FREE Downloa Stuy Package from website: wwwtekoclassescom & wwwmathsbysuhagcom Get Solution of These Packages & Learn by Vieo Tutorials on wwwmathsbysuhagcom PRINCIPLE OF SUPERPOSITION When two or more waves simultaneously pass through a point, the isturbance of the point is given by the sum of the isturbances each wave woul prouce in absence of the other wave(s In case of wave on string isturbance means isplacement, in case of soun wave it means pressure change, in case of EMW it is electric fiel or magnetic fiel Superposition of two light travelling in almost same irection results in moification in the istribution of intensity of light in the region of superposition This phenomenon is calle interference Superposition of two sinusoial waves : Consier superposition of two sinusoial waves (having same frequency, at a particular point Let, x (t a sin ωt an, x (t a sin (ωt + φ represent the isplacement prouce by each of the isturbances Here we are assuming the isplacements to be in the same irection Now accoring to superposition principle, the resultant isplacement will be given by, x(t x (t + x (t a sin ωt + a sin (ωt + φ A sin (ωt + φ 0 where A a + a + a a cos φ ( an tan φ 0 a sinφ a + a cos φ Ex If i 3sin ωt an i 4 cos ωt, fin i 3 Sol from kirchoff s current law, i 3 i + i Ex 3 sin ωt + 4 sin (ωt + π 5 sin (ωt + tan ( i Figure i 3 S an S are two source of light which prouce iniviually isturbance at point P given by E 3sin ωt, E 4 cos ωt Assuming E & E to be along the same line, fin the result of their superposition Sol S S Figure E E + E ( E E + E 5sin ( ωt + tan 4/3 SUPERPOSITION OF PROGRESSIVE WAVES; PATH DIFFERENCE : Let S an S be two sources proucing progressive waves (isturbance travelling in space given by y an y At point P, S x y a sin (ωt kx + θ y a sin (ωt kx + θ P y y +y A sin(ωt + φ S x Here, the phase ifference, φ (ωt kx + θ (ωt kx + θ Figure: 3 k(x x + (θ θ k p + θ Here p x is the path ifference Clearly, phase ifference ue to path ifference k (path ifference π where k WAVE OPTICS i Teko Classes, Maths : Suhag R Kariya (S R K Sir, Bhopal Phone : 0 903 903 7779, 0 98930 5888 page Successful People Replace the wors like; "wish", "try" & "shoul" with "I Will" Ineffective People on't
Get Solution of These Packages & Learn by Vieo Tutorials on wwwmathsbysuhagcom π φ k p x (3 FREE Downloa Stuy Package from website: wwwtekoclassescom & wwwmathsbysuhagcom Ex 3 Sol For Constructive Interference : φ nπ, n 0,, or, x n A max A + A Intensity, max + max ( For Destructive interference : φ (n + π, n 0,, or, x (n + / A min A A Intensity, min + (4 max ( (5 Light from two sources, each of same frequency an travelling in same irection, but with intensity in the ratio 4 : interfere Fin ratio of maximum to minimum intensity max min WAVEFR VEFRONT ONTS + + + 9 : Consier a wave spreaing out on the surface of water after a stone is thrown in Every point on the surface oscillates At any time, a photograph of the surface woul show circular rings on which the isturbance is maximum Clearly, all points on such a circle are oscillating in phase because they are at the same istance from the source Such a locus of points which oscillate in phase is an example of a wavefront A wavefront is efine as a surface of constant phase The spee with which the wavefront moves outwars from the source is calle the phase spee The energy of the wave travels in a irection perpenicular to the wavefront Figure (a shows light waves from a point source forming a spherical wavefront in three imensional space The energy travels outwars along straight lines emerging from the source ie raii of the spherical wavefront These lines are the rays Notice that when we measure the spacing between a pair of wavefronts along any ray, the result is a constant This example illustrates two important general principles which we will use later: (i (ii Rays are perpenicular to wavefronts The time taken by light to travel from one wavefront to another is the same along any ray If