Interference

Superposition PrincipleFor all linear systems, the net response at a given place and time caused by two or more stimuli is the sum of the responses which would have been caused by each stimulus individually. So if input A produces response X and input B produces Y than input (A+B) produces response (X+Y).InterferenceWhen the light from two different sources with same frequency and having a constant phase difference move in the same direction, then these light wave trains superimpose upon each other. This results in the modification of distribution of intensity of light. This modification of the intensity of light resulting from the superposition of wave is called interference. Coherent SourcesTwo sources are said to be coherent if they emit light of same frequency and always having a constant phase difference between them.

At some points the resultant intensity is greater than the sum of the intensities of the waves.Destructive InterferenceConstructive InterferenceAt some points the resultant intensity is smaller than the sum of the intensities of the waves.This is called Constructive Interference.This is called Destructive Interference.

The wave front originating from a common source is divided into two parts by using mirrors, prisms or lenses and the two wave fronts thus separated travels and finally brought together to produce interference.In this type sources are small like a point source.Classification of Interference1. Division of Wave front2. Division of AmplitudeThe amplitude of the incoming beam is divided into two parts either by partial reflection or refraction. These two parts travel in different paths and finally brought together to produce interference.In this type broad sources are required.

Youngs Double Slit Experiment:crestTrough S (Coherent Source)S1

S2 (Slit) (Screen)

Let S be a narrow slit illuminated by a monochromatic light of wavelength . S1 and S2 are two narrow slits close to each other and equidistant from S. Suppose is the frequency of the waves. Let a1 and a2 be the amplitudes of the two wave coming out of S1 and S2.Analytical treatment of InterferenceThe displacement y1 due to one wave from S1 at any instant t isThe displacement y2 due to other wave from S2 at any instant t isWhere is the constant phase difference between the two waves.The resultant displacement at P is the algebraic sum of the individual displacements

Squaring (i) and (ii) and then adding (i) (ii) Let

For maximum intensity orPhase Difference Path Difference Path Difference = n

For minimum intensity orPhase Difference Path Difference Path Difference For good contrast and

Average intensity When a1 = a2

Conditions for sustained interference:

Two light sources must be coherent.

Two coherent sources must be narrow, otherwise a single source will act as a multi sources.

The amplitude of two waves should be equal so that we can get good contrast between bright and dark fringes.

The distance between two coherent sources must be small.

The distance between two coherent sources and screen should be reasonable. The large distances of screen reduce to intensity.

Calculation of the fringe width:To determine the spacing between the bands/ fringes and the intensity at point P.

D S1

2d

S2 S (Coherent Source) Slit Screen O P N M

Path difference () = S2P-S1P

To calculate S2P, consider the S2NP

orExpending by binomial theorem

Therefore, higher power term of D can be neglected. Then we getorHereSimilarly, we can calculate S1P, consider the S1MP

Then the path difference is () = S2P - S1P For the nth fringe the path difference =

Bright Fringes:

The path difference should be equal to.where n = 0, 1, 2, 3, 4, ---------------- The distance between two consecutive fringes is also known as fringe width.

Dark Fringes: The path difference should be equal toPoint P to be darkwhere n = 0, 1, 2, 3, 4, --------------Fringe width

From the above equations, it is clear that fringe width depends on1. It is directly proportional to the distance between two coherent sources and screen

2. It is directly proportional to the wavelength of light

3. It is inversely proportional to the spacing between two coherent sources.

Fresnels Biprism:Fresnels biprism is a device to produce two coherent sources by division of wave front. SS2S1OHPGQ

Construction:

A biprism consists of a combination of two acute angled prisms placed base to base.The obtuse angle of the biprism is 179 and other two acute angles are 30.179

(b) Determination of the distance between two virtual sources:

Displacement method is one of the methods to calculate the distance between two virtual coherent sources:

UvuvAccording to the linear magnification produced by the lens:

Further the lens moves towards the eyepiece and a focused image of virtual sources S1 and S2 is visible in eyepiece again. This time the image separation of S1 and S2 should be appear different (d2) so that:

