To find the position of maximum and minimum energy, consider two sources of light A and B, as in the following figure (1), lying close to one another.

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1 INTERFERENCE OF LIGHT When two or more wave trains of light pass through the same medium and cross one another, then the net effect [displacement] produced in the medium is the sum of the effects due to all the waves. At any instant, the resultant displacement of the particles of the medium and hence the intensity of light depends upon (1) the phase difference between the waves and () the algebraic sum of the displacement due to the individual waves. Hence there is a modification in the intensity of light in the region of superposition of the two waves and is known as interference of light. Thus, modification in the distribution of light energy due to the superposition of two or more waves is called interference of light. With a single source, the distribution of energy in the medium is uniform. But, when there are two sources nearby, giving out continuous waves of the same wavelength and amplitude and having the same phase or a constant phase difference, the distribution of the energy in the medium is no longer uniform. At some points where the crest of one wave falls on the crest of the other wave (or trough of one wave falls on the trough of the other wave), the resultant displacement doubles and the intensity at that point become four times [Intensity is directly proportional to the square of the amplitude] and it is called constructive interference. On the other hand, if the crest of one wave falls on the trough of the other wave and vice versa, the resultant displacement is zero and hence the intensity becomes zero. This is known as destructive interference. Due to interference of light, there is only a redistribution of the energy in the medium and there is no creation or destruction of energy in the medium. The energy is only transferred from some points to neighboring points in the medium. Interference of light obeys the law of conservation of energy: We can prove that interference obeys law of conservation of energy. If A is the amplitude of each of individual interfering waves, then the sum of intensity in the absence of interference is I = A + A = A (1) On the other hand, in interfering waves, at points of constructive interference, I1 = (A) = 4 A and at adjacent points of destructive interference, I = (A- A) = 0. Hence the average intensity in the I I 4A 0 region of superposition is 1 1 I A () From equations (1) and () it follows that the law of conservation of energy is obeyed. 1 To find the position of maximum and minimum energy, consider two sources of light A and B, as in the following figure (1), lying close to one another. Let the two sources A and B emit continuous waves of monochromatic light of wavelength and of the same amplitude and in the same phase. At some point O equidistant from A and B, the two wave trains always arrive in the same phase and reinforce each other producing maximum intensity at that point. At some other point P, where AP and BP are not equal, there will be a difference in phase, which depends upon the path difference [BP AP]. If this path difference is an even multiple of /, [BP AP = n(/) = n] the waves arrive at the point in the same phase and reinforce each other producing maximum intensity. When the path difference [BP AP] = [n + 1](/) [odd multiple of /] the waves arrive at P in opposite phase and one wave cancel the effect of the other wave resulting in zero intensity at such points. CONDITIONS FOR PERMANENT INTERFERENCE OF LIGHT - COHERENT SOURCES: To obtain sustained interference of light waves at a point in the medium the following conditions should be satisfied. 1. Two sources should continuously emit waves of the same wavelength and frequency. Interference is a result of millions of waves crossing a point in the medium and hence the resultant displacement is a constant only if the periodic time is same for both the interfering waves.

2 . For observing interference fringes, the amplitude of the two interfering wave trains should be equal or very nearly equal. This will result in distinct visibility of bright and dark fringes. 3. Two sets of interfering wave trains should either have the same phase or a constant phase difference. This means that the two waves must originate from a single parent source. The phase difference between waves arriving at any point in the medium depends upon (i) path difference of the point from the two sources and (ii) the phase relation between the waves at the time of emission from the source. The path difference is constant for a given point in the medium as it depends only on the distance, while the latter rapidly changes with time not only in different sources, but even in different parts of the same source. Due to this reason it is not possible to observe interference phenomenon using two independent sources. In addition to the above three conditions, the following conditions should be maintained to observe clear, well defined fringe pattern in the region of superposition. 4. Two sources should be very narrow. A broad source is equivalent to a large number of narrow sources lying side by side. Each set of these sources produces their own interference pattern, which overlap on one another to such an extent that the interference pattern is replaced by general illumination. 5. Two sources emitting a set of interfering beams should be very close to each other. That means the two interfering wavefronts must intersect at a very small angle. Otherwise, the path difference between the interfering waves is large and owing to small wavelength of the waves the interference bands are formed so close to one another that the fringes may not be distinctly visible. Coherent sources: Two sources of light are said to be coherent if they emit (1)waves continuously of same frequency and wavelength () amplitude of the two waves must be equal or nearly equal and (3) they must emit waves of same phase or waves having a constant phase difference. Methods for producing coherent sources: There are two methods for producing coherent sources. They are (1) by the method of Division of wavefront and ( ) by method of division of amplitude. Division of wave front: In division of wavefront, the wavefront is divided into two parts, such that each part act as independent source emitting light having the same phase or constant phase difference. Examples: (1) In Young s double slit method, a single wavefront is divided in to two parts by allowing light from a single slit to pass through two slits () In Lloyd s method, the light is reflected from a plane mirror at grazing incidence. The source and its reflected image act as two coherent sources. In this case the waves emitted from the source and its virtual image emit waves of constant phase difference of Π. (3) Fresnel s Bi-prism method, light from a narrow source is passed through the bi-prism. The two refracted images act as two virtual coherent sources Division of Amplitude: In division of amplitude, the wavefront is divided in amplitude and the divided amplitude is made to interfere. The divided amplitudes act as two coherent sources. Examples: (1) In thin films, the light falling on the film is partly reflected form the upper and lower surfaces. These reflected waves from the two surfaces act as two coherent sources. () In Newton s rings, the light reflected from the upper and lower surfaces of the air film formed between the lens and the glass plate act as two coherent sources. (3) In Michelson interferometer, the beam is divided in amplitude by using a glass plate and again they are re-united. The re-united light act as coherent sources.

