INTERFERENCE.

INTERFERENCE

Phase difference and path difference
The phase difference between two signals of the same frequency can be thought of as a delay or advance in the start of one signal's cycle with respect to another. Phase difference is expressed in degrees from 0 to 360. If the difference is 180 degrees then the two signals are said to be in antiphase: they are equal but opposite, and if added together will sum to zero. If the phase difference is 90 degrees then the signals are said to be in quadrature. If the phase difference between two waves is 2 then the path difference between that two waves is . Let for a path difference x, the phase difference is . For a path difference , phase difference = 2  so, for path difference x, the phase difference = So, phase difference  =

Two ways to get coherent source
Division of wave front. Young’s double slit Loyd Mirror Division of amplitude : Thin film interference Two ways to get coherent source Division of wave front : A narrow source and its virtual image or two virtual images can be used as coherent sources. Two slits illuminated by the light coming through a single slit can be used as coherent sources e.g. Young’s double slit , Fresnel bi-prism etc. Division of amplitude : the amplitude( Intensity) of a light wave is divided into two parts reflected and transmitted components, by partial reflection at two surfaces. Two reflected beams so produced interfere due to different paths traveled by the two beams leading to an additional path difference. E.g. thin films, Newton’s rings etc Note : No two independent sources are coherent

There are two types of interference
Interference is the superposition of two or more coherent waves resulting in a new wave pattern. There are two types of interference Constructive Destructive Two waves in phase Two waves 180° out of phase

Superposition of two waves: Intensity distribution
Let y1 and y2 are the displacements of two waves coming from S1 and S2 S2

 Resultant Intensity = (Resultant Amplitude) 2 I  cosine square

MAXIMA  Imax > I1+I2 Imax=4a2 MINIMA Imin =0  Imin < I1+I2

Law of conservation of energy
Iav = 2I Iav = I1+I2  Imax > I1+I2  Imin < I1+I2 This establishes that in the interference pattern, intensity of light is simply being redistributed i.e. energy is being transferred from regions of destructive interference to the regions of constructive interference. No energy is being created or destroyed in the process. Thus the principle of energy conservation is being obeyed in the process of interference of light

For incoherent light = sum of intensity of constituent waves.

Visibility of the fringes (V)

THE INTERFERENCE FRINGES
At point P for maxima we must have S2P – S1P = n, n = 0,1,2,3…

   D  1/d Position of nth bright fringe on the screen
Position of nth dark fringe on the screen Distance between any two consecutive bright or dark fringes (Fringe width)     D  1/d 

Separation between dark and bright fringes
/2 O is equidistant from S1 and S2 so light waves superposed at O are in phase so light intensity at O will be maximum. At O, we observe the central bright fringe. For this fringe n=0 y = 0. So central bright fringe will be referred as zeroth order bright fringe. O S1 S2

White Light in Young’s Experiment
Nature of the fringes????

Coherence time: the average time during which the wave remains sinusoidal and phase of the wave packet can be predicted reliably. Coherent length : the length of the wave packet over which it may be assumed to be sinusoidal and has predicted phase. Q2/Tut 1: Coherence length of a light is  10-2 m and its wavelength is 5896 Å. Calculate the coherence time and number of oscillations corresponding to the coherence length.

Fresnel biprism O Screen Displacement method Deviation method b a c d
Fresnel biprism Displacement method Deviation method b a c O d Fringes of equal width It consists of two thin acute angled prisms joined at the bases. It is constructed as a single prism of obtuse angle of 179º. The acute angle  on both side is about 30´. The prism is placed with the refracting edge in a way such that Sa is normal to the face bc of prism S is source S’ and S” are coherent sources (Virtual) obtained through ‘ab’ and ‘ac’ surface of biprism D= distance of screen from source. d = distance between the sources S’ and S’’. Z1 = distance of biprism from source Z2 = distance of screen from biprism D= Z1+Z2 Z1 Z2 D  Screen  179o  Ques: Can we determine Wavelength of source through this setup ? Ans: Yes 

