The Principle of Superposition
We have learnt about the superposition of waves in Chapter 6. According to this principle, when two or more waves overlap, the resultant displacement of a particle of the medium at a given point is the sum of the displacements of the particle produced by individual waves, as if each wave is the only one which is present.
The last part of that statement is what makes the principle powerful. Each wave carries on exactly as it would have done alone; the medium simply adds their effects together at every point.
(Principle of superposition: when two or more waves overlap at a point, the resultant displacement of the particle at that point is the vector sum of the displacements that each wave would have produced individually at that point.)
Because of this, particles in the medium present where the crests (or troughs) of the two waves coincide will have larger displacements, while for particles present where the crest of one wave coincides with the trough of the other, the displacement will be minimum.
Take the special case of equal amplitudes, which makes the effect as sharp as possible. If the amplitudes of the two waves are equal, then for the first set of particles, the displacement will be twice the amplitude of the individual wave, and for the second set of particles, the displacement will be zero — the two waves cancel each other completely.
Thus, the intensity of the wave, which is proportional to the square of the amplitude of the wave, will be nonuniform, being larger at some places and smaller at others. This is called interference.
(Interference is the redistribution of intensity that results when two or more waves superpose, producing a nonuniform pattern with intensity maxima at some points and minima at others.)
Q3. Can two beams of light be added together in such a way that the result is darkness?
A. Yes. If two light waves of the same wavelength arrive at a point such that the crest of one falls exactly on the trough of the other, their displacements cancel at every instant, and if their amplitudes are equal the resultant is exactly zero. This is called destructive interference, and it is why a screen showing an interference pattern has genuinely dark bands where light from both sources is arriving. No energy is destroyed — it is simply redistributed into the bright bands.
Interference Seen in Water Waves
Interference is shown in Fig. 7.9 for water waves. S\(_1\) and S\(_2\) are sources of water waves of the same wavelength and amplitude, and are in phase with each other, i.e., at any given instant of time, the phases of the waves emitted by both sources are equal.
In the figure, the crests are shown by continuous circles while the troughs are shown by dotted circles. This simple convention lets us find at a glance what is happening at any point.
Points where the crests of one wave coincide with the crests of another wave, and where the troughs of one wave coincide with the troughs of another wave, are shown by blue dots. At these points the displacement is maximum and is twice that for each wave. These are points of constructive interference.
The points where the crests of one wave coincide with the troughs of another are shown by red dots. The displacement is zero at these points. These are points of destructive interference.
(Constructive interference occurs where the superposing waves arrive in phase, giving a resultant of maximum amplitude; destructive interference occurs where they arrive exactly out of phase, giving a resultant of minimum amplitude.)
Thus, along some straight lines radially diverging from the midpoint of S\(_1\)S\(_2\), there is constructive interference (along the radial lines connecting the blue dots) and destructive interference (along the radial lines connecting the red dots). The pattern is not a random scatter of bright and dark points but a set of well-organised lines fanning outwards.
Interference had been observed in the case of water waves and sound waves. It was observed for light waves in the laboratory for the first time by Thomas Young (1773–1829) in the year 1801. As noted above, this was the first proof of the wave nature of light. We will discuss this below.
Young’s demonstration settled a debate that had run for more than a century. Particles cannot cancel each other — two streams of corpuscles arriving at a point can only add up to more, never to nothing. By producing genuine dark bands from two beams of light, Young showed that light must be capable of cancellation, which is a property only waves possess.
7.8.1 Coherent Sources of Light
Two sources which emit waves of the same frequency having a constant phase difference, independent of time, are called coherent sources.
(Coherent sources are two sources that emit waves of the same frequency with a phase difference between them that stays constant in time.)
Now here is a puzzle worth thinking about. At any given point in space, at every instant of time, there are light waves from multiple sources overlapping one another. These must be interfering, and we should be able to see interference all around us at all times. However, we see no interference pattern anywhere in an ordinary room.