3 3 Young s double slit experiment: - Thomas Young demonstrated the interference phenomenon in The arrangement is as shown in figure (1). Light from a monochromatic source is incident on a screen having a narrow slit S in it. The cylindrical waves emerging from it are allowed to fall on another screen containing two close, narrow and parallel slits S1 and S. The slits S1 and S are parallel to the slit S. The cylindrical waves emerging from the two slits S1 and S super-impose and the resultant intensity of light is studied by arranging another screen at a suitable distance as shown in figure (1). On the screen bright and dark bands known as interference bands are seen. The waves arriving in phase at a point on the screen form a bright band and the waves arriving out of phase at a point on the screen form a dark band at that point. Theory of interference: - Let us consider two light waves of same angular frequency traveling in a medium along the same direction. Let a and b represents the amplitudes of the individual waves. The displacement of any particle of the medium due to the two waves at any instant t are given by, y1 = a sin t and (1) y = b sin (t +) () where is the phase difference between the waves at the point where they meet each other. According to the principle of superposition, the net displacement is given by, y = y1 + y (3) = a sin t + b sin (t +) = a sin t +b (sin t cos + cos t sin) = (a + b cos) sin t + (b sin) cos t = R sin t cos + R cost sin (4) = R {sin (t +)} (5) Where, a + b cos = R cos, (6) b sin = R sin (7) Squaring and adding equations (6) and (7), we get R = a + b + ab cos (8) and, tan = b sin /(a + b cos ) (9) Where R is the resultant amplitude and θ is the phase difference between the resultant and first wave.

4 Conditions for constructive interference: - The resultant amplitude R has maximum value when cos = +1 or when =n where n is 0, 1,, 3, 4. The value of R max = a + b + ab = (a+b) Rmax = (a + b) (10) The corresponding path difference for maximum value of R is given by, = (/) = (/) n = n (11) If a = b, then Rmax = a. The intensity of light Imax = R max.= 4 a. The two waves interfere constructively and hence the resultant intensity is 4 times the intensity of the individual waves. Condition for Destructive interference:- The resultant amplitude R has minimum value when cos = zero or when = (n+1) where n is 0, 1,, 3, 4. The value of R min = a + b - ab = (a-b) Rmin = (a - b) (1) The corresponding path difference for maximum value of R is given by, = (/) = (/) (n + 1) = (n +1)/ (13) If a = b, then Rmin = zero. The intensity of light Imin = zero. The two waves interfere destructively and hence the resultant intensity is zero. NOTE: - During interference the two waves does not destroy each other, but, their energy is only redistributed in the region of superposition. The average of the intensity of maximum and the minimum is equal to Iav = (Imax+ Imin)/ = {(a) +0 }/ = a, which is equal to the sum of the intensity of the two individual waves (a +a ). In other words, this phenomenon is supporting the law of conservation of energy. Expression for fringe width: - 4 Let d be the distance between the two slits A & B and D is the distance between the screen and the plane of the slits. Let O be a point on the screen such that AO=BO. The waves from A and B arrive in phase at O since they travel the same path and hence a bright band is formed at O. It is called central bright band (centre of the fringe pattern). Figure () Let P be a point on the screen at a small distance x from O as shown in figure. The path difference between the waves from A and B reaching the point P is BP AP = From BFP, BP = BF +FP = D + (x + d/) ---- (1) From AEP, AP = AE +EP = D + (x - d/) ---- ()

