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WAVE OPTICS & E.M. WAVE INTERFERENCE 1. PRINCIPLE OF SUPERPOSITION Two, or move Progressive waves can travel simultaneously ina medium without affecting the mation of one another, Therefore, the resultant displacement of each particle of the medium at any instant is equal to the vector sum of the displacements produced by the two waves separately. This principle is called principle of superposition. Resultant displacement : Y=¥it¥o Meaning of principle of superposition : The principle of superposition means that if a number of waves are travelling in a medium, then each one travels independently as if the other waves were not present at all; the shape and other characteristics of any wave not changed. 2. INTERFERENCE OF TWO WAVES When two waves of same frequency travel in a medium simultaneously in the same direction then, due to their superposition, the resultant intensity at any point of the medium is different from the sum of intensities of the two waves. At some points the intensity of the resultant wave is very large while at some other points it is very small or zero . 2.1. Mathematical interpretation of interference of two waves : Let us consider two simple harmonic progressive waves of the same frequency travelling in the same direction, Let a, and a, be the amplitudes of the waves at that point be y, and y, then. yy, = a SIN ote. (i) and y, = asin (it +) ...... (ii) av2n is the frequency of each wave. By the principle of superposition, the resultant displacement at the point is given by y= y¥, 7 ¥2 = a, sin mt + a, sin (tot + ) a, sin @t + a, sin @t cos > + a, cos at sin db sin @t (a, + a, cos }) + cos mt (a, sin >) Let a, + a, cos > = R cos @............ (iii) . (iv) where R and 0 are new constants. Then 1 2 and a, sin @ =R sin 0 y = R sin wt cos? + R cos mt sin 0 or y= R sin (mt + 0) This equation is similar to eq. (i) and (ii) Hence the resultant displacement at that point is changing sinosudeally with amplitude R. To determine R, we square eq(iii) and (iv) and then add: R? cos? § + R? sin? = (a, + a, cos)? + (a, sin o) 2 2 ye 2 o Re =a? +a’ + 2a,a, cosh WAVE OPTICS & E.M. WAVE The intensity is directly proportional to the square of the amplitude. Hence the resultant intensity | is given by a, sing a, +a, cos Thus the resultant intensity at any point depends upon the phase difference > between the two waves at that point. Comment : If $ phase diff is equivalent of x path difference or At time difference, then remember | @ (a,? + a,? + 2a,a, cos) and phase angle } = tan” a ee 2n AT 3. TWO TYPES OF INTERFERENCE (i) Constructive (ii) Destructive Law of conservation of energy in constructive & Destructive interference : Energy redistribution takes place in interference . Energy will slift from the position of distructive interference to position of contructive interference, total energy remain constant. CONSTRUCTIVE {I > (I, + I,)} DISTRUCTIVE {I < (I, + I,)} * The phase difference between two waves an The phase difference between two waves is anis even multiple of m.i.e odd multiple of 7. 6 = 2nn where n= 0,1, 2 6 = (2n —1)n where n= 0,1, 2 * The path difference between two waves The path difference between two waves is an odd is an even multiple of 2/2. multiple of 4/2. A= 2n (A /2) wheren=0, 1,2 A=(2n-1)A/2, where n= 1, 2 * = Thetime interval between two wavesis even The time interval between two waves is an odd an multiple of T/2 multiple of T/2 6 =(2n)T/2, n=0,1,2 8 =(2n-1)T/2,n=1, 2,3 * Theresultant amplitude of wave is equal to the The resultant amplitude of wave is equal to the sum of amplitudes of individual waves difference of amplitudes of two waves A=a,t+a, A=a,-a, f @ =0, 2, 4n, ... 