Displacement Method from magnification formula,
In this method, the distance between the slit and the eye-piece is kept more than four times the focal length of a convex lens of short focal length used so that the two positions of the lens to form the images of S1 and S2 are found. A convex lens of short focal length is placed between the biprism and the eye-piece (as shown in fig. ). By moving the lens along the bed of the bench, two positions L1 and L2 are obtained such that the real images of S1 and S2 are obtained in the eye-piece. Let d1 and d2 be the separations between real images in the positions L1 and L2 of the lens, respectively. If ‘u’ and ‘v’ are the distances of the slit and the eye-piece from the lens in position L1 then from magnification formula, we have As the two positions of the lens are conjugate, for the second position L2 of the lens, we have Comparing eqns. (I) and (2), we get Thus, by measuring d1 and d2, d can be calculated. For accurate measurement of d, the procedure of determining d1 and d2 is repeated at least three times by moving eye-piece into different positions. from magnification formula, for L1 we have Multiplying (1) and (2) we get Similarly, for L2 Substituting the values of , d and D we can calculate the value of wavelength () of given monochromatic light.

In a biprism experiment with sodium light, bands of width 0
In a biprism experiment with sodium light, bands of width mm are observed at 100 cm from the slit. On introducing a convex lens 30 cm away from the slit, two images of the slit are seen 0.7 cm apart, at 100 cm distance from the slit. Calculate the wavelength of sodium light. Ans: Å

Deviation Method For small angles δ is angle of deviation
From right angle triangle and equation (1). In this method, d can be determined by using the fact that for a prism of very small refracting angle, the deviation  produced is given by z1 Hence (d1+d2) α1 D z1 α2 If base angles are different, then

Numerical In a certain region of interference 45th order maximum for the wavelength  = 5893 Å are obtained. What will be the order of interference at the same place for (a)  = 4820 Å, (b)  = 7576 Å. Ans: (a) 55th (b)35th The inclined faces of a glass prism of refractive index 1.5 make an angle of 1o with the base of the prism. The slit is 10 cm from the biprism and is illuminated by light of  = 5900 Å. Find the fringe width observed at a distance of 1 meter from the biprism. Ans: Fringe width= mm (Note: Use angle α in radian)

In a Fresnel's bi-prism experiment the refracting angles of the prism were 1.5o and the refractive index of the glass was 1.5. With the single slit 5 cm from the bi-prism and using light of wavelength 580 nm, fringes were formed on a screen 1 m from the single slit. Calculate the fringe width. For a thin prism: deviation (θ) = (n - 1)A = ( )1.5π/180 (in this case) Therefore: d= 7.5π/180 = cm therefore the fringe width is given by: x = λD/d = [580x 10-7 x 100]/0.131 = cm

WHAT WILL HAPPEN? In Double slit experiment:
Insert a thin transparent glass sheet of thickness t and refractive index  in the path of one beam. Fringe pattern will remain same WHAT WILL HAPPEN? or S1 S2 D d P y Any change in Fringe pattern t At n = 0 the shift y0 of central bright fringe is

D, d,  and t can be calculated
Knowing the shift in central fringe D, d,  and t can be calculated We use white light to determine the thickness of the material. For monochromatic light central fringe will similar to other bright fringe. For white light central fringe is white. WHY??????

Displacement of fringes
Determine condition of net path difference. µ1,t1 C’ S1 C S2 µ2,t2 Screen Case 1: If µ1= µ2=µ and t1>t2 , yo is positive ( upward shift) or t2>t1 , yo is negative ( downward shift) Case 2: If t1=t2= t and 1>2 , yo is positive ( upward shift) or 2>1 , yo is negative ( downward shift)

Principle of optical reversibility
In the absence of any absorption, a light ray that is reflected or refracted will retrace its original path if its direction is reversed. a  amplitude of incident ray r1  reflection coefficient t1,  transmission coefficient ar1  amplitude of reflected ray at1  amplitude of refracted ray Note r1, r2  reflection coefficients t1, t2  transmission coefficients n1, n2  refractive index of two media

Phase change  occurs when light gets reflected from denser medium.
According to principle of optical reversibility the two rays of amplitudes ar12 and at1t2 must combine to give the incident ray of fig 1. So, ar12 + at1t2 = a Stokes’ relation  t1t2 = 1- r12 ………(1) The two rays of amplitudes at1r1 and at1r2 must cancel each other. So, at1r1 + at1r2 = 0 Stokes’ relation  r2 = -r1 ……………(2) The –ve sign in the amplitude of reflected wave indicates the phase difference of π b/w two rays The wave looses a half wave on reflection at the boundary of rarer to denser medium: π phase or λ/2 path difference Phase change  occurs when light gets reflected from denser medium.  LLOYD’S MIRROR no phase change will occur when light gets Reflected by a rarer medium.