5 5 Now, BP AP = (BP+AP) (BP AP) Therefore, the path difference is given by, = [D + (x + d/) ] [D + (x - d/) ] = x + d /4 + xd - x - d /4 + xd = xd = BP AP = xd/ (BP+AP) = xd/ (D+D) [BP AP D] = xd/d (3) For bright fringe = xd/d = n (4) The distance of the n th bright fringe is given by, xn = nd/d The width of a dark fringe = xn xn-1 = [n (n -1)] D/d = D/d----- (5) For dark fringe = xd/d = (n+1)/ (6) The distance of the n th dark fringe is given by, xn = (n+1) D/d The width of a bright fringe = xn xn-1 = [(n+1)-({n-1}+1)] D/d = [n+1 n+1] D/d = D/d (7) Equations (4) and (5) show that, the width of a bright or a dark fringe (band) is a constant and is independent of the fringe number. The width of a bright or a dark fringe is known as fringe-width and is denoted by. Thus, = D/d. --- (8) Note : - 1. Interference pattern consists of bright and dark fringes of equal width.. If white light sources are used, the fringe pattern is coloured with the central bright fringe white and the other fringes are coloured in the order of VIBGYOR. BIPRISM A biprism consists of two thin prisms with their bases joined and their two faces making an obtuse angle of about so that the other angles are each of about In practice the biprism is constructed from a single glass plate so that it is ground to have the required angles. When biprism is held symmetrically at a short distance from an illuminated slit, the wavefront is divided into two parts. Each half of the Bi-prism intercepted by one half of the wavefront produces a virtual image of the source S by refraction. The distance between the source S and the Biprism P is so adjusted that the two virtual images S1 & S are quite close to one another. The waves from the two virtual sources superimpose and produce interference fringes in the region of superposition AB on the screen as shown in the following figure (3)

6 6 A closely spaced fringe system is produced only in the region AB while the wide set of fringes are formed at the edges of the pattern on account of diffraction [in regions AE & BF]. Since interference fringes are narrow they are generally observed using a low power microscope so that the fringes formed in the focal plane of the microscope are observed. Determination of wavelength of monochromatic light: Theory of Bi-prism is same as the theory of interference discussed in Young s method. The fringe width is given by the formula, = D/d (1) Where, is the wavelength of the radiation, D is the distance between the slit and the screen & d is the distance between the two virtual sources. Procedure: The slit S, the Bi-prism P and a micrometer eyepiece E are mounted at the same height along a straight line using three uprights capable of vertical and transverse adjustments, fitted on an optical bench. The slit is illuminated by the monochromatic light of wavelength (), which is to be determined. The bi-prism is moved at right angels to the optical bench, so that on looking through the eyepiece holder, two bright virtual slits S1 & S are observed. The eyepiece is moved perpendicular to the bench so that overlapping region AB is brought in the field of view and the fringes are obtained in the focal plane. Well-defined fringes are obtained by arranging the refracting edge of the bi-prism parallel to the slit by rotating the bi-prism suitably. (1) Measurement of fringe width : The centre of the particular bright fringe is brought on the cross-wire of the micrometer eyepiece and the reading for this fringe is noted using micrometer eye piece. The position of the cross wire is moved to one side until 5 fringes cross the field of view using slow motion screw and the reading for the 5 th bring fringe is noted. The same procedure is repeated to note the readings for 10 th, 15 th, 0 th, 5 th and 30 th fringes. The readings are tabulated and the width of 0 fringes and hence with its mean value is calculated. From the mean width (β 1 ) of 0 1 fringes, the fringe width () is calculated using the relation, 0 Trial No Fringe Number Reading for Fringes Width of 0 fringes \ Mean width of 0 fringes 1 Mean fringe width, 0 = mm

7 () Measurement of distance D: The distance D between the slit and the focal plane of the eyepiece is measured with the help of the scale on optical bench. (3) Measurement of distance between the virtual sources: A convex lens of suitable focal length is placed between the eyepiece and the biprism without disturbing their positions. The position of lens is adjusted to get magnified image of the slits. The separation of the two magnified images x is noted. If, d is the distance between the two slits, then the magnification is given by, m1 = x/d = v/u () 7 The position of the lens is again adjusted to get diminished image of the slits. The distance between the two diminished images y is determined. The magnification of the diminished image is given by, m = y/d = u/v (3) Therefore, m1 x m = [x/d] [y/d] =[v/u] [u/v] = 1 And hence, xy = d or d = [xy] 1/ (4) Knowing x & y the distance between the slits can be calculated using the equation (4) Thus by knowing fringe width (), d and D the wavelength of the monochromatic source is calculated using equation, = d/d (5) LLOYD S SINGLE MIRROR Lloyd s mirror is an optical device for producing interference fringes devised by Dr. H. Lloyd in It is a single mirror of length about 30 cm and breadth about 5 cm made of either polished metal or highly polished black glass so that reflection occurs only from the front surface. When light from a slit S [with the slit parallel to the breadth of the mirror] is made to fall at grazing incidence, the light reflected from the mirror appears to come from the virtual image S 1. Thus the direct light from S and the light from the virtual slit S 1 superimpose in the region AB on the screen to produce interference pattern with straight fringes parallel to the slit as shown in figure (4). Note 1: Fringes are formed only on one side of the central fringe [not visible] corresponding to point C of equal path difference from the two sources. Central fringe can be made visible by introducing a thin mica sheet of a certain thickness t so that the extra path difference of ( - 1) t is introduced in the path one beam. Note: The central fringe in this case is dark, which according to theoretical consideration must be bright. The reason for central dark fringe is due to a phase change of on reflection from the denser glass plate. Thus the waves from the two sources have an initial phase difference of at the time of emission itself. Note: Due to reason mentioned in note (), the condition for constructive interference to get bright fringe is = xd/d = (n+1)/ and the condition for destructive interference is = xd/d = n