The reason is this. Different sources emit waves of different frequencies, and even if they emit waves of the same frequency, they are not in phase. Thus, the interference pattern changes every instant of time, and no pattern is sustained over a significant length of time for us to see. Bright and dark regions are indeed being formed, but they move about far too rapidly for the eye to register anything but a uniform average.
For interference to be seen over sustained periods, we need two sources of light which emit waves of the same frequency, and the waves emitted by them are in phase or have a constant phase difference between them, i.e., we need coherent sources of light.
This criterion cannot be satisfied by two independent primary sources, as they emit waves independently and there need not be a constant phase relation between them. Thus, to obtain a sustained interference pattern, one usually obtains two secondary sources from the same primary source, as is done in Young’s double slit experiment below. Since both secondary sources are fed by the same original wave, any change in that wave reaches both of them together, and their phase relationship never drifts.
7.8.2 Young’s Double Slit Experiment
The Experimental Arrangement
In this experiment, a plane wavefront is made to fall on an opaque screen AB having two similar narrow slits S\(_1\) and S\(_2\).
The plane wavefront can be either obtained by placing a linear source S far away from the screen, or by placing it at the focus of a convex lens kept close to AB. The rays coming out of the lens will be parallel rays and the wavefront will be a plane wavefront. Both methods work because a plane wavefront is simply a spherical wavefront from a very distant source, as we saw in Section 7.4.2.
Figure 7.10 shows a cross section of the experimental set up, and the slits have their lengths perpendicular to the plane of the paper. For better results, the slits should be about 2–4 mm apart from each other. An observing screen PQ is placed behind AB.
For simplicity we assume that the slits S\(_1\) and S\(_2\) are equidistant from S, so that the wavefronts starting from S and reaching S\(_1\) and S\(_2\) at every instant of time are in phase.
How the Two Slits Become Coherent Sources
When the rays fall on S\(_1\) and S\(_2\), the two slits act as secondary sources of light emitting cylindrical wavelets (with axis along the slit length) to the right of AB. This is Huygens’ principle in action: every point of the wavefront that reaches a slit becomes a fresh source, and because the slit is a narrow line, the wavelets it sends out are cylindrical.
The two secondary sources emit waves in phase with each other at all times, as the waves falling on them are in phase, coming from a plane wavefront. This is precisely the coherence condition of Section 7.8.1, achieved by deriving both sources from one original source.
The crests and troughs of the secondary wavelets superpose as shown in the figure and interfere constructively along straight lines joining the blue dots as shown. The points where these lines meet the screen have high intensity and are bright. The images of the slits are also perpendicular to the plane of the paper. The image plane, rotated through \(90^\circ\), is shown on the right side of the figure.
Midway between these lines are lines joining the red dots, along which the crest of one wave coincides with the trough of the other, causing zero intensity and producing dark images of the slits on the screen PQ. These are the dark regions on the screen as shown on the right.
The dark and bright regions are called fringes and the whole pattern is called the interference pattern.
(Fringes are the alternate bright and dark bands formed on the screen by interference; the complete set of such bands is called the interference pattern.)
Setting Up the Geometry
Let us now determine the positions and intensities of the fringes mathematically.
Let us take the direction of propagation of the plane waves incident on AB as the \(x\)-axis. The screen is along the \(y\)-\(z\) plane, the \(y\)-axis being along the plane of the paper. The origin of the axes can be taken to be O, the central point of S\(_1\)S\(_2\), as shown in Fig. 7.11, which shows the \(x\)-\(y\) cross section of the experimental setup.
Let the distance between the two slits S\(_1\) and S\(_2\) be \(d\) and that between O and O\('\) be \(D\), both measured in metres. For better results, \(D\) should be about a metre for the slit separation mentioned above.
The Central Bright Fringe
The point O\('\) on the screen is equidistant from S\(_1\) and S\(_2\). Thus, the distances travelled by the wavelets starting from S\(_1\) and S\(_2\) to reach O\('\) will be equal.