Interference in thin films due (Division of Amplitude)
to reflection (Division of Amplitude) Colors of oil film on water Colors of soap bubble Antireflection thin films

Interference by Division of Amplitude
Examples Interference by thin films Newton’s Ring Michelson’s Interferometer

( tthin-film order of the  visible light)
If plane wave falls on a thin film then the wave reflected from the upper surface interferes with the wave reflected from the lower surface. For thin-film optics: tthin-film 1 µm ( tthin-film order of the  visible light)

Interference by the film of Uniform thickness ‘t’: Reflected rays
Path difference due to optical path b/w beam 1 & 2 Δ1 = Path ACD(in thin film) – path AB (in air) Δ1 = μ(AC+CD) – AB = μ( d/cos(r)+ d/cos(r)) – AB Solve for AB = AD sin (i)= 2AE sin(i) = 2d tan(r) sin(i) = 2dtan(r) μ sin(r) =2 μ d sin2(r)/cos(r) Δ1 = 2μdcos(r) Total path difference b/w 1 & Δ = Δ1 -λ/2 (due to stoke’s law) =2μtcos(r) - λ/2 Condition for maxima Δ = 2μdcos(r) - λ/2 = n λ 2 dcos(r) = (2n+1) /2 n = 0,1,2,3… Condition for minima Δ = 2μdcos(r) - λ/2 = (2n-1) λ/2 2 dcos(r) = n  n = 0,1,2,3…

Do it your self….. 2 tcos(r) = n n = 0,1,2,3…
Interference by the film of Uniform thickness ‘t’: Transmitted rays: Do it your self….. Condition for maxima 2 tcos(r) = n n = 0,1,2,3… Condition for minima 2 tcos(r) = (2n+1) /2 n = 0,1,2,3…

Colors in thick films A thick film shows no color in reflected system when illuminated with an extended source of white light. An excessively thin film appears black in reflected system when illuminated with an extended source of white light

Problem: A man whose eyes are 150 cm above the oil film on water surface observes the greenish color at a distance of 100 cm from his feet. Calculate the probable thickness of the film. ( =500nm, oil=1.4 , water=1.33) 9.725×10-6(2n-1) cm Problem: A parallel film of sodium light of wavelength 5880Å is incident on a thin glass plate of =1.5 such that the angle of refraction in the plate is 60o. Calculate the smallest thickness of the plate which will make it appear dark at reflection. 3920Å

Numerical Light of wavelength 589.3nm is reflected at nearly normal incidence from a soap film of refractive index what is the least thickness of the film that will appear (I) black, (II) bright? Ans: nm, nm. A parallel beam of sodium light (589nm) is incidenting on an oil film on water. When viewed at an angle of 300 from normal, 8th dark band is seen. What is the thickness of oil film? Ans: 1.7 m

[Anti reflection coatings (AR)]
Anti reflection films [Anti reflection coatings (AR)]

Anti reflection films [Anti reflection coatings (AR)] Phase condition -wave reflected from the top and bottom surfaces of the thin films are in opposite phase. Overlapping leads to destructive interference . Amplitude Condition – Waves have equal amplitude

Assuming normal incidence of light i.e. cos r = 1
Δ between 1 and 2 = (2n-1)/2 ; where n=1,2,3... Assuming normal incidence of light i.e. cos r = 1 For the film to be transparent, the thickness of the film should be minimum which is possible for n = 1.

Anti reflection films [Anti reflection coatings (AR)] Phase condition -wave reflected from the top and bottom surfaces of the thin films are in opposite phase. Overlapping leads to destructive interference . Amplitude Condition – Waves have equal amplitude

Non reflecting films When films are coated on lens or prism surface the reflectivity of these surfaces is appreciably reduced. No light is destroyed by non reflecting film, but there is redistribution means decrease in reflection results increase in transmission. Thickness ‘t’ and refractive index ‘µ’ are important parameters for the fabrication of non-reflecting films.