8 8 Thickness of a transparent plate using Biprism: When light from the source passes through the glass plate, an extra path difference is introduced by the glass plate. The distance travelled in the glass plate = t. The distance travelled in air = AP t If c and v are the velocities of light in air and glass respectively, then the time taken to travel the distance from A to P AP t t AP t t AP ( 1) t c v c c c Due to the introduction of the glass plate, the ray from A travels a path = AP + (μ 1) t. In the absence of the plate, the central maximum is formed at O. Due to the introduction of the plate, the central maximum shifts to a point P at which the optical paths travelled by the rays from A and B are equal. x d In the absence of the plate, the path difference = BP AP. D x d In presence of the glass plate, the path difference = BP AP 1t 1t D If P corresponds to the position of n th maximum, then, x d D nd D 1t 1 t n x n 1 t D d d d In the absence of the plate, the n th nd maximum is at a distance = d The shift in the position of n th nd D 1t nd D 1t maximum is x d d d d It is evident that the shift in the n th fringe is independent of the fringe number and is same for all fringes. Thus the introduction of glass plate does not change the fringewidth. The thickness of glass plate is related to the shift in fringe pattern by the relation, x d t D 1 Thus, by determining x, d & D from experiment and by assuming the value of μ, it is possible to determine the thickness of a thin glass plate by using interference phenomenon. Alternately, if thickness is known, the refractive index of the material of the medium can be calculated. INTERFERENCE BY DIVISION OF AMPLITUDE Beautiful colour observed in thin films of oil spread on water, soap bubbles and heated metal surface is a result of interference phenomenon. In all these cases the division of amplitude produces interference phenomenon. The divided amplitude reunites in the region of superposition to produce interference.

9 Interference in thin film of uniform thickness [plane parallel thin films] Let monochromatic light of wavelength from an extended source S fall on the surface XY of the thin film of uniform thickness t and refractive index as shown in figure (1). The ray SA of the incident wavefront is partly reflected along AP and partly refracted along AE. At E, the ray is partly reflected along EB and partly refracted along the emergent ray EP1. Similar reflection and refraction occurs at B, F, C, G, D and so on as shown in figure. This successive reflection and refraction of the incident light results in two system of parallel beam. One of the beams is due to reflection of light & the other is due to transmission of light. In both the system of parallel beam intensity of successive rays falls off rapidly from one ray to the next ray. These reflected and transmitted parallel rays produce interference phenomenon. [A] CONDITION FOR CONSTRUCTIVE & DESTRUCTIVE INTERFERENCE FOR REFLECTED SYSTEM OF RAYS: Figure () ABN = i BAM = r Also, AGM = r AG = t Let us consider two reflected rays AP and BQ in reflected system. AP is reflected from the first surface so that the reflection occurs from denser medium of refractive index and BQ, after reflection from the lower surface from a rarer medium is transmitted from the first surface. Draw BN perpendicular to AP and AM perpendicular to EB. Draw AL normal to the first surface at the point A & produce it to meet BE produced at G. From figure, it follows that, ABN = i, BAM =AGM = r. Also, the triangles ALE & GLE are congruent. Therefore, EA = EG & AL = LG If is the refractive index of the material of the film, then the path difference measured in air between the two reflected rays AP & BQ is given by, Path difference, = [AE +EB] AN = [GE +EB] AN = GB AN 9 Therefore, path difference, = GB - MB = [GB BM] = GM In triangle AGM, Cos r = GM/AG GM = AG Cos r = t Cos r Thus, the path difference between the reflected rays, = t Cos r (1) Since the ray AP suffers reflection from denser medium, it undergoes a phase change of or a path increase of [/], Where is the wavelength of light in air. The ray AE, EB and BQ do not undergo any phase change as it is reflected from a rarer medium. Thus, the effective path difference between AP and BQ = t Cos r - / Consider the reflected rays BQ and CR. These have a path difference of t Cos r as they have only internal reflection. The same path difference exists between successive pairs of reflected rays.