we look at a small portion of a spherical wave, far away from the source, then the wavefronts are like parallel planes The rays are parallel lines perpenicular to the wavefronts This is calle a plane wave an is also sketche in Figure (b A linear source such as a slit Illuminate by another source behin it will give rise to cylinrical wavefronts Again, at larger istance from the source, these wave fronts may be regare as planar (a (b Figure : : Wavefronts an the corresponing rays in two cases: (a iverging spherical wave (b plane wave Teko Classes, Maths : Suhag R Kariya (S R K Sir, Bhopal Phone : 0 903 903 7779, 0 98930 5888 page Successful People Replace the wors like; "wish", "try" & "shoul" with "I Will" Ineffective People on't
Get Solution of These Packages & Learn by Vieo Tutorials on wwwmathsbysuhagcom The figure on the left shows a wave (eg light in three imensions The figure on the right shows a wave in two imensions (a water surface FREE Downloa Stuy Package from website: wwwtekoclassescom & wwwmathsbysuhagcom 3 COHERENCE : Two sources which vibrate with a fixe phase ifference between them are sai to be coherent The phase ifferences between light coming form such sources oes not epen on time In a conventional light source, however, light comes from a large number of iniviual atoms, each atom emitting a pulse lasting for about ns Even if atoms were emitting uner similar conitions, waves from ifferent atoms woul iffer in their initial phases Consequently light coming from two such sources have a fixe phase relationship for about ns, hence interference pattern will keep changing every billionth of a secon The eye can notice intensity changes which lasts at least one tenth of a secon Hence we will observe uniform intensity on the screen which is the sum of the two iniviual intensities Such sources are sai to be incoherent Light beam coming from two such inepenent sources o not have any fixe phase relationship an they o not prouce any stationary interference pattern For such sources, resultant intensity at any point is given by + (3 4 YOUNG S DOUBLE SLIT EXPERIMENT (YDSE In 80 Thomas Young evise a metho to prouce a stationary interference pattern This was base upon ivision of a single wavefront into two; these two wavefronts acte as if they emanate from two sources having a fixe phase relationship Hence when they were allowe to interfere, stationary interference pattern was observe S 0 A S S B D Central Figure : 4 : Young s Arrangement to prouce stationary interference pattern by ivision of wave front S 0 into S an S C Teko Classes, Maths : Suhag R Kariya (S R K Sir, Bhopal Phone : 0 903 903 7779, 0 98930 5888 page 3 Successful People Replace the wors like; "wish", "try" & "shoul" with "I Will" Ineffective People on't
Get Solution of These Packages & Learn by Vieo Tutorials on wwwmathsbysuhagcom FREE Downloa Stuy Package from website: wwwtekoclassescom & wwwmathsbysuhagcom Incient wave S 0 A Figure 4 : In Young's interference experiment, light iffracte from pinhole S 0 excounters pinholes S an S in screen B Light iffracte from these two pinholes overlaps in the region between screen B an viewing screen C, proucing an interference pattern on screen C 4 Analysis of Interference Pattern We have insure in the above arrangement that the light wave passing through S is in phase with that passing through S However the wave reaching P from S may not be in phase with the wave reaching P from S, because the latter must travel a longer path to reach P than the former We have alreay iscusse the phase-ifference arising ue to path ifference If the path ifference is equal to zero or is an integral multiple of wavelengths, the arriving waves are exactly in phase an unergo constructive s0 interference If the path ifference is an o multiple of half a wavelength, the arriving waves are out of phase an unergo fully estructive interference Thus, it is the path ifference x, which etermines the intensity at a point P Figure : 43 Path ifference p S P S P y + + D y + D (4 Approximation I : For D >>, we can approximate rays r parallel, at angle θ to the principle axis S S B an r as being approximately Now, S P S P S A S S sin θ path ifference sin θ (4 s s A θ r D s s C θ r Figure : 44 θ A r r P y Teko Classes, Maths : Suhag R Kariya (S R K Sir, Bhopal Phone : 0 903 903 7779, 0 98930 5888 page 4 Successful People Replace the wors like; "wish", "try" & "shoul" with "I Will" Ineffective People on't
