflected back, but we are not interested in that. The angle through which Ray 2 is refracted is q_{1}.
The incident ray and Ray 1 are in red, while the path followed by Ray 2 is shown in orange. The geometric path difference of Ray 2 can be seen to be 2d. We can also express d in terms of the thickness of the film, t:
For the sake of simplicity, we will keep this factor out of the argument until the end. However we must also take into account the phase change which is ½ a wavelength:
Let us consider what would happen for constructive interference. We need a whole number of wavelengths (or even number of half wavelengths) for the optical path difference. Therefore:
We can rewrite this as:
Hence:
Rearranging:
It is far more likely that we will know the thickness, t. No problem; we have an expression above linking d to t. Therefore:
For destructive interference (a dark region on the film) we need the optical path difference to be an odd number of half wavelengths:
We know that:
Therefore:
Since the l/2 terms cancel, we can therefore write:
And we can rearrange to give:
We can now bring in the expression linking d and t:
A thin layer of oil of refractive index 1.51 is illuminated with red light of wavelength 600 nm, which is normally incident on the surface. The surface appears dark. The path difference has been calculated as 10 wavelengths. What is the thickness of the film? |
If you have a good quality camera, you will see that it has a thin coat of magnesium fluoride (MgF_{2}) on the lens. The idea is to stop light reflecting from the lens. In the picture, you can see how the reflection from the bright flash has been reduced.
The reason for the coating is to reduce flare and ghost images, which can spoil the picture. This camera also has a daylight filter, primarily to protect the lens. So let's look at how this works. Remember:
If the light goes from an optically less dense material (lower refractive index) to an optically more dense material (higher refractive index), there is a phase change of p radians when the light is reflected.
If the light goes from an optically more dense material (higher refractive index) to an optically less dense material (lower refractive index), there is no phase change when the light is reflected.
If the lens of a camera is pointed towards a very bright source, you can see repeated reflections of the iris diaphragm.
Consider a ray that strikes the air to magnesium fluoride coating at an angle of q_{1}.
As before most of the ray is refracted to an angle q_{2} and passes into the magnesium fluoride coating. The weak reflected ray undergoes a phase change of p radians.
When the refracted ray reaches the magnesium fluoride-glass boundary, most is transmitted into the glass. We are not interested in this transmitted ray. We are interested in the weak reflected ray. There is a second phase change of p radians.
Why is there a second phase change of p radians? |
Therefore the two reflected rays are in phase.
In the diagram, we showed the incident ray coming in at an angle. If we were to take into account the angle, we would need to bring in the relationship we saw with the oil film:
However for simplicity, we are going to make the incident ray strike the air-magnesium fluoride boundary normally, i.e. with an incident angle of zero. The transmitted ray is not deviated. Therefore:
d = t
The optical path difference is the geometrical path difference multiplied by the refractive index of the magnesium fluoride coating (n_{coating}):
We want destructive interference, so:
Therefore:
Remember that m is a whole number, 0, 1, 2...
If we want the thinnest coating, we need to have the thickness as half a wavelength, i.e. when m = 0. Therefore we can write:
And we can rewrite this as:
You may be asked to derive this equation in the exam.
Green light of wavelength 530 nm is normally incident on a lens coated with magnesium fluoride of refractive index 1.38. What is the thickness of magnesium fluoride that need to be applied to a camera lens to prevent the reflection of the green light from the lens? |
The thickness of the coating is designed to stop green light (the middle of the visible spectrum) from being reflected. Blue and red light is reflected as their wavelengths are different. Therefore a coated lens appears to be magenta (the mixture between blue and red).
If the ray comes in at an angle, we need to take that into account as we did above:
So we write:
If you put a couple of clean pieces of thin glass together, you may see interference patterns that look like this:
If you press on the pattern you can see it change. In a school lab, you will most likely see this with microscope slides or cover slips. Be careful, though. Microscope slides and cover-slips can very easily break and leave nasty shards of glass that can puncture your fingers.
The reason for this is that the surfaces of the slides are not perfectly smooth, and there may be small bits of dirt. Therefore there is an air gap, leading to interference patterns. We will look at this in more detail. In the diagram below, we can see two thin pieces of glass with a wedge of air between the two, as there is thin piece of paper between them at one end. Not shown are the paper clips that hold the slides in place. Monochromatic light is shone onto the slides.
The angles and thickness of the piece of paper are exaggerated to show what is happening. (The angle in the diagram is 2 degrees while in reality, the angle is a small fraction of a degree.)
Notice also that we ignore the weak reflected ray, which does not take part in the interference pattern.
We know that as light passes from glass to air, the reflected ray is subject to zero phase change. This is because it's at a boundary from a high refractive index to a lower refractive index. When the refracted ray strikes the boundary between the air and the glass of the bottom slide, it undergoes a phase change of p radians. This is because the glass has a higher refractive index than the air gap.
The optical path difference is the air gap between the two slides. It is also the geometric path difference. So we can write:
Why are the geometric path difference and the optical path difference both the same? |
For constructive interference to occur:
Rearranging:
Therefore:
For destructive interference:
Rearranging and tidying up, we get:
Remember that m is any whole number. In this equation, we can include m = 0.