The two waves will be in phase at O\('\), resulting in constructive interference. Thus, there will be a bright spot at O\('\) and a bright fringe at the centre of the screen, perpendicular to the plane of the paper, as shown at the right in Fig. 7.10. This central bright fringe is present for every wavelength, since equal paths mean zero path difference regardless of colour.
Positions of the Other Fringes
Now let us determine the positions of other bright fringes on the screen. Consider any point P on the screen. The two wavelets from S\(_1\) and S\(_2\) travel different distances to reach P, and so the phases of the waves reaching P will not be the same.
If the path difference (\(\Delta l\)) between S\(_1\)P and S\(_2\)P is an integral multiple of \(\lambda\), the two waves arriving there will interfere constructively, producing a bright fringe at P. A whole number of wavelengths of extra path brings the wave back into step.
If the path difference between S\(_1\)P and S\(_2\)P is a half integral multiple of \(\lambda\), there will be destructive interference and a dark fringe will be located at P. Half a wavelength of extra path puts a crest exactly where a trough is.
Considering triangles S\(_1\)S\(_1'\)P and S\(_2\)S\(_2'\)P, we can write
\[(S_2P)^2 - (S_1P)^2 = \left[ D^2 + \left(y + \frac{d}{2}\right)^2 \right] - \left[ D^2 + \left(y - \frac{d}{2}\right)^2 \right] = 2yd\]
The two \(D^2\) terms cancel, and expanding the two squares leaves only the cross terms, which is why the result is so simple. Writing the left side as a product of sum and difference gives
\[S_2P - S_1P = \Delta l = \frac{2yd}{S_2P + S_1P}\]
For \(d/D \ll 1\), we can write \(S_2P + S_1P \approx 2D\), giving
\[\Delta l = \frac{2yd}{2D} = y\,\frac{d}{D}\]
This approximation is worth understanding rather than memorising. When the slits are very close together compared with the distance to the screen, both S\(_1\)P and S\(_2\)P are very nearly equal to \(D\), so their sum is very nearly \(2D\). The whole derivation depends on this condition, which is why the experiment is always set up with \(D \gg d\).
Thus, the condition for constructive interference at P can be written as
\[\Delta l = y_n \frac{d}{D} = n\lambda \qquad \text{--- (7.16)}\]
\(y_n\) being the position (\(y\)-coordinate) of the \(n^{\text{th}}\) bright fringe (\(n = 0, \pm 1, \pm 2, \ldots\)). It is given by
\[y_n = n \lambda D / d \qquad \text{--- (7.17)}\]
Here \(y_n\) and \(D\) are in metres, \(d\) in metres, \(\lambda\) in metres, and \(n\) is a pure number. The value \(n = 0\) gives \(y_0 = 0\), which is the central bright fringe we found above.
Similarly, the position of the \(n^{\text{th}}\) (\(n = \pm 1, \pm 2, \ldots\)) dark fringe (destructive interference) is given by
\[S_2P - S_1P = \Delta l = y_n \frac{d}{D} = \left(n - \frac{1}{2}\right)\lambda, \text{ giving}\]
\[y_n = \left(n - \tfrac{1}{2}\right) \lambda D / d \qquad \text{--- (7.18)}\]
Fringe Width
The distance between any two successive dark or any two successive bright fringes \((y_{n+1} - y_n)\) is equal. This is called the fringe width and is given by
\[\text{Fringe width} = W = \Delta y = y_{n+1} - y_n\]
\[W = \lambda D / d \qquad \text{--- (7.19)}\]
Subtracting Eq. (7.17) for \(n\) from the same equation for \(n+1\) leaves exactly one factor of \(\lambda D/d\), and the same subtraction applied to Eq. (7.18) gives the identical result. Thus, both dark and bright fringes are equidistant and have equal widths.