10 CASE (1): When the thickness of the film is negligibly small compared to. If t Cos r = n, except the first two reflected rays all other successive pair of rays will be in phase, but their amplitudes rapidly decreases. If the thickness t of the film is negligibly small, then the total path difference between AP & BQ equals / and hence they interfere destructively and darkness will result. [due to the effect of AP cancelled by all other rays taken together]. CASE (): When the thickness of the film is not negligible compared to. (a) The film appears bright if t Cos r - / = n, where, n = 0, 1,, 3, Or, t Cos r = [n +1] / () When the above condition is satisfied, the first and the second rays interfere constructively. The third, fifth, seventh etc., rays are out of phase with the first two rays, where as, the fourth, sixth, eighth etc., rays are in phase with the second ray. The fourth has larger intensity than the fifth, sixth has larger intensity than the seventh and so on. Due to this the rays of stronger intensity in the series combine with the first ray to give maximum intensity in the field of view of reflected beam. (b) The film appears dark if t Cos r - / = [n 1] / Or, t Cos r = [n 1 +1]/ = n (3) In this case the first ray is out ;of phase with the second, while the other rays are in phase with the second ray. The second and the other reflected rays cancel the effect of the more intense first ray to produce darkness in the field of view of the reflected beam. As the source is extended, the rays of light are incident at various angles and hence the path difference has variable values for different set of reflected rays. Hence the field of view in the reflected beam consists of alternate bright and dark band. Also, the fringe pattern is not very well defined as the intensity of minima is not equal to zero. [B] CONDITION FOR INTERFERENCE IN THE CASE OF TRANSMITTED SYSTEM OF FRINGES: The rays emerging from the lower surface of the film may be brought together by a lens to produce interference. The path difference between the rays EP 1 & FQ 1 is given by, 10 = [EB + BF] EN = [EB + BG] EN = EG - EM = [EG EM] = GM = GF Cos r = t Cos r = t Cos r ----(4) figure (3) EN = EM Since all reflections occurs from the rarer medium, no extra path difference is introduced. Case (i) If t is very small compared to, t Cos r is negligible and the two waves interfere constructively so that the film appears bright in transmitted light. Case (ii) If t is not negligible compared to, [t Cos r] increases with the increase in the thickness of the film. (a) When t Cos r = (n+1)/, the film appears dark (b) When t Cos r = n, the film appears bright. As the source is extended, the rays of light are incident at various angles and hence the path difference has variable values for different set of transmitted rays. Hence the field of view in the transmitted beam consists of alternate bright and dark band. Also, the fringe pattern is not very well defined as the intensity of minima is not equal to zero.

11 FRINGES PRODUCED BY WEDGE SHAPED FILM: Consider two optically flat similar glass plates in contact at one end using rubber band and inclined to one another at an angle using a blade or wire kept in between the glass plates at the other end. If a liquid drop of refractive index is kept in between the plates, then it encloses liquid film of variable thickness with zero thickness at one end and a thickness equal to the thickness of the object kept at the other end as shown in the following figure (1). Let the film is illuminated by monochromatic light from a slit arranged parallel to the edge of the wedge. Interference occurs between the rays reflected at the upper and the lower surfaces of the film respectively and parallel interference bands result. These bands are alternately dark and bright. Let t is the thickness of the film at a distance x from the edge. The geometric path difference between the two rays at the this position is given by Fig. (1) Interference using wedge shaped film = t Cos r = t --[because r 0 & Cos r =1] Since the angle of the wedge is very small, we can write, t = x Therefore, the path difference = x. 11 Condition for bright fringe: At the position x n th bright fringe if formed when the following condition is satisfied. x = [n + ½ ] The distance of the n th bright maximum is, Condition for dark fringe: At the position x n th dark fringe if formed when the following condition is satisfied. x = n The distance of the n th bright maximum is, Fringe width: If x1 is the distance of the n th dark band and If x is the distance of the [n+m] th dark band, then Therefore, the width of m fringes is given by, If d is the thickness of the object used to form the wedge shaped film and l is the length of the air wedge, then = [d/l] or, [1/] = [l/d] Thus by finding experimentally and by knowing, & l, the thickness of the object to form the wedge shaped film can be calculated. By finding d by using screw gauge and by knowing, & l, the refractive index of the liquid used as wedge shaped film can be calculated. Also, by finding d and knowing and l, the wavelength of monochromatic source can be calculated.