FREE Downloa Stuy Package from website: wwwtekoclassescom & wwwmathsbysuhagcom Get Solution of These Packages & Learn by Vieo Tutorials on wwwmathsbysuhagcom Approximation II : further if θ is small, iey << D, sin θ tan θ D y an hence, path ifference D y for maxima (constructive interference, y p D n (43 n D y, n 0, ±, ±, ± 3 (44 Here n 0 correspons to the central maxima n ± correspon to the st maxima n ± correspon to the n maxima an so on for minima (estructive interference p consequently, y Here 3 5 p ±, ± ± (n (n + D (n D (n + n,, 3 n -, -, - 3 n,, 3 n -, -, - 3 n ± correspons to first minima, n ± correspons to secon minima an so on (45 4 Fringe with : It is the istance between two maxima of successive orer on one sie of the central maxima This is also equal to istance between two successive minima fringe with β D Notice that it is irectly proportional to wavelength an inversely proportional to the istance between the two slits 43 imum orer of Interference Fringes : In section 4 we obtaine, S S (46 maxima maxima D 3 D D D D B B B 0 B B First maxima Central maxima D 3 B 3 Figure : 45 fringe pattern in YDSE n D y, n 0, ±, ± for interference maxima, but n cannot take infinitely large values, as that woul violate the approximation (II ie, θ is small or y << D β β Teko Classes, Maths : Suhag R Kariya (S R K Sir, Bhopal Phone : 0 903 903 7779, 0 98930 5888 page 5 Successful People Replace the wors like; "wish", "try" & "shoul" with "I Will" Ineffective People on't
Get Solution of These Packages & Learn by Vieo Tutorials on wwwmathsbysuhagcom FREE Downloa Stuy Package from website: wwwtekoclassescom & wwwmathsbysuhagcom Aliter y n << D Hence the above formula (44 & 45 for interference maxima/minima are applicable when n << when n becomes comparable to path ifference can no longer be given by equation (43 but by (4 Hence for maxima p n sinθ n n Hence highest orer of interference maxima, n max where [ ] represents the greatest integer function Similarly highest orer of interference minima, n min + (48 p S P S P s p p max (3r sie of a triangle is always greater than s the ifference in length of the other two sies 44 Intensity : sinθ Figure : 46 (47 Suppose the electric fiel components of the light waves arriving at point P(in the Figure : 43 from the two slits S an S vary with time as E E 0 sin ωt an E E 0 sin (ωt + φ π Here φ k x x an we have assume that intensity of the two slits S an S are same (say 0 ; hence waves have same amplitue E 0 then the resultant electric fiel at point P is given by, E E + E E 0 sin ωt + E 0 sin (ωt + φ E 0 sin (ωt +φ where E 0 E 0 + E 0 + E 0 E 0 cos φ 4 E 0 cos φ/ Hence the resultant intensity at point P, φ 4 0 cos (49 max 4 0 when φ nπ, n 0, ±, ±,, φ min 0 when n π n 0, ±, ± Here φ k x π x If D >>, π φ sin θ If D >> & π y << D, φ y D However if the two slits were of ifferent intensities an, say E E 0, sin ωt an E E 0, sin (ωt + φ then resultant fiel at point P, E E + E E 0 sin (ωt + φ Teko Classes, Maths : Suhag R Kariya (S R K Sir, Bhopal Phone : 0 903 903 7779, 0 98930 5888 page 6 Successful People Replace the wors like; "wish", "try" & "shoul" with "I Will" Ineffective People on't