2nd lf d= 2, 37, 5m, ... (2n-1)n * The resultant intensity is more than the sum The resultant intensity is less than the sum of of intensities of individual waves. intensities due individual waves. I=, +,+2/q, = (h+ Vu) [=1+-2/hb = (J, -y~h) * Both waves reach at any point in same phase Both waves reach at any point in opposite phase =1,+1,+20L = +6 phase diff jad a phase: cif. aa a os oka a Za « pohon, Sete anes A 4, = 2) « pathdift, Intensity of resultant wave (interference) Intensity of resultant wave (interference pattern) pattern when the waves have equal when the intensity of the two waves are not equal intensities note that the average intensity is linea 1 |= mex Ae l, WAVE OPTICS & E.M. WAVE Example based on intensity of interference pattern Ex.1 The two coherent sources of intensity that ratio 2 : 8 produce an interference pattern. The values of maximum and minimum intensities will be respectively. (1) 1, and 9 |, (2) SI, and 1, (3) 21, and 8l, (4) 81, and 21, Where |, is the intensity of first source Sol. Ina = 1 + 1p + 2yvll essen 1) According to question W_2_1 1 8 4 y= 4h cesses 2) From eqs. (1) and (2) : Imax =; +41, +2y417 = 51, + 4l, lea = 9h, seven) lain = | + ly — 2,411, vessese) From eqs. (2) and (4) lain = y + 41, — 24/412 Vrain = I, Hence the correct answer will be (2) 4. COHERENT SOURCE The two sources of light, whose frequencies are same and the phase difference between the waves emitted by which remains constant with respect to time are defined as coherent sources. There are two independent concepts of coherence namely (i) temporal coherence (ii) spatial coherence Temporal coherence: In a typical light source (like sodium lamp) a light wave (photon) is produced when an excited atom goes to the ground state and emits light the duration of this ~ 10°? to 10-' sec. Thus the electromagnetic (light) wave remains sinusoidal for this much time. This time known as coherence time it is denoted by 2 Spatial coherence: Two waves at different point in space are said to be space coherent if they preserve a constant phase difference over any time t . Note: Laser is a source of manochromatic light waves of high degree of coherence. The entire wave front of the laser beam is spatially coherent. Main points :- 1. They are obtained from the same single source. 2. Their state of polarization is the same Note: 1. Laser light is highly coherent & monochromatic 2. The light emitted by two independent sources (candles, bulbs, ect.) is non coherent and interference phenomenon can not be produced by such two sources, Method of obtain coherent light source : (1) Division of wavefront (2) Divsion of Amplitude WAVE OPTICS & E.M. WAVE CONDITIONS FOR SUSTAINED INTERFERENCE PATTERN The source of light must the monochromatic. Two sources of light must be coherent. Frequencies of two waves must be same. The amplitudes of two waves must be nearly equal. The distance between two light sources must be small narrow. The two coherent sources must be narrow. If the two light waves are polarized then their state of polarizations must be same. The two light waves must travel in the same direction . The vibrations of two waves must be in the same direction . The distance between the source & the screen must be large. EXPERIMENT OF YOUNG'S DUAL-SLIT This experiment shows interference of light s, & s, slit behaves like two coherent source. On Screen, Bright & Dark portion are altenatively found. Bright portion is called Bright fringe & dark portion is called dark fringe. Central fringe is always bright. Energy conserves in interference of light. It is explained on the basis of hygen principal . @eeeeese85e@esseresecssesseseegees#9ku#8§ This experiment verifies the wave nature of light. Second dark 'e—Second bright : 3 First bright ‘ = First dark ee (Central / bright Light : First dark: ene 5, q }e— First bright Second dare - Second bright ‘Screen (a) At a point on screen to find dark or bright fringe, it depends upon path difference between s,P & s,P light waves. (b) Two types of path difference between light waves : (i) Geometrical path difference. (ii) Optical path difference. (c) In above experiment optical difference s,P & s,P has geometrical path difference