12 1 Determination of thickness of a blade/diameter of a wire using interference of light formed in air wedge: The experimental arrangement to get interference fringes using air wedge formed by a wire is shown in figure (). The position of crosswire in microscope is adjusted to one of the dark fringe and treating this fringe number as zero, the microscope reading is noted. The position of the cross wire is shifted to the 5 th, 10 th......, 55 th fringe and the readings are tabulated. Fringe no. n Microscope reading R in cm Fringe no. n Microscope reading R in cm Width of 30 fringes β 1 Mean width of 30 fringes 0 30 Mean [β 1 ] 5 35 = cm The fringe width is calculated using the relation, 1 Mean 30 The length of the air wedge is measured by taking the reading for the edge of the glass plate [R 0 ] and the reading for the object [R] and taking the difference l = [R R 0 ]. Knowing β, l and the wavelength of the source of light, the thickness of the object is l calculated using the relation, d NEWTON S RINGS Newton has observed circular interference fringes in the light reflected from the air film enclosed between the slightly curved surface of a convex lens and a plane glass plate in contact with it and made a careful study of the diameter of the rings thus formed and hence they are known as NEWTON S RINGS. An arrangement to study Newton s rings is as shown in the following figure (1). Light from a monochromatic source S rendered parallel by a convex lens L1 and reflected by a glass plate G held at 45 0 to the rays. These reflected rays fall on a thin air film enclosed by a long focus plano-convex lens L and a plane glass plate. The light reflected from the upper and lower surface of the thin air film superimpose on one another to produce interference rings as shown in figure () Theory of Newton s rings: Figure (1)

13 The interference is produced due to the rays reflected from the upper and lower surface of the film. If t is the thickness of the air film at AA1 or BB1, the path difference between the two rays, one reflected from A and the other reflected from A1 will be x =t cos (1) Where is the angle of refraction. For air, = 1 & since is nearly zero, Cos = 1. Hence x =t. Since the ray reflected from the upper surface of the air film is reflected from air [a rarer medium] and the other ray reflected from the lower surface of the air film [a denser medium], the second ray experiences a further path change of [/]. Due to this, the effective path difference between the interfering rays will be x =t + /. Therefore, the points A and B lie on a bright ring of diameter AB = d, if the path difference satisfies the following condition. μt + /. = n OR, t = [n 1] / () Let us consider the vertical section ACBE of the plano-convex lens through the centre of curvature of the lens surface as shown in figure (3). Let R be the radius of curvature of the convex surface of the lens and C be the point of contact of the lens with the plane glass plate such that the points A and B are equidistant from C. Draw AA1 and BB1 perpendicular to the glass plate P. Let AA1 = BB1 = t be the thickness of the film at distance r = CA1 = CB1 from the point of contact C of the lens with the plate. Figure (3) From the geometry of the circle, we have DE x CD = AD x DB OR (CE CD) CD = AD, [because AD = DB] OR (R t) t = r (3) Where, r is the radius of the ring. Since the radius of curvature R of the lens is very large compared to t, we can neglect t compared to R. Rt = r = [AB/] = [d/] = d /4 t = d /4R (4) Where d is the diameter of the ring. Comparing equations () & (4), the diameter of the n th bright ring is given by the following equation. d n 1 n or n 1 4R R d n or n 1 R d n (5) Where n is an integer taking values 1,,3,..for first, second, third..rings respectively. Since, R and are constants dn [n 1] 1/. Thus, the diameter of the bright rings is proportional to the square root of odd natural numbers. Also for the n th dark ring, d n μ n 4R or 4nR d n or 4nR d n n d n, where, n = 1,, 3 etc. Thus the diameter of the dark rings is proportional to the square root of the natural numbers. Experimental determination of refractive index of liquids: 13

14 14 Figure (4) Graph 1 Graph A few drops of liquid whose RI is to be found is placed between lens and glass plate to form a thin film of liquid and arranged as in figure (4) to get Newton s rings. The vertical cross wire of travelling microscope is arranged to the position of 18 th dark ring, on left side, so that it is tangent to the ring. The reading of the microscope is noted. The readings are also noted successively for 16 th, 14 th th dark ring on left side and on the right side, the readings for 8 th, 10 th, th dark ring are noted. The diameter [D] of the ring is calculated for each ring by finding difference of reading for each ring. The square of the diameter [d ] is calculated and tabulated as in the table. A graph of d versus ring number is plotted. The graph is a straight line. The slope 1 of the graph is calculated. Ring number n Microscope reading [x 10 - m] Left edge Right edge Diameter [D n,l] of the ring [x 10 - m] D n,l [x 10-4 m ] The experiment is repeated in a similar way with air film formed in between the lens and the glass plate. The slope of the second graph is calculated. Ring number n Microscope reading [x 10 - m] Left edge Right edge Diameter [D n,a] of the ring [x 10 - m] D n,a [x 10-4 m ] 4( n m) R 4( n m) R Since, Slope 1 and Slope as μ = 1 for air. In these equations, (n-m) is 1 Slope difference in ring number. Therefore, the gives the refractive index of the liquid. Thus, the Slope1 Slope refractive index of the liquid is calculated using the relation, Slope1