(Fringe width is the distance between the centres of any two successive bright fringes, or of any two successive dark fringes, in the interference pattern; it is given by \(W = \lambda D/d\).)
Read Eq. (7.19) as a piece of physics, not just a formula. The fringes get wider if you use light of longer wavelength, if you move the screen further away, or if you bring the slits closer together. Since \(\lambda\) for visible light is only a few hundred nanometres, \(d\) must be kept very small — a few millimetres — for the fringes to be wide enough to see at all.
(In a Young’s double slit experiment the slits are 1.0 mm apart and the screen is 1.5 m away. Light of wavelength 600 nm is used. Calculate the fringe width, and find the distance of the centre of the 3rd dark fringe from the central bright fringe.)
The Same Conditions in Terms of Phase Difference
We can also write Eq. (7.16) to Eq. (7.19) in terms of phase difference between the two waves as follows. Path difference and phase difference are two ways of saying the same thing, and different problems find one or the other more convenient.
The relation between path difference (\(\Delta l\)) and phase difference \(\Delta \phi\) is given by
\[\Delta \phi = \left(\frac{2\pi}{\lambda}\right) \Delta l \qquad \text{--- (7.20)}\]
The factor \(2\pi/\lambda\) simply converts a distance into an angle: one whole wavelength of extra path corresponds to a phase change of \(2\pi\) radians.
Thus, the phase difference between the two waves reaching P, from S\(_1\) and S\(_2\), is given by
\[\Delta \phi = y \frac{d}{D} \left(\frac{2\pi}{\lambda}\right) \qquad \text{--- (7.21)}\]
The condition for constructive interference in terms of phase difference is given by
\[\Delta \phi_n = n\,2\pi, \qquad n = 0, \pm 1, \pm 2, \ldots \qquad \text{--- (7.22)}\]
The condition for destructive interference in terms of phase difference is given by
\[\Delta \phi_n = \left(n - \frac{1}{2}\right) 2\pi, \qquad n = \pm 1, \pm 2, \ldots \qquad \text{--- (7.23)}\]
- For the interference pattern to be clearly visible on the screen, the distance (\(D\)) between the slits and the screen should be much larger than the distance (\(d\)) between two slits, i.e., \(D \gg d\). We have already used this condition while deriving Eq. (7.16).
- Conditions given by Eq. (7.16–7.19), and hence the locations of the fringes, are derived assuming that the two sources S\(_1\) and S\(_2\) are in phase. If there is a nonzero phase difference between them, it should be added appropriately. This will shift the entire fringe pattern but will not change the fringe widths.
Intensity Distribution
As there is constructive interference at the centre of a bright fringe, the amplitude of the wave is twice that of the original wave incident on AB, and the intensity, \(I\), being proportional to the square of the amplitude, is four times the intensity of the incident wave \(I_0\) say.
Pause on that factor of four, because it is a favourite examination point. Doubling the amplitude quadruples the intensity, so a bright fringe is four times as intense as one wave alone would produce — not twice.
At the centres of the dark fringes, the intensity is zero. At in-between points, the intensity gradually changes from zero to \(4I_0\) and vice versa.
Thus, the interference fringes are equally bright and equally spaced. This is, however, valid only in the limit of vanishing widths of the slits. For wide slits, the waves reaching a given point on the screen from different points along a single slit differ in path lengths travelled, and the intensity pattern changes, resulting in a blurred interference pattern with poor contrast.
Calculating the Intensity at Any Point
We can calculate the intensity at a point P on the screen where the phase difference between the two waves is \(\phi\), as follows.