so, Total path difference = Geometrical difference xd ap (note - sind = tan = Q) WAVE OPTICS & E.M. WAVE e Bright Fringes: If n“ bright fringe forms at point P, then for Bright fringe : g . fork X= AO S,P -S,P = so distance between central fringe & n'" bright fringe : nD Ki - d n = 1 first bright fringe > n = 2 Second bright fringe s Dark Fringe : If n dark fringe forms at a point P, then for dark fringe. Xd (2n-1) i D 2 = distance between central fringe & n™ fringe. x, = (2n=1AD 2d n=1 First bright fringe. n = 2 Second bright fringe. Distance between dark & bright fringe which are in coming order is called fringe width. 4D In Interference Fringe width of dark & bright fringes are same [}= a e Angular width of fringe S 7 a D=B a = pid = al\> Comments on young's interference experiment : (1) Energy is conserved in interference. This indicated that energy is redistributed from destructive interference region to the constructive interference region . (2) If the entire arrangement of young's double slit experiment is immersed in water then fringe width decreases B h 1 1 water _ "*water _ Bar ev aly (4/3) (3) If white light is used in placed of monochromatic light in young's double slit experiment. (a) central fringe is white (b) Coloured fringe around the central white fringe (c) Inner edge of the dark fringe is red. While the outer edge is violet (or blue) (d) Inner edge of bright fringe is violet (or blue) and the outer edge is red. (4) If a filter allowing only 4,,, (Of 4,) is placed in front of slit s, and filter allowing only 4,,,. (or 4,) is placed is fornt of slit s,. then there is no interference pattern. (refer to point no.3 of conditions) (5) If a thin glass plate or mica sheet is placed in front of one of the slit, then central fringe shifts towards that slit, refractive index of glass is 1 and the thickness of sheet t, then the optical path = ut so extra path difference (u - 1)t WAVE OPTICS & E.M. WAVE lf the central fringe now appears at the location of previously formed nth bright fringe then (u -1)t = nd. if the central fringe appears at the position of previously formed nth dark fringe then (uu — 1)t (20-1) (6) If the width of the slit S increased then degree of spatial coherence decreases. As a result the interference pattern gradually disappears similar occurs if distance between s, and s, is increased. (7) The fringe visibility : V =)" ae max rin Vinaes Wf OL= I, = I or lig = 0 If widths of slits s, and s, are unequal the brightness of the bright fringe and the darkness of the dark fringe decreases. if |, > > I, then Ia = Iain (8) When waves from two coherent sources S, are S, interfere in space the shape of the fringe is hyperbolic with foci at S, and S, . _Examples based on young double slit experiment Ex.1 In young’s double slit experiment, the distance between two slits is made three times then the fringe width will become - (1) 9 times (2)1/9 times (3) 3 times (4) 1/3 times 1 Sol. Bx a On increasing d three times fi will become 1/3 times. Hence the correct answer will be (4) Ex.2 In Young's slit experiment 10th order maximum is obtained at the point of observation in the interference pattern for } = 7000 A. If the source is replaced by another one of wavelength 5000 A then the order of maximum at the same point will be - (1) 12 th (2) 14 th (3) 16 th (4) 18 th Sol. nj, A, = nm, A, 10 x 7000 =n, x 5000 n, = 14 Ex.3 In Young's double slit experiment the two slits are illuminated by light of wavelengh 5890 A and the distance between the fringes obtained on the screen is 0.2°. If the whole apparatus is immersed in water then the angular fringe width will be, it the refractive index of water is 4/3. (1) 0.30° (2) 0.15° (3) 15° (4) 30° Sol. w, = Ald a (©) water is A water 0, *A=> 2 a A a (p tee A —S—_——- = SS (o = tall (0 )water = 0,15° WAVE OPTICS & E.M. WAVE Ex.4 The intensities of two sources are I and 9 I respectively. If the phase difference between the waves emitted by them is