State whether there is a light or dark fringe where the two slides touch. Explain your answer in terms of thickness and path difference. |
Two glass slides are l m long and are touching at the left end, and are separated by y m at the right-hand end. Consider a ray of monochromatic light striking the top slide x m from the left hand side. The separation of the slides is t m. This is shown in the diagram below:
There are m dark fringes in the distance x. We can see that there are two similar triangles. One has a base x and height t, while the other has a base l and height y. Therefore:
Rearranging:
Since:
We can substitute to give:
We measure the distance between the m^{th} and the (m+1)^{th} dark fringe to give us a distance Dx. This needs to be done with a travelling microscope as the fringe spacing is in the order of about 1 mm. From this we can write:
Which gives us:
In carrying out some measurements to study wedge fringes, a student is too hippy lazy to use the travelling microscope. Instead he measures the length x and counts the number of dark fringes. He works out the Dx term by dividing x by the number of of fringes. Explain whether or not this is wrong. |
Worked Example Two thin pieces of glass are 10.0 cm long. At one end, a piece of thin paper separates the ends of the glass slide. Monochromatic light of wavelength 612 nm is shone onto the glass. At a certain point, the separation between two adjacent fringes is found to be 1.51 mm. Work out the thickness of the paper. |
Answer Use:
y = (612 × 10^{-9} m × 0.100 m) ÷ (2 × 1.51 × 10^{-3} m)
y = 20.3 × 10^{-6} m = 20 mm |
Two microscope slides are 8.0 cm long and touch on one end. At the other end, there is a piece of paper 0.030 mm thick. The set-up is exposed to monochromatic light of wavelength 633 nm. Calculate the fringe spacing. |
The fringe separation decreases as the wavelength of the light decreases.
Interference by Division of Wavefront
In the process of interference by division of amplitude, a single wave was split and recombined. In interference by division of wavefront, we are combining two separate waves. The processes described in Waves 7 and Waves 8 arise due to the division of wavefronts.
For any interference effect to be observed, remember that the waves have to be coherent by:
having the same wavelength;
having a constant phase relationship.
The easiest way to achieve this is with a laser.
To study interference effects, we need to have a point source. A laser is close to a point source, which makes things much easier. Other light sources are extended sources, and getting coherence from the whole source is impossible. This is because photon production is random, and photons consist of short trains of waves.
To explain this, we will think about an example from sound waves. Consider the unpleasant experience of being close to an earth-strike in a thunderstorm. The lightning bolt is an extended source, with an irregular shape. The initial noise is an intensely loud high-pitched bang. All up the length of the stroke, the same would be heard. However as the sound waves propagate away from the source, the higher frequency sounds are absorbed, leaving only the low frequency sounds, which we hear as a rumble. As sounds reach you from higher up the stroke, you hear them as a rumble of varying intensity, as the lightning bolt has (or had) an irregular shape.
To get coherence from an extended light source, we can screen off the extended source, and allow the light to escape though a narrow hole, which makes the source act as a point source.
The basic treatment of this is to be found in Waves 7. The experiment can be carried out using apparatus as in the diagram above, but it is very difficult to see anything convincing. It is much more easily carried out using a laser.
On the screen we see the pattern in the diagram, along with the intensity of the bright spots:
The dark fringes occur when the path difference is an odd number of half wave lengths. The bright fringes happen when there is an even number of half wavelengths, or a whole number of wavelengths. The central bright spot occurs when the two waves have zero path difference. The first bright fringe occurs when the path difference is 1 whole wavelength, and so on. The pattern is symmetrical.
In this tutorial, we use the physics code m for number of wavelengths, rather than n. We will rerun the derivation from Waves 7, using the Physics codes used in this tutorial:
Wavelength is the same (l);
Distance to the screen is the same (D);
Slit separation is d here (instead of s);
Fringe separation is Dx (instead of w);
Number of wavelengths or orders is m (instead of n).
Consider a ray of monochromatic and coherent light of wavelength l m falling onto a double slit. The spacing between the centres of the slits is d m. The distance from the slit to the screen is D m, where D >> d. The fringe separation is Dx m. This is shown below:
For the central bright fringe, the path difference is 0.
S_{2}O
- S_{1}O = 0.
For the first bright fringe
S_{2}Q
- S_{1}Q =
l
Notice that in the triangle S_{1}S_{2}Z, S_{2}Z is l.
We know that S_{1}S_{2} is d.
Therefore:
sin q = l/d
For the triangle OPQ,
tan
f
= Dx/D.
Although in this diagram, it is clear that
q
¹
f, in the real thing, we can assume that
q =
f,
as the real set up is very much longer.
We know that for small angles
sin
q = tan
q.
Therefore:
Rearranging:
Worked Example A laser is shone onto a double slit of which the slit separation is 0.050 mm. The distance from the screen to the slits is 4.5 m. If the wavelength is 615 nm, what is the separation of the fringes m = 0 and m = 1? |
Answer Formula:
Dx = (615 × 10^{-9} m × 4.5 m) ÷ 0.050 × 10^{-3} m = 0.055 m = 5.5 cm |
A Young's slits experiment is set up with a slit separation of 0.400 mm. The fringes are viewed on a screen placed 1.00 m from the slits. The separation between the m = 0 and m = 10 bright fringes is 1.40 cm. What is the wavelength of the monochromatic light used? Give your answer to an appropriate number of significant figures. |