We can write the equations of the two waves coming from S\(_1\) and S\(_2\), at the point P on the screen, as
\[E_1 = E_0 \sin \omega t \quad \text{and} \quad E_2 = E_0 \sin (\omega t + \phi)\]
The second wave is written with an extra phase \(\phi\) because it has travelled a slightly different distance to reach P. The incident intensity is
\[I_0 = |E_0|^2\]
The resultant electric field at P will be given by adding the two, using the trigonometric identity for the sum of two sines:
\[E = E_0 \sin \omega t + E_0 \sin (\omega t + \phi) = 2E_0 \cos (\phi/2) \, \sin (\omega t + \phi/2)\]
The amplitude of the wave is \(2E_0 \cos (\phi/2)\). Notice the structure of this result: the \(\sin(\omega t + \phi/2)\) factor describes the oscillation in time, while the \(2E_0\cos(\phi/2)\) factor in front is the amplitude, which depends only on the phase difference and therefore only on the position of P.
Thus, the intensity at P will be proportional to \(|2E_0 \cos (\phi/2)|^2\) and it will be equal to
\[4I_0 \cos^2 (\phi/2)\]
Check this against what we already know. When \(\phi = 0\), \(\cos^2 0 = 1\) and the intensity is \(4I_0\), a bright fringe. When \(\phi = \pi\), \(\cos^2(\pi/2) = 0\) and the intensity is zero, a dark fringe. The formula contains both earlier conditions as special cases.
When the Two Amplitudes Are Unequal
Let us consider the case when the amplitudes of the waves coming from the two slits are different, \(E_{10}\) and \(E_{20}\), say. In this case, the intensities of the bright and dark bands will be different from what is discussed above.
At the centre of the bright fringes, the amplitude will be \(E_{10} + E_{20}\) and hence the intensity will be proportional to \(|E_{10} + E_{20}|^2\), while at the centres of the dark fringes, the intensities will not be zero but will be proportional to \(|E_{10} - E_{20}|^2\).
This is why unequal amplitudes spoil the contrast. The dark fringes are no longer truly dark, since the smaller wave cannot completely cancel the larger one, and the pattern looks washed out.
7.8.5 Optical Path
Phase Change and the Idea of Optical Path
The phase of a light wave having angular frequency \(\omega\) and wave vector \(k\), travelling in vacuum along the \(x\) direction, is given by \((kx - \omega t)\) (Eq. (7.11)). Remember that wave vector \(k\) and the angular frequency \(\omega\) are related as \(k = \omega/v\), \(v\) being the speed of the wave, which is \(c\) in vacuum.
If the light wave travels a distance \(\Delta x\), its phase changes by
\[\Delta \phi = k \Delta x = \omega \Delta x / v\]
If the wave is travelling in vacuum, \(k = \omega/c\) and \(\Delta \phi = \omega \Delta x / c\).
In case the wave is travelling in a medium having a refractive index \(n\), then its wave vector \(k'\) and angular frequency are related by \(k' = \omega/v = \omega/(c/n)\), \(v\) being the speed of the wave in this medium. Since the wave is slower, the same distance now contains more oscillations, so \(k'\) is larger by exactly the factor \(n\).
Thus, if the wave travels a distance \(\Delta x\) in this medium, the phase difference generated will be
\[\Delta \phi' = k' \Delta x = \omega n \Delta x / c = \omega \Delta x' / c \qquad \text{--- (7.24)}\]
where
\[\Delta x' = n \Delta x \qquad \text{--- (7.25)}\]
Thus, when a wave travels a distance \(\Delta x\) through a medium having a refractive index of \(n\), its phase changes by the same amount as it would if the wave had travelled a distance \(n\Delta x\) in vacuum.
We can say that a path length of \(\Delta x\) in a medium of refractive index \(n\) is equivalent to a path length of \(n \Delta x\) in vacuum. \(n\Delta x\) is called the optical path travelled by a wave.
(The optical path travelled by light through a medium is the effective path it would have to travel in vacuum to generate the same phase difference; for a distance \(d\) in a medium of refractive index \(n\), the optical path is \(nd\).)
Thus, the optical path through a medium is the effective path travelled by light in vacuum to generate the same phase difference. In vacuum, the optical path is equal to the actual path travelled, as \(n = 1\).