Numerical A camera lens of refractive index 1.47 is to be coated with magnesium fluoride of refractive index 1.38 to obtain high transmission at infrared radiation of wavelength 900 nm determine the minimum thickness of coating.

Reflection and transmission coefficient
Normal incidence µ1 > µ 2 ai ai t µ1 µ2 ai r On reversing the condition ai t’ µ1 µ2 ai ai r’ Note: r, t , r’ and t’ obey Stoke’s relation.

Wedge shaped thin film No shift tcos(r+) shift of l/2

Wedge shaped thin film 2 tcos(r+)=n Condition for
constructive interference 2 tcos(r+) - /2 = n 2 tcos(r+) = (2n+1) /2 destructive interference 2 tcos(r+)=n So bright and dark fringes of different orders will be observed at different thickness of the film.

Note: In thin film interference 1. When  is very small ,
Cos(r+ )  Cos(r). 2. For nearly normal incidence on film, r = 0 Cos(r)=1 1. A parallel beam of light ( = 6000 Å) is incident on a thin glass plate of refractive index 1.5 such that the angle of refraction is 60o. Calculate the smallest thickness of the plate, which will appear dark by reflection. Ans: 0.4 µm. [2µtcos(r)=λ]

Fringe at Apex is Dark Straight and parallel fringes Locus of points having same thickness lie along lines parallel to the contact edge Equidistant fringes Localized fringes Fringes formed close to the top surface of wedge. Fringes of equal thickness

Determination of Wedge Angle
Using microscopes the positions of dark fringes at two distant points Q and R , OQ = x1, OR = x2 Thickness of the wedge t1 at Q, t2 at R For dark fringes at Q Fringe width: if N=1 then (x2-x1)=

The two glass surfaces are in contact at one end and separated at the other end by a thin wire. If the sodium light is incident normally, 20 interference fringes are observed between these edges. What is the thickness of the wire? t = 10 = 5.893x10-6m. Explain what happens to the fringe width (i) if wedge angle is decreased and (ii) oil is introduced between surfaces of the glass plates in a wedge shaped films.

Determination of the thickness of the spacer
Spacer forms the wedge shaped air film between the glass slides For thickness t of the spacer where l = length of the air wedge

The two glass surfaces are in contact at one end and separated at the other end by a thin wire. If the sodium light is incident normally, 20 interference fringes are observed between these edges. What is the thickness of the wire? Ans: For thin wedge shaped films, 2μtcos(r+) = n; t = xtan and  20fringe width) = x/n  x/n =  = /(2μtancos(r+)); for normal incidence and small value of ,  = /(2μ); tan =  = t/x = t/20;   = /2(t/20)  t = 10 = 5.893x10-6m. Explain what happens to the fringe width (i) if wedge angle is decreased and (ii) oil is introduced between surfaces of the glass plates in a wedge shaped films.

Newton’s Rings Special case of interference in air film of variable thickness

Newton’s Rings due to Reflected Light Interference is maximum
bright ring is produced. interference is minimum A dark ring is produced.

The experimental set up consists of a monochromatic light from an extended source S rendered parallel by a lens L. It is incident on a glass plate inclined at an angle 45o to the horizontal and is reflected normally down onto a plano-convex lens placed on a flat glass plate. Part of the light incident on the system is reflected from the glass-to –air boundary, say from point D. The remainder of the light is transmitted through the air film. It is again reflected from the air-to-glass boundary say at point J. The two rays reflected from the top and bottom of the air film are derived through the division of amplitude from the same incident ray CD and are therefore, coherent. The rays 1 and 2 are close to each other and interfere to produce dark and bright fringes. The condition of brightness or darkness depends on the path difference between the two reflected light rays. This in turn depends on the thickness of the air film at the point of incidence.

 radius of the circular fringe
RADII of DARK FRINGES  radius of curvature of the lens thickness of the air film  dark fringe be located  R2 =rn2 + (R-t)2 As R >> t rn2 = 2Rt – t2 , since 2Rt >> t2 rn2  2Rt  radius of the circular fringe

bright fringe at Q is dark fringe at Q is

Numerical 1. CENTER IS DARK
A thin Plano-convex lens of focal length 50 cm and µ=1.5 rests on an optically flat glass plate. Using light of wavelength 5890Å, Newton’s rings are viewed normally by reflection. Calculate the diameter of the 8th bright ring. Ans: mm. 1. CENTER IS DARK

This causes the rings to be unevenly spaced.
Each maximum and minimum → a locus of constant film thickness fringes of equal thickness. But  the rings are unevenly spaced. The diameter of the dark rings is Hence, the rings get closer and closer as the order of rings i.e. ‘n’ increases. This causes the rings to be unevenly spaced.