(1)(2)d1d2L2L1S1S2d

From equation (1) and (3), we get

or But

and

Thus equation (2) becomes

(3)

Applications of Fresnels Biprism:

Determination of thickness of thin sheet of transparent material like glass or mica. orHow to calculate the displacement of fringes when a mica sheet is introduced in the path of interfering rays? S1

2d

S2txDPmO

The time taken by light to reach P from S1 isButThe path difference S2P and S1P will then be given by

The path difference between S2P and S1P isorWe have already calculated that orLet the point P is the center of the nth bright fringe if the path difference is equal to nWhere xnis the distance of the nth bright fringe from the central fringe in the absence of mica.The position of the central bright fringe when the mica sheet is placed in the path S1P is obtained by putting n=0 in equation (1) we get(1)

Since >1 so that is positive. (2)The fringe width is Using equation (1)It means the fringe width is not affected by introduce of mica sheet.

Put these values in equation (2) we get,orThus we can calculate the thickness of mica sheet.

Light Reflection From Denser Media:l/2 Shift In Position of WaveInversion with l/2 shift in positionNo inversion

Change of Phase on Reflection When a wave of light is reflected at the surface of denser medium, it always gives a phase change of or path difference of /2 iAirGlassaratrNMDCBAa Here r and t are the reflection and the transmission coefficients when wave is travelling from rarer to denser medium.

If we reverse the direction of reflected and transmitted light then according to the Principle of reversibility , the original wave of amplitude a is produced, provided that there is no absorption of energyiAirGlassaratrNMDCBAar2+attart+atr Here r and t are the reflection and the transmission coefficients when wave is travelling from denser to rarer medium. The reversal of ar and at must reproduce the amplitude a. The sum of components along BE should be zero.E

Interference due to Reflection:SourcetiirrR1R2T1T2ABCDMN(Reflected rays)(Transmitted rays)The path difference between the reflected rays

(1)

BM = t and also Now, for AN

AC = AM + CM (because AM = CM)

or

So that,So that the actual path difference: As the ray is reflected from a denser medium, so an addition of path difference of /2 will be there.

So for Maximum Intensity, path difference should be equal to Where n = 0,1,2,3,4,5.. So for Minimum Intensity, path difference should be equal to Where n = 0,1,2,3,4,5.. Interference will not be perfect as there is difference in the amplitude of the reflected rays.

Production of colors in thin films: When a thin film of oil on water, or a soap bubble, exposed to white light (such as sunlight) is observed under the reflected light. The brilliant colors are seen due to the following reasonsThe path difference depend on the wavelength. It means the path difference will be different for different colors, so that with the white light the film shows various colors from violet to red.

The path difference also varies with the thickness of film so that various colors appear for the same angle of incidence when seen in white light.

The path difference changes with the angle r and angle r changes with angle i. So that the films assumes various colors when viewed from different directions with white light.

Newtons Rings: SourceActually the path difference between the interfering rays is The effective path difference for large radius of curvature or for small angle

For normal incidence cosr =1, then the path difference For maximaFor minima

Why Central Ring is dark in Newtons rings experiment At point of contact t = 0Path difference for dark ringFor n = 0,Path differenceThats why Central Ring is dark

How to calculate the radius or diameter of the nth fringe:Let be the radius of bright ring at point C and t is the thickness of air film at that point. Let R be the radius of plano-convex lens. In triangle OACso that the higher power terms are neglected. Therefore, we haveRRR-tOABCDorButt

For constructive interference: We have,Diameter

Now, for the air film the refractive index Therefore, Therefore, Diameter of the nth bright ring is proportional to the square root of the odd natural numbersFor Dark rings But for air film So,

Therefore, Diameter of the nth dark ring is proportional to the square root of the natural numbers

Spacing between successive rings This shows that the spacing decreases with increase in the order of the rings.

Determination of wavelength of light by Newtons Rings methodTherefore,

For Air film, Determination of refractive index of unknown liquid by Newtons Rings method For unknown liquid with refractive index

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