Optical Path as an Equal-Time Path
Optical path in a medium can also be defined as the corresponding path in vacuum that the light travels in the same time as it takes in the given medium. This second definition is equivalent to the first and is often quicker to apply.
Now, time \(= \dfrac{\text{distance}}{\text{speed}}\), so equating the time taken in the medium to the time taken over the equivalent vacuum path,
\[\frac{d_{\text{medium}}}{v_{\text{medium}}} = \frac{d_{\text{vacuum}}}{v_{\text{vacuum}}}\]
\[\therefore \text{Optical path} = d_{\text{vacuum}} = d_{\text{medium}} \frac{v_{\text{vacuum}}}{v_{\text{medium}}} = n\, d_{\text{medium}}\]
Thus, a distance \(d\) travelled in a medium of refractive index \(n\) introduces a path difference
\[= nd - d = d(n - 1)\]
over a ray travelling an equal distance through vacuum. This extra amount \(d(n-1)\) is the quantity that appears in every problem about a thin plate placed in front of one slit, so it is worth remembering in this form.
Two waves interfere constructively when their optical path lengths are equal or differ by integral multiples of the wavelength.
(A ray of light of wavelength 500 nm travels 4 cm through a medium of refractive index 1.5. Calculate its optical path, and find the extra path difference it has gained compared with an identical ray that travelled 4 cm through vacuum.)
Shifting the Fringe Pattern with a Thin Plate
If we introduce a transparent plate of thickness \(t\) and refractive index \(n\) in front of slit S\(_1\) (Fig. 7.11), then the optical path travelled by the wave along S\(_1\)P is higher than that travelled by the wave along S\(_1\)P in the absence of the plate by \((n-1)t\).
Thus, the optical path lengths, and therefore the phases, of the rays reaching the midpoint O\('\) from S\(_1\) and S\(_2\) will not be equal. The centre of the screen is no longer a place of zero path difference.
The path lengths will be equal at a point different from O\('\), and so the bright fringe will not occur at O\('\) but at a different point where the two optical path lengths are equal. The dark and bright fringes will be situated symmetrically on both sides of this central fringe. Thus, the whole interference pattern will shift in one direction.
Note carefully what does not change. The fringe width \(W = \lambda D/d\) contains no reference to the plate, so the spacing of the fringes is unaffected — the entire pattern simply slides sideways, towards the slit covered by the plate.
What must be the thickness of a thin film which, when kept near one of the slits, shifts the central fringe by 5 mm for incident light of wavelength 5400 Å in a Young’s double slit interference experiment? The refractive index of the material of the film is 1.1 and the fringe width is 0.5 mm.
Given \(\lambda\) = 5400 Å, the refractive index of the material of the film = 1.1, and the shift of the central bright fringe = 5 mm.
Let \(t\) be the thickness of the film and P be the point on the screen where the central fringe has shifted. Due to the film kept in front of slit S\(_1\) say, the optical path travelled by the light passing through it increases by
\[t(1.1 - 1) = 0.1t\]
using the result \(d(n-1)\) derived above.
Thus, the optical paths between the two beams passing through the two slits are not equal at the midpoint of the screen, but are equal at the point P, 5 mm away from the centre. At this point, the distance travelled by light from the other slit S\(_2\) to the screen is larger than that from S\(_1\) by \(0.1t\) — this extra geometrical distance is what compensates for the extra optical path introduced by the film.
The difference in distances is
\[S_2P - S_1P = yd/D = y\lambda/W,\]
where \(y\) is the distance along the screen = 5 mm = 0.005 m, and \(W\) is given to be 0.5 mm = 0.0005 m. The substitution \(d/D = \lambda/W\) comes straight from the fringe width formula.
This has to be equal to the difference in optical paths introduced by the film. Thus,
\[0.1t = 0.005 \times 5400 \times 10^{-10}/0.0005\]
\[t = 5.4 \times 10^{-5}\text{ m} = 0.054\text{ mm}\]