DETERMINATION OF WAVELENGTH OF LIGHT
Diameter of the mth dark ring Dm2 = 4 mλR Diameter of the (m+p)th dark ring Dm+p2 = 4 (m+p) λR  Dm+p2 - Dm2 = 4 pλR The slope is, Thus, R may be determined by using a spherometer  λ is calculated. Dm2 m+p m Dm+p2-Dm2

Refractive index of a Liquid
The liquid whose refractive index is to be determined is filled between the lens and glass plate. Now air film is replaced by liquid. The condition for interference then (for darkness) For normal incidence Therefore but  Following the above relation the diameter of mth dark ring is Similarly diameter of (m+p)th ring is given by

Newton’s rings in transmitted light
The condition for maxima (brightness) is 2tcos r =m And for minima (darkness) is For air (µ = 1) and normal incidence , for maxima (brightness) is for minima (darkness) is CENTRE IS BRIGHT

Newton’s rings formed by two curved surfaces
Case I: Lower surface concave Case II: Lower surface convex

Case I: Lower surface concave t=t1-t2 r

Case II: Lower surface convex r

→ 2[t + (λ/4)] = nλ+ λ/2=(2n+1)λ/2 (bright)
How can we make centre bright in reflected rays? Two ways: By using a liquid film with refractive index µliquid with condition µconvex lens< µliquid < µplate. Ex: crown glass=1.45, flint glass=1.63 Liquid with 1.45 < µ <1.63 Liquid film 2. By lifting convex lens upward with a distance λ/4. Because 2t=nλ (dark) → 2[t + (λ/4)] = nλ+ λ/2=(2n+1)λ/2 (bright)

Numerical: refractive index
In a Newton’s ring experiment the diameter of the 12th dark ring changes from 1.40 cm to 1.27 cm when a liquid is introduced between the lens and the plate. Calculate the refractive index of the liquid. =1.215 In Newton’s ring exp., the diameter of 4th and 12th dark rings are 0.4 and 0.7 cm, what will be the diameter of 20th dark ring. D20=0.905cm If the diameter of nth ring change from 0.3cm to 0.25 cm after filling a liquid b/w the lens and plate, find out the refractive index of liquid. = 1.44

Numerical: Two curved surfaces
The convex surface of radius 40 cm of a plano-convex lens rests on the concave spherical surface of radius 60 cm. If the Newton’s rings are viewed with reflected light of wavelength 6000 Å, calculate the radius of 4th dark ring. D4 = mm Newton’s rings by reflection are formed between two plano-convex lenses having equal radii of curvature being 100 cm each. Calculate the distance between 5th and 15th dark rings for monochromatic light of wavelength 5400 Å in use. D15 - D5 = 1.701mm

MICHELSON INTERFEROMETER

Interferometer An interferometer works on the principle of superposition. Michelson designed an interferometer to determine the wavelength of light. Here the basic building blocks are monochromatic source (emitting light waves) detector two mirrors one semitransparent mirror (often called beam splitter) Compensating plate

Experimental set up Interferometer produces interference fringes by splitting a beam of monochromatic light so that one beam strikes a fixed mirror and the other a movable mirror. When the reflected beams are brought back together, an interference pattern results.

M1’ Image of M1 M2 fix x2 x1 x2 M1 Where d = x1 ~ x2 movable

2dcos = n variable fix M2 S S’ M’1 S’’ d 2d θ 2dcosθ Source observer
path difference in between two rays (travel) 2dcos = n n is maximum at the centre variable fix Circular fringes: Because of constant (equal) inclination Haidinger Fringes

For d=0.3mm and =610-5 cm the bright fringes occur at  = cos-1 (m/1000) = 0, 2.56, 3.62, 4.44, 5.13, 5.73, 6.28, . . . m = 1000, 999, 998, 997, 996, 995, For d=0.15mm, the angles at which the bright rings occur will  = cos-1 (m/500) = 0, 3.62, 5.13, 6.28, 7.25, . . . m = 500, 499, 498, 497, 496, 495, . . Thus as ‘d’ reduces, the fringes will appear to collapse at the center and the fringes become less closely placed in figure (b).

