Interference

a InterferenceCoherent Sources and Sustained Interferencenterference is a natural phenomenon that happens at every place and at every moment. Yet we don’t see interference patterns everywhere. Interference is the phenomenon in which two waves superpose to form the resultant wave of the lower, higher or same amplitude. The most commonly seen interference is the optical interference or light interference. This is because light waves are randomly generated every which way by most sources. This means that light waves coming out of a source do not have a constant amplitude, frequency or phase. The most common example of interference of light is the soap bubble which reflects wide colours when illuminated by a light source.Coherent SourcesTwo sources are said to be coherent when the waves emitted from them have the same frequency and constant phase difference.Interference from such waves happen all the time, the randomly phased light waves constantly produce bright and dark fringes at every point. But, we cannot see them since they occur randomly. A point that has a dark fringe at one moment may have a bright fringe at the next moment. This cancels out the effect of the interference effect, and we see only an average brightness value. The interference is not said to be sustained since we cannot observe it.Characteristics of Coherent SourcesCoherent sources have the following characteristics:• The waves generated have a constant phase difference• The waves are of a single frequencyCoherent Source Example• Laser light is an example of coherent source of light. The light emitted by the laser light has the same frequency and phase.• Sound waves are another example of coherent sources. The electrical signals from the sound waves travel with the same frequency and phase.Types of InterferenceInterference of light waves can be either constructive interference or destructive interference. Constructive interference: Constructive interference takes place when the crest of one wave falls on the crest of another wave such that the amplitude is maximum. These waves will have the same displacement and are in the same phase. for constructive interference, the path difference between the waves from two sources is path difference, QX–PX=n𝜆$$\text{path difference,}QX–PX=n𝜆$$ where n is the integer number.Destructive interference: In destructive interference the crest of one wave falls on the trough of another wave such that the amplitude is minimum. The displacement and phase of these waves are not the same. For destructive interference, we have path difference, QY–PY=(n+1

2)𝜆$$\text{path difference,}QY–PY=(n+\frac{1}{2})𝜆$$ where n is the integer number.Conditions for Interference of Light Waves For sustained interference of light to occur, the following conditions must be met:• Coherent sources of light are needed.• Amplitudes and intensities must be nearly equal to produce sufficient contrast between maxima and minima.• The source must be small enough that it can be considered a point source of light.• The interfering sources must be near enough to produce wide fringes.• The source and screen must be far enough to produce wide fringes.• The sources must emit light in the same state of polarization.• The sources must be monochromatic. Path Difference and Phase DifferenceWhen a wave passes through a medium, the particles of the medium vibrate. When the particles completer one to and fro motion, the wave advances by a distance equal to its wavelength λ. For a complete wave, the wavelength varies in λ and the phase is changed through 2𝜋${\displaystyle 2𝜋}$.. Let there be two waves with a path difference of λ. Then, the phase difference them will be 2𝜋${\displaystyle 2𝜋}$. If the path difference is x, then path difference = 2𝜋

𝜆 × x${\displaystyle =\frac{2𝜋}{𝜆}\times x}$ hence, phase difference=2𝜋

𝜆×path difference$$\text{hence, phase difference}=\frac{2𝜋}{𝜆}\times \text{path difference}$$Optical PathThe product of the distance travelled by the light in a medium and the refractive index of that medium is called the optical path. If d be the distance travelled by the light in a medium of refractive index µ, then by definition, Optical Path=𝜇d$$\text{Optical Path}=𝜇d$$ Young’s Double Slit Experiment

Simulation

S is a narrow vertical slit (of width about 1 mm) illuminated by a monochromatic source of light. At a suitable distance (about 10 cm ) from S, there are two fine slits S1${\displaystyle S_1}$ and S2${\displaystyle S_2}$ about 0.5 mm apart at equidistant from S. when a screen is placed at a larger (about 2m) from the slits S1${\displaystyle S_1}$ and S2${\displaystyle S_2}$, alternate bright and dark bands appear on the screen. The appearance of bright and dark bands are called the fringes.Theory of Interference of Light Suppose S1${\displaystyle S_1}$ and S2${\displaystyle S_2}$ be two fine slits at a small distance d apart in the figure. Let slits are illuminated by monochromatic light from a strong source S of wavelength λ and MN is a screen at a distance D from the double slits. The two waves starting from S1${\displaystyle S_1}$ and S2${\displaystyle S_2}$ superimpose upon each other resulting an interference pattern on the screen placed parallel to the double slit as in the figure.

Theory of interference fringesLet O be the centre between the slits S1${\displaystyle S_1}$ and S2${\displaystyle S_2}$. Draw S1P, S2P ${\displaystyle S_{1}P,S_{2}P}$and OC perpendicular to MN. The intensity of light at a point on the screen will depend upon the path difference between the two waves arriving at the point. The point C on the screen lies on the perpendicular bisector of S1${\displaystyle S_1}$ and S2${\displaystyle S_2}$. Therefore, the path difference between two waves reaching C is zero and hence, they are in phase. So, the point C is the position of maximum intensity. It is called central maximum. Consider a point P at a distance x from C. The path difference between two waves arriving at P is given by path difference=S2P–S1P$$\text{path difference}=S_{2}P–S_{1}P$$ From the geometry in figure, it is found that