x then the resultant intensity at the point of observation will be - (1) 31 (2) 41 (3) 10 1 (4) 821 Sol. P=1, +1, 42h 1, cosh L=LL=9h gen I'=1+91+2/9P cose = 101-61=41 7. FRESNEL'S BIPRISM EXPERIMENT (1) Itis optical device to obtain two coherent sources by refraction of lights. Screen (2). The angle of biprism is 179° & refracting angle is a = 1/2°. (3) Distance between source & screen D =a +b. Distance between two coherent source = d = 2a (u—1)ca Where a = distance between source & Biprism b = distance between screen & Biprism Fringe Pattern uu = refractive index of the material of prism. a ,, 9B _ 2a(u-)op Vo 8 a) (a+b) (a+b) 180 calculate by convex lens. Note- « is in radian a°=ax . Suppose refracting angle & refractive index is not known then d can be One convex lens whose focal length (f) and 4f < D. First convex lens is kept near biprism & d, is calculated then it is kept near eyepiece & d, is calculated, d=,/d,d, Application : With the help of this experiment the wavelength of monochromatic light, thickness of thin films and their refractive index & distance between apparent coherent sources can be determined. When Fresnel's arrangement is immersed in water (1) Effect on d d between the two virtual sources decreases but in young's double slit experiment it does not change. water “~~ aj Thus when the Fresnel's biprism experiment is immersed in water, then the separation (2) In young's double slit experiment fi decrease and in fresnel's biprism experiment [) increases. 7 WAVE OPTICS & E.M. WAVE Example based on fresnel's biprism experiment _ Ex.1 In Fresnel's biprism experiment the width of 10 fringes is 2em which are formed at a distance of two 2 meter from the slit. If the wavelength of light is 5100 A then the distance between two coherent sources will be (1) 5.1 x 104 m (2) 5.1 * 104 cm. (3) 5.1. * 104 mm — (4) 10.1 * 10-4 em Di. Sol. ae ciel (1) According to question 2% = 5100 x 10-7 m 2 age =—x10™m inet B 10 * (2) D=2m qd=?7 From eqs. (1) and (2) g—2%51x10* ; = 5.1 * 10¢*m 2x107 8. DIFFERENCE BETWEEN MONOCHROMATIC LIGHT FRINGE & WHITE LIGHT FRINGES Monochromatic light fringe White light frin “* Central fringe is always bright Central fringe is always white The width of all fringes is same The fringe width of different colors is different * Fringes of higher order are also obtained In the region of higher order fringes of uniform * The number of fringes obtained is more The number of fringes obtained is less. There are alternate bright and dark When path difference is small then some coloured finges on both sides of central fringes fringes are obtained on both sides of the central fringe. venodromat 3, source of light | Bright or White light S, dark fringe source 9. TO DETERMINE THICKNESS OF THIN FILM On placing the film, the whole interference pattern gets shifted on that side where the film is placed. When a thin film of thickness t & refractive index u is placed in the path of one of the waves then the fringe pattern gets shifted by a distance x and if the shift is equivalent to n fringes, then nai g ,_ (n= 1tD a d the path difference increases by (4 —1)t on placing the plate. WAVE OPTICS & E.M. WAVE Example based on a plate of thickness of t in the way of interference pattern Ex.1 When a mica sheet of thickness 7 microns and p = 1.6 is placed in the path of one of interfering beams in the biprism experiment then the central fringe gets at the position of seventh bright fringe. The wavelength of light used will be (1) 4000 A (2) 5000 A (3) 6000 A (4) 7000 A , _ (u- jt Sol, A= 7 see(1) According to question n=7.p216, t=7* 10% meter 000 2 2 2 2 Oe, (2) From eqs, (1) and (2) kK = 6 * 107 meter Hence the correct answer is (3) Ex. When a plastic thin film of refractive index 1.45 is placed in the path of one of the interfering waves then the central fringe is displaced through width of five fringes. The thinckness of the film will be, if the wavelength of light is 5890A. (1) 6.544 x 10 em (2) 6.544% 10% m (3) 654 * 104 cm (4) 6.5 * 104 om B p(0.45)t 2X) = (u-1t> 5p = Sol. Xo = 5 (u- Mt SB= Been 40-0 _ 5x 5890 x10" - 0.45 INTERFERENCE BY THIN FILMS The interference cause by thin films is due to the interference between the waves reflected from the upper and lower surfaces. These two reflected coherent waves are obtained from the same incident wave by division of amplitude. A film of thickness t and refractive index 1 produces a path difference of 2ut cos r between the two at = 6.544 x 107% cm A reflected waves and an additional path difference 2 or phase difference of m is produced due to reflection ( A of one wave from a denser medium. Thus the total path difference in reflected waves is | aeteaer 4) ih For maxima 2ut cos > = nA A or 2 wt cos r= (2n - 1)5 ES M and for minma 2ut cos r + 2 = (2n + ND x o 2 ut cos r = (2n - 1) > Here r is angle of refraction. For normal incidence or near normal incidence r = 0 ,so that for maxima 2ut = (2n + 5 for minima 2ut = nd If a film is very thin t = 0, then the condition of minima is satisfied and the film appears dark in reflected light. 9 WAVE OPTICS & E.M. WAVE a In transmitted waves sililar interference is oberved but now the additional path difference of — is absent. 2 So in transmitted waves. 2ut cos r = n& for maxima aA : = and 2ut cos r = n+ 15 for minima. POINTS TO REMEMBER e The fringe width increases with increase of distance between the source & the screen e Fringe width decreases by increasing distance between two slits s, & s, @ ~=sif the experiment is repeated in water instead of air then Bf decreases . e When one of the slits of s, & s, is close then interference does not take place 1 e When the two slits are illuminated by two independent sources then interference fringes are not obtained. @ When one of the slit is closed & width of another is made of the order of A, then diffraction fringes are observed e When slit is illuminated with different colours then fringes are obtained of the same colour but their fringes width is different. e In young's double slit experiment light waves undergo diffraction at both the slits and the diffracted waves superimpose to produce interference. e If biprism experiment is a liquid instead of air, then the fringe width increases (where as in young's double slit experiment it decreases) e The wavelength undergoing destructive interference, the colour of that wavelength will be absent. e The wavelength for which the condition of constructive interference is fulfilled that colour will be visible maximum consequently the fringes will be coloured. 10 WAVE OPTICS & E.M. WAVES DIFFRACTION 1. MEANING OF DIFFRACTION It is the spreading of waves round the comers of an obstacle, of the order of wave length. 2. DEFINITION OF DIFFRACTION The phenomenon of bending of light waves around the sharp edges of opaque obstacles or aperture and their encroachment in the geometrical shadow of obstacle or aperture is defined as diffraction of light. 3. NECESSARY CONDITIONS OF DIFFRACTION OF WAVES The size of the obstacle (a) must be of the order of the wavelength of the waves ().). a4 A Note : Greater the wave length of wave higher will be its degree of diffraction. This is the reason that diffraction of sound & radio waves is easily observed but for diffraction of light, additional arrangement have to be arrange. > } Meound 7 right Wave length of sound is nearly equal to size of obstacle. If size of obstactle is a & wavelength of light is i then, S.No. a VIS 2 Diffraction [1] ass i Not possible [2] a>> i Not possible [3] an~i Possible 4. INTERPRETATION OF DIFFRACTION As a result of diffraction, maxima & minima of light intensities are found which has unequal intensities. Diffraction is the result of superposing of waves from infinite number of coherent sources on the same wavefront after the wavefront has been distorted by the obstacle. 