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CoherenceDefinition: - A predictable correlation of the amplitude and phase at any one point with other point is called coherence. Coherence is a property of waves. Two produce interference pattern coherent sources are required. Coherent light sources cannot be obtained from the separate light sources, but they are obtained from the single light source.Two waves are said to be coherent, the waves must have Same wavelengthSame amplitude andConstant phase differenceThere are two types of coherence Temporal coherence (or longitudinal coherence)Spatial coherence (or transverse coherence)Temporal coherence (or longitudinal coherence):- Definition:- It is possible to predict the amplitude and phase at a one point on the wave w.r.t another point on the same wave, then the wave is in temporal coherence. To understand this, let us consider two points and on the same wave as in shown in figure. Suppose the phase and amplitude at any one point is known, then we can easily calculate the amplitude and phase for any other point on the same wave by using the wave equation y=asin2πλc t-xWhere ‘a’ is the amplitude of the wave and ‘x’ is the displacement of the wave at any instant of time‘t’P1. P2Spatial coherence (or transverse coherence)Definition:- It is possible to predict the amplitude and phase at a one point on the wave w.r.t another point on the second wave, then the wave is in spatial coherence. Interference of lightDefinition:- The modification in the distribution of intensity in the region of superposition is known as interferenceThe best evidence for the wave nature of light is interference phenomenon. Interference concept is explained on the basis of superposition of wave’s concept.When two light waves superimpose, then the resultant amplitude or intensity in the region of superposition is different than the amplitude of individual waves. In case of interference pattern we observe two casesConstructive interferenceDestructive interferenceConstructive interferenceThe waves are reaching at a point are in phase constructive interference occursIn constructive interference, the resultant amplitude is always equal to the sum of the amplitudes of two individual waves. Condition The path difference between the two waves is equal to the integral multiple of wave length λ the constructive interference occurs. path diference=nλ Where n = 0, 1, 2, 3, 4 …Resultant WaveWave 2Wave 1Destructive interferenceThe waves are reaching at a point are in out of phase destructive interference occursIn Destructive interference, the resultant amplitude is always equal to the difference of the amplitudes of two individual waves.ConditionThe path difference between the two waves is equal to the odd half integral multiple of λ destructive interference occurspath diference=2n±1λ2 Where n = 0, 1, 2, 3, Resultant WaveWave 1Wave 2Conditions for interferenceTwo light sources of emitting light waves should be coherent.Two sources must emit continuous light waves of same wavelengths or frequency.The separation between the two sources should be small.The distance between the two sources and the screen should be large.To view interference fringes, the background should be dark.The amplitude of light waves should be equal or nearly equal.The sources should be narrow. The sources should be monochromatic. Interference in thin films by reflectionPrinciple:-The formation of colours (or bright or dark fringes) on thin films can explained as due to the phenomenon of interference. In this example, the formation of interference pattern is obtained by the division of amplitude.. Consider a thin film of uniform thickness ’t’ and refractive index ‘μ’. Let a monochromatic light ray AB is incident is on the upper surface of the film with an angle ‘i’. The Incident light ray AB is divided into two light rays BC and EF by the division of amplitude principle. The two light rays BC (ray 1) and EF (ray 2) are parallel and superimpose and produce interference. The intensity of interference fringe depends up on the path difference between the ray 1 and ray 2.Ray 1To find out the path difference between the light rays (1) and (2), draw one normal DG to BE and from point E to BC. From point E and H onwards the two light rays 1 and 2 travels equal distances. By the time the light ray 1 travels from B to H, the transmitted ray has to travel from B to C and D to E.The path difference between the light rays (1) and (2) ispath difference= μBD+DE in film-μairBH ( μair=1) (1)From ?BDG cosr=DGBD=tBDBD=tcos rSimilarly from ?DEG cosr=DGDE=tDEDE=tcos r∴BD=DE=tcos r (2)From ?BEH sini =BHBE=BHBG+GE∴BH=BG+GE.sini From ?BDG and ?DEG BG=GE=ttanrBH=2ttanr. sini From Snell’s law at point Bsini= μsinr ∴BH=2μttanr.sinr(3)Substituting the equations (2) and (3) in equation (1), we getPath difference =2μtcos r - 2μttanr.sinr =2μtcos