	PQ=x-d 2;PR=x+d 2
	And(S2P)2–(S1P)2 =[D2+(x+d 2)2]-[D2+(x-d 2)2]=2xd
	or, (S2P–S1P)(S 2P+S1P)=2xd
	or, (S2P–S1P)=2xd BP+AP
	In practice, point P lies very close to C.
	So S2P≈S1P≈D
	S2P+S1P=D+D=2D
	Path difference=S2P–S1P=2xd 2D=xd D

$$\begin{array}{c}PQ=x-\frac{d}{2};PR=x+\frac{d}{2}\\ \text{And}(S_{2}P{)}^2–(S_{1}P{)}^2=[D^2+(x+\frac{d}{2}{)}^2]-[D^2+(x-\frac{d}{2}{)}^2]=2xd\\ \text{or,}(S_{2}P–S_{1}P)(S_{2}P+S_{1}P)=2xd\\ \text{or,}(S_{2}P–S_{1}P)=\frac{2xd}{BP+AP}\\ \text{In practice, point P lies very close to C.}\\ \text{So}{S}_{2}P\approx S_{1}P\approx D\\ S_{2}P+S_{1}P=D+D=2D\\ \text{Path difference}=S_{2}P–S_{1}P=\frac{2xd}{2D}=\frac{xd}{D}\end{array}$$ The waves from S1${\displaystyle S_1}$ and S2${\displaystyle S_2}$ arriving at a point on the screen will interfere constructively or destructively depending upon this path difference. The phase difference for this path difference is given by Phase difference, 𝜙=2𝜋

𝜆(xd

D )$$\text{Phase difference,}𝜙=\frac{2𝜋}{𝜆}\left(\frac{xd}{D}\right)$$ Bright fringesIf the path difference is an integral is an integral multiple of wavelength λ, then point P is bright. Therefore, for bright fringes

	xd D	=n𝜆
	x	=n𝜆D d…(i)

$$\begin{array}{rl}\frac{xd}{D}& =n𝜆\\ x& =n𝜆\frac{D}{d}\dots (i)\end{array}$$ where n = 0, 1, 2, 3, … The distance of the various bright fringes from the central maximum at C can be found as follows:

	For n=0,x0=0… central bright fringes
	For n=1,x1=𝜆D d…first bright fringes
	For n=2,x2=2𝜆D d…second bright fringes
	For n=n,xn=n𝜆D d…nth bright fringes

$$\begin{array}{c}\text{For}n=0,x_0=0\dots \text{central bright fringes}\\ \text{For}n=1,x_1=\frac{𝜆D}{d}\dots \text{first bright fringes}\\ \text{For}n=2,x_2=\frac{2𝜆D}{d}\dots \text{second bright fringes}\\ \text{For}n=n,x_n=n𝜆\frac{D}{d}\dots n^{th}\text{ bright fringes}\end{array}$$The distance between any two consecutive bright fringes is called fringe width, denoted by β.

	Fringe width, 𝛽=x2–x1=2𝜆D d-𝜆D d=𝜆D d
	∴𝛽=𝜆D d…(ii)

$$\begin{array}{c}\text{Fringe width,}𝛽=x_2–x_1=\frac{2𝜆D}{d}-\frac{𝜆D}{d}=\frac{𝜆D}{d}\\ \therefore 𝛽=\frac{𝜆D}{d}\dots (ii)\end{array}$$Dark fringesIf the path difference is an odd integral multiple of half wavelength λ, then point P is dark. Therefore, for dark fringes;

	xd D	=(2n -1)𝜆 2 where n=1,2,3,…
	or, x	=(2n-1)𝜆D 2d…(iii)

$$\begin{array}{rl}\frac{xd}{D}& =(2n-1)\frac{𝜆}{2}\text{where}n=1,2,3,\dots \\ \text{or,}x& =(2n-1)\frac{𝜆D}{2d}\dots (iii)\end{array}$$Equation (iii) gives the distance of the dark fringes from point C. the distance of the various dark fringes from point C can be calculated as below:

	For n=1,x1=𝜆D 2d…first dark fringe
	For n=2,x2= 3𝜆D 2d…second dark fringes
	For n=n,xn=(2n-1)𝜆𝜆D 2d…nth dark fringes

$$\begin{array}{c}\text{For}n=1,x_1=\frac{𝜆D}{2d}\dots \text{first dark fringe}\\ \text{For}n=2,x_2=\frac{3𝜆D}{2d}\dots \text{second dark fringes}\\ \text{For}n=n,x_n=(2n-1)𝜆\frac{𝜆D}{2d}\dots n^{th}\text{ dark fringes}\end{array}$$The distance between any two consecutive dark fringes is called fringe width β, given as

	Fringe width, 𝛽=x2–x1=3𝜆D 2d-𝜆D 2d=𝜆D d
	∴𝛽=𝜆D d…(iv)

$$\begin{array}{c}\text{Fringe width,}𝛽=x_2–x_1=\frac{3𝜆D}{2d}-\frac{𝜆D}{2d}=\frac{𝜆D}{d}\\ \therefore 𝛽=\frac{𝜆D}{d}\dots (iv)\end{array}$$from equation (ii) and (iv), it is clear that width of the bright is equal to the width of the dark fringe. From these two equations it is clear that fringe width increases as the1. Wavelength increases.2. Distance D of the screen from the sources increases3. Distance between the sources decreases.