Now, if ‘d’ is slightly decreased, from 0. 15 to 0
Now, if ‘d’ is slightly decreased, from 0.15 to mm, then the bright central spot corresponding to m =500 would disappear and the central fringe will become dark. Thus, as ‘d’ decreases, the fringe pattern tends to collapse toward the center. Conversely, if d is increased, the fringe pattern will expand.)

WHITE-LIGHT FRINGES G R
If a source of white light is used, no fringes will be seen at all except for a path difference so small that it does not exceed a few wavelengths. In observing these fringes, the mirrors are tilted slightly as for localized fringes, and the position of M1 is found where it intersects M1. With white light there will then be observed a central bright fringe, bordered on either side by 8 or 10colored fringes. This position is often rather troublesome to find using white light only. It is best located approximately beforehand by finding the place where the localized fringes in monochromatic light become straight. Then a very slow motion of M1 through this region, using white light, will bring these fringes into view. The fact that only a few fringes are observed with white light is easily accounted for when we remember that such light contains all wavelengths between 400 and 500 nm. The fringes for a given color are more widely spaced the greater the wavelength. Thus the fringes in different colors will only coincide for d = 0, as indicated in Fig. 13R. The solid curve represents the intensity distribution in the fringes for green light, and the broken curve that for red light. Clearly, only the central fringe will be uncolored, and the fringes of different colors will begin to separate at once on either side, producing various impure colors which are not the saturated spectral colors. After 8 or 10 fringes, so many colors are present at a given point that the resultant color is essentially white. Interference is still occurring in this region, however, because a spectroscope will show a continuous spectrum with dark bands at those wavelengths for which the condition for destructive interference is fulfilled. White-light fringes are also observed in all the other methods of producing interference described above, if white light is substituted for monochromatic light. They are particularly important in the Michelson interferometer, where they may be used to locate the position of zero path difference.

To obtain circular fringes---The planes of mirrors M1 and M2 should be made perfectly perpendicular to each other.

Michelson’s Interferometer
Newton’s Ring Michelson’s Interferometer Circular fringes because of locus of equal thickness of film Newton’s ring have minimum order at the centre Circular fringes because of equal inclination of light of incidence. Michelson’s interferometer have maximum order at the centre. 2dcos = n 2µt cos r=nλ Variable (angle of incidence) fix variable fix

Applications 1: Determination of wavelength
2dcos = n Bright = 2 ∆d / ∆n For Normal incidence cosθ =1, If we shift the position of variable mirror by ∆d (d2-d1= ), number of fringes ∆n (n2-n1=1) will appear or disappear on the screen accordingly. Finally we can decide the wavelength of source.

2. Determination of refractive index or thickness of a plate
Inserting a plate in the optical path increases path difference by (µ-1)t. Hence, 2(µ-1)t=∆n λ Where ∆n :number of fringes appeared or disappeared because of the insertion of thin glass plate in the path of any one ray. µ :refractive index of the plate

3: Difference between two neighboring spectral lines
Where, = wavelength d = separation between two position of distinctness

15.9 In the Michelson interferometer arrangement, if one of the mirrors is moved by a distance 0.08 mm, 250 fringes cross the field of view. Calculate the wavelength. [Ans: 6400 Å] 15.10 The Michelson interferometer experiment is performed with a source which consists of two wavelengths of 4882 and 4886 Å. Through what distance does the mirror have to be moved between two positions of the disappearance of the fringes? [Ans: mm]

15.11 In the Michelson interferometer experiment, calculate the various values of ̍ (corresponding to bright rings) for d = 5 ×10–3 cm. Show that if d is decreased to ×10–3cm, the fringe corresponding to m = 200 disappears. What will be the corresponding values of ̍ ? Assume  = 5 ×10–5 cm.

INTERFERENCE.

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