r - 2μt.sin 2rcosr =2μtcos r 1- sin 2r=2μtcos r cos 2r =2μtcosr At point B the light ray (1) is reflected at the surface of thin film (denser medium). So the light ray (1) undergoes a phase change π or an additional path difference λ/2. Total path difference =2μtcosr-λ2Constructive interference (or Bright fridge)General condition; pathdifference=nλ2μtcosr-λ2=nλ2μtcosr=nλ+λ22μtcosr=(2n+1)λ2Destructive interference (or Dark fridge)General condition: pathdifference=2n-1λ22μtcosr-λ2=2n-1λ22μtcosr=nλNewton’s ringsPrinciple:-The formation of Newton’s rings can explained as due to the phenomenon of interference. In this example, the formation of interference pattern is obtained by the division of amplitude.Experimental arrangementThe experimental arrangement of Newton’s rings is shown in figure. The Plano -convex lens (L) of large radius of curvature (R) is placed with its convex surface on the glass plate (P). The lens touches the glass plate at O. A monochromatic light is allowed to fall normally on the lens with the help of glass plate M kept at 450 to the incident monochromatic light. OMThese reflected rays from the lower surface of the lens and the upper surface of the glass plate (or top and bottom surfaces of the air film) are interfere and give rise to an interference pattern in the form of concentric circular fringes. These rings are seen through microscope. Explanation of formation of Newton’s ringsLet R be the radius of curvature of the lens and t be the thickness of the air film at point P. At point P a part of the incident monochromatic light is reflected from the bottom surface of the glass plate without phase change. The other part of the light is refracted from the upper surface of glass plate with additional phase change of π or path difference λ/2. The path difference between the two rays is 2μtcosr+λ2For air film μ=1 and for normal incidence r=0, soThe path difference =2t+λ2Constructive interference (or Bright fridge)General condition: pathdifference=nλ2t+λ2=nλ2t=(2n-1)λ2Where n= 1, 2, ………..Destructive interference (or Dark fridge)General condition: pathdifference=2n+1λ22t+λ2=2n+1λ22t=n λWhere n=0, 1, 2, ………..At the point of contact t=0, path difference is λ2 i.e., the reflected and incidence light are out of phase and destructive interference occur. So the center fringe is always dark. Theory of Newton’s rings3938270351155To find the diameters of a dark and bright fringes construct a circle with the radius of curvature R of a lens L and let us choose a point P at a distance ‘r’ from the center of lens and t be the thickness of air film at point P.From the property of a circle NP.NB=NO.NDr.r=T.(2r-t) r2=2Rt-t2If t is small t2 is negligible.r2=2Rt t=r22RBright ringsFor bright ring, the condition is 2t=(2n-1)λ2 2 r22R=(2n-1)λ2r2=(2n-1)λR2By replacing r by D/2 the diameter of the bright ring is D24=(2n-1)λR2D2=2(2n-1)λRD=2(2n-1)λRD=2n-12λRD∝2n-1D∝odd natural numberDark rings416941060960For dark rings, the condition is 2t=n λ2 r22R=n λr2=n λRBy replacing r by D/2 the diameter of the nth dark ring is D24=n λRD=4n λRD=2n λRD∝nD∝natural numberNote: suppose a liquid is taken in between the lens and glass plate having refractive index μ, then the diameter of the dark nth dark ring can be written asD=4n λRμ Determination of wave length of sodium light using Newton’s ringsBy forming Newton’s rings and measuring the radii of the rings formed, we can calculate the wavelength of the light used if the radius of curvature of the lens is known. Let R be the radius of curvature of the lens and λ is the wavelength of the light used. Let Dm and Dn are the diameters of mth and nth dark rings. So the diameter of the nth dark ring can be written asDn2=4 n λ R (1)Similarly the diameter of the mth dark ring is Dm2=4 m λ R(2)Subtracting equation (1) from (2) we getDn2-Dm2=4 nλ R-4 m λ RDn+m2-Dn2=4 (n-m) λ Rλ= Dn+m2-Dn24(n-m)RUsing the above relation wavelength can be calculatedDetermination of refractive index of a liquid using Newton’s ringsBy forming Newton’s rings and measuring the diameter of the rings formed, we can calculate the refractive index of the liquid. In air film, the diameter of the nth and mth dark rings are Dn and Dm are measured with the help of travelling microscope.The diameter of the nth dark ring is Dn2=4 n λ R(1)The diameter of the mth dark ring is Dm2=4 mλ R (2)So Dm2-Dn2=4 mλ R-4 n λ RDm2-Dn2=4 (m-n) λ R (3)The Newton’s rings setup is taken in a liquid. Now the air film is replaced by liquid film. In liquid film, the diameters of the same nth and mth dark rings are D'n and D'm are measured with the help of travelling microscope.D'n2=4 n λ Rμ andD'm2=4 m λ RμSo D'm2-D'n2=4 (m-n) λ Rμ (4)Dividing equation (3) by (4)Dm2-Dn2D'm2-D'n2=4 (m-n) λ R4 (m-n) λ RμDm2-Dn2D'm2-D'n2=μUsing the above relation μ can be calculated. ................
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