FREE Downloa Stuy Package from website: wwwtekoclassescom & wwwmathsbysuhagcom Get Solution of These Packages & Learn by Vieo Tutorials on wwwmathsbysuhagcom Ex 4 Sol Ex 5 where E 0 E 0 + E O + E 0 E 0 cos φ Hence resultant intensity at point P, + + cos φ (40 In a YDSE, D m, mm an / mm (i Fin the istance between the first an central maxima on the screen (ii Fin the no of maxima an minima obtaine on the screen D >> Hence P sin θ, clearly, n << is not possible for any value of n Hence p D y cannot be use for Ist maxima, p sin θ sin θ Hence, y D tan θ (ii meter 3 imum path ifference P max mm Highest orer maxima, n max θ 30º an highest orer minima n min + Total no of maxima n max + * 5 *(central maxima Total no of minima n min 4 S S θ Figure 47 Monochromatic light of wavelength 5000 Aº is use in YDSE, with slit-with, mm, istance between screen an slits, D m If intensity at the two slits are, 4 0, 0, fin (i fringe with β (ii istance of 5th minima from the central maxima on the screen (iii (iv (v Sol (i β Intensity at y 3 mm Distance of the 000 th maxima Distance of the 5000 th maxima D (ii y (n (iii 5000 0 0 3 0 At y 3 mm, y << D y Hence p D π y φ p π D Now resultant intensity 05 mm D, n 5 y 5 mm 4π 3 + + cos φ D y Teko Classes, Maths : Suhag R Kariya (S R K Sir, Bhopal Phone : 0 903 903 7779, 0 98930 5888 page 7 Successful People Replace the wors like; "wish", "try" & "shoul" with "I Will" Ineffective People on't
Get Solution of These Packages & Learn by Vieo Tutorials on wwwmathsbysuhagcom 4π 4 0 + 0 + 4 cos φ 5I 0 0 + 4I 0 cos 30 3 FREE Downloa Stuy Package from website: wwwtekoclassescom & wwwmathsbysuhagcom (iv (v 3 0 6 000 05 0 n 000 is not << 000 Hence now p sin θ must be use Hence, sin θ n 000 sin θ 000 y D tan θ meter 3 Highest orer maxima n max 000 Hence, n 5000 is not possible θ 30º 5 SHAPE OF INTERFERENCE FRINGES IN YDSE We iscuss the shape of fringes when two pinholes are use instea of the two slits in YDSE Fringes are locus of points which move in such a way that its path ifference from the two slits remains constant S P S P constant (5 If ±, the fringe represents st minima 3 If ± it represents n minima If 0 it represents central maxima, If ±, it represents st maxima etc Figure : 5 Y S S 3 0 - - Equation (5 represents a hyperbola with its two foci at S an S The interference pattern which we get on screen is the section of hyperboloi of revolution when we revolve the hyperbola about the axis S S A If the screen is er to the X axis, ie in the YZ plane, as is generally the case, firinges are hyperbolic with a straight central section B If the screen is in the XY plane, again fringes are hyperbolic C If screen is er to Y axis (along S S, ie in the XZ plane, fringes are concentric circles with center on the axis S S ; the central fringe is bright if S S n an ark if S S (n X Teko Classes, Maths : Suhag R Kariya (S R K Sir, Bhopal Phone : 0 903 903 7779, 0 98930 5888 page 8 Successful People Replace the wors like; "wish", "try" & "shoul" with "I Will" Ineffective People on't
Get Solution of These Packages & Learn by Vieo Tutorials on wwwmathsbysuhagcom Y FREE Downloa Stuy Package from website: wwwtekoclassescom & wwwmathsbysuhagcom A Figure : 5 Z Y B Y 6 YDSE WITH WHITE LIGHT The central maxima will be white because all wavelengths will constructively interference here However slightly below (or above the position of central maxima fringes will be coloure for example if P is a point on the screen such that Ex 6 Sol β S P S P violet 90 nm, completely estructive interference will occur for violet light Hence we will have a line evoi of violet colour that will appear reish An if S P S P re 350 nm, completely estructive interference for re light results an the line at this position will be violet The coloure fringes isappear at points far away from the central white fringe; for these points there are so many wavelengths which interfere constructively, that we obtain a uniform white illumination for example if S P S P 3000 nm, 3000 then constructive interference will occur for wavelengths nm In the visible region these wavelength n are 750 nm (re, 600 nm (yellow, 500 nm (greenish yellow, 430 nm (violet Clearly such a light will appear white to the unaie eye Thus with white light we get a white central fringe at the point of zero path ifference, followe by a few coloure fringes