15 MICHELSON S INTERFEROMETER Michelson s interferometer consists of two highly polished mirrors M1 and M and two plane glass plates A and C arranged as shown in the following figure (1). The rear side of the mirror A is half silvered so that light from the source S is equally reflected and transmitted by it. Light from a monochromatic source rendered in to a parallel beam by passing it through a lens L is made to fall on the plate A inclined at an angle of 45 0 to the incident light. The light is partially reflected along AM1 and partly transmitted along AM. These two beams moving mutually perpendicular to each other are reflected back by the mirrors M1 and M and retrace their path and reach the plate A. The ray reflected from M is again reflected from A and the ray from M1 is transmitted through A and these two rays travel along the same direction and are received by the eye. The ray reflected from M1 passes through A thrice, while the reflected ray from M passes only once through A. To compensate this optical path, a glass plate C of same material Michelson s Interferometer and same thickness is arranged parallel to A in the path of the second beam. The mirror M1 is fixed on a carriage and can be moved with the help of a handle H. The distance through which the mirror M1 is moved can be measured with the help of micrometer screw attached to the Handle. The plane of the two mirrors can be made perfectly perpendicular to one another with the help of the screws attached to them. The plate C is necessary for white light fringes and can be dispensed with while using monochromatic light. If the mirrors M1 and M are perfectly perpendicular, the observer s eye will see the image of M1 and M through A. There will be an air film between the two images and the distance between them can be varied with the help of the handle H. In this case, the fringes will be perfectly circular. If the path traveled by the two rays is same, then the field of view will be completely dark. If the two images of M1 and M are inclined, the air film between them has wedge shape and straight fringes are observed. When the handle H is rotated, the fringes cross the centre of the field of view of the observer s eye. If M1is moved through /, one fringe will cross the field of view. 15 TYPES OF FRINGES: Fig () Formation of Circular fringes Circular fringes: When the mirror M1 and the virtual image M of the mirror M are exactly parallel, in general circular fringes are formed. In this case the source is extended one and S1 and S are the virtual images of the source due to the mirrors M1 and M. If the distance between M M 1 = d, then the distance between S1 and S = d. The path difference between the beam will be d Cos. When d Cos = n, the beam will reinforce to produce maximum. These fringes are circular and are formed due to phase difference determined by the inclination, they are known as the fringes of equal inclination or Haidinger s fringes. When M1 and M coincide, the path difference

16 become zero and the field of view is perfectly dark [due to path change occurs for one of the beam due to reflection at denser surface]. Localized fringes: When the mirror M1 and the virtual image M of M are inclined, the air film enclosed is wedge shaped. The fringes formed depends on the nature of inclination of the two mirror images. If the two mirrors intersect at the centre, the fringes are straight and parallel as shown in figure [3(ii)] If the mirrors are inclined to one another and does not intersect, then the fringes are curved with the convex towards the thin edge of the wedge. White light fringes: With white light, the fringes are observed only when the path difference is small. The different fringes overlap on one another and only a few coloured fringes are visible. The central fringe is dark and a few fringes near the central dark fringe are coloured. After about 8 to 10 fringes a number of colours overlap at a point and as a result general illumination is observed. White light fringes are useful for the determination of zero path difference between the interfering rays. It is useful in the standardization of a metre scale. Applications of Michelson s Interferometer: (1)Determination of wavelength of monochromatic light: When the two mirrors are equidistant from A, the field of view is perfectly dark. The mirror M is fixed at this position and M1 is moved to get circular fringe pattern [or straight and parallel fringes]. The reading of the micrometer eyepiece is noted [R1]. The handle H is rotated until 50 fringes appear or disappear at the centre [50 fringes cross the reference cross wire in the case of parallel fringes]. The reading of the micrometer eyepiece is noted again [R]. If d = R R1 is the displacement of the mirror position for the crossing of 50 fringes, then, d = 50. Thus knowing d, the wavelength of the source is calculated by using the relation, d 50 ()To determine the wavelength difference between two monochromatic, closely lying spectral lines : [Resolution of spectral lines] In the case of Sodium source, we get two spectral lines D1 and Dhaving wavelengths 1 and such that the difference in wavelength is very small. When observed in a Michelson s interferometer, each wavelength forms its own fringes, which overlap on one another. By adjusting the position of mirror m1, a single fringe pattern having maximum brightness can be obtained. In this position, the bright fringe due to D1 coincides with the bright fringe due to D line. A reading [R1] of micrometer screw is noted. When the handle is turned to move the mirror M1, the fringes go out of step and on further moving the mirror, the fringes with maximum brightness is again formed. The reading [R] for this position is noted. This is possible only when (n + 1) th order of shorter wavelength coincides with the n th order of longer wavelength. If d is the distance between two positions of M1 for maximum brightness, then d = n1 1 = (n1+1) n (1) 1 16