5. EXAMPLE OF DIFFRACTION ae When an intense source of light is viewes with the partially opened eye, colours are observed in the light. He Sound produced in one room can be heard in the nearby room. * Appearance of a shining circle around the section of sun just before sun rise. Es Coloured spectrum is observed if a light source at far distance is seen through a thin cloth. 6. TWO TYPE OF DIFFRACTION Fresnel Diffraction : Fresenel diffraction which involves non-plane (spherical) wavefronts, so that the sources and the point p (where diffraction effect is to be observed) are to be at a finite distance from the diffracting obstacle. WAVE OPTICS & E.M. WAVES Fraunhoffer Diffraction : Fraunhofer diffraction deals with wavefronts that are plane on arrival and an effective viewing distance of infinity. If follows that fraunhofer diffraction is an important special case of fresnel diffraction. In youngs double slit experiment, we assume the screen to be relatively distance, that we have fraunhofer conditions. [1] Fresnel Diffraction : According to fresnel principal to determine the intensity of light at any point, a wavefront can be divided into a number of small parts which are known as fresnel's half period zones. Each point on the wavefront is a source of secondary wavelets, so that the wave from two consecutive zones reach the point of observation in opposite phase corresponding to a path difference of i/2. Mathematical Interpretation of Half Period Zones (HPZ) : ABCD is a plane wavefront emitted by a monochromatic source of light. We want to determine its effect at a point of observation P distance b from its centre. Taking P as centre and distance b, b + 4/2, b + 2i/ Dy maanies etc as radii we draw a large number of cocentric circles. The area of first circle is known as the first half period zone. The area between first & second circle is known as second HPZ. Thus, the peripheral area enclosed between the n" circle & (n — 1)" circles is defined as the n™ HPZ. me Radius of nth Half Period Zone (HPZ) Radius of n* HPZ = Radius of n** circle. A Cc ling na oF bh? r, ={(b4n./ 27 —b? y 0, = a mee ae GH Ce Pp en 2o2 b r, = {nba +2 \ D Cc If 4? is very small we can neglect it. r, depends on- (i), a vn (ii) , a vb (ii) ra WR If b & % remains constant for wavefront itt V4: v2: V8 Radius of HPZ is proportional to square root of natural numbers. * Average Distant of Point P From nth HPZ _1 Distance of n™ circle from point P + 7 ~ 2| Distance of (n-1)" circle from point P | 1 ((n-1 (2n-1) 0, -3|(0+nnw)+{+ al ~% d, =b+"——A 4 WAVE OPTICS & E.M. WAVES Be Area of n" Half Period Zone (HPZ) A, = Area of n" HPZ = Area of n" circle — Area of (n — 1)" circle = a 2 A = aurn?- a _, ? * . a A, == | \ rte If — <<< ba 4 << A A. = mba. n Note : Area of HPZ does not depend on value of n AY BAg! Ay ccuesccwl P 84 1 OF A, = Ay Ag = ceteceeee A, = mba, a2 if x <<< ba, area of HPZ increases with increasing value of n * Amplitude at Point P due to n™ HPZ : . : . 1 (i) Inversely proportional to distant => R, a — nv (ii) proportional to area > R, a A, (iii) proportional to obliquity = R, a (1 + cos 6.) A R, « —(1+cos0, ) n The phase difference between the wavelets originating from two consecutive HPZ is 1 radian and reaching the point P is p(or path difference 1/2 & time difference T/2). The wavelets originating fram two consecutive HPZ meet at point P in the opposite phase i.e. the amplitudes of any two consecutive wavelets are of opposite signs Amplitude of Light of Consecutive HPZ may be shown by Vector Diagram : R, R, R, ae Resultant Amplitude : R=R,-R,+R,-R, + RR, ..-. ee (1977 R, WAVE OPTICS & E.M. WAVES It is Clear from figure : _R,+R, R,+R R, LR, =e 4 Re Re Re _ Re 2 4 R. R. Re Roo = Constant (< 1) 1 2 a a Resultant amplitude R at Point P due to Whole Wavefront : RR = RoR = Ryne 2 Bi. KBE 4 4 R’' >l=k jx R* => (8 Resultant intensity due to whole wavefront is equal to 25% of intensity due to first HPZ. Note 1: Alternative HPZ has phase difference 2x & path difference 4. They remain in same phase at point P. 