17 17 Substituting for n1 in the above equation, we get, 1 d () (3) d d Where, λ is the mean wavelength of Sodium light. Thus, by knowing the distance [d] moved by mirror for positions of distinct fringe pattern, wavelength difference can be calculated. 1 FABRI PEROT INTERFEROMETER It consists of two glass plates E1 and E separated by a distance d. The inner surfaces of the two plates are parallel and thinly silvered to reflect 70% of the incident light. The outer surfaces of the plates are also parallel to each other, but, inclined to their respective inner faces. This is to avoid interference effects due to multiple reflections and refractions that all faces are not made parallel. Light from a broad monochromatic source S1 is made parallel using a collimating lens L1. Each parallel beam suffers multiple reflections in the air film between the silvered surfaces of the plates E1 and E and emerges from E as a parallel beam, so that for interference the rays must be brought to focus at P by a convex lens L. The fringes formed are seen in the focal plane of L. The fringes are circular and due to path difference of equal inclination known as Haidinger s fringes. If is the angle of incidence on the silvered surface of E1 and d is the distance between E1 and E, then the path difference between successive rays = d Cos. The condition for the rays to produce maximum is given by, d Cos = n (1) In equation n is an integer giving the order of interference for the particular bright fringe. All the points on a circle passing through the point P with the centre at O on the axis O1O satisfy this condition and hence the maxima consists of a series of concentric rings on the screen with O as centre. In the interferometer one the plate is fixed, while, the other is capable of motion with the help of slow motion screw, so that the thickness of the air film can be changed. When the thickness changes by /, one fringe disappear at the centre. Hence by knowing the number of fringes disappearing at the centre when the mirror moves through a distance d, the wavelength of the monochromatic source can be calculated using the equation (1). In this case the radius of a ring is a function of and hence two sets of fringe pattern is obtained if there are two different wavelengths in a source. Uses: (1) It can be used to determine the wavelength of a monochromatic source of light. () It can be used to determine the wavelength difference between two monochromatic sources of light. (3) It can be used for the calibration of a metre scale.

18 18 Advantages over Michelson s interferometer:: (1) Fringes obtained are very sharp as compared to the fringes formed in Michelson s interferometer. () The resolving power of the microscope is very large. (3) Clear and distinct fringes are formed even when the separation between the plates is large. INTERFERENCE FILTERS Interference filters are the optical devices, which work on the principle of interference of transmitted light in thin films, and transmit a narrow band of about A around a chosen wavelength. In simplest form the arrangement is very much similar to a Fabri-Perot etalon as shown in the following figure. In this arrangement, two thin transparent layers a and b of good reflecting material like silver separated by an optical thickness d equal to one or more half wavelengths of a dielectric material like magnesium fluoride. To construct such a filter one layer of silver is deposited on a solid transparent substance like glass. Then a coating of the dielectric (acts as a spacer layer) is deposited on the silver and finally the second layer of silver is added. The thin sandwich of dielectric metal constitutes the filter. A second glass plate is added to the silver coating to give mechanical protection. When the path difference between the successive transmitted pairs of rays is equal to d Cos or simply d [as is nearly equal to zero, Cos = 1]. Therefore, the emergent rays are all in phase when d is equal to one wavelength 0 or an integral multiple of 0. Therefore, the condition for maximum is, d = m 0 With white light the intensity of the transmitted beam will be maximum for only those wavelengths, which satisfy the above equation. Therefore, a spectrum of white light consists of a series of sharp bright bands separated by wide dark regions. Bright bands are obtained for those wavelengths satisfying the above equation and a dark region for all those, which do not satisfy the above equation. d The maxima will occur at wavelengths given by the equation, 0, where m is any integer m value. If the thickness of dielectric is such that d = 0 and 0 = A, then only a single narrow band in this region is transmitted, because the other peaks in the spectrum would occur at A, A and these are in Ultra Violet region and hence the device acts as optical filter. In a similar way, if the filter is constructed to transmit a wavelength 0 = A, then it also transmits A and A also and hence it allows a band of wavelength around these wavelengths. Thus, the filtering action mainly depends on the thickness of dielectric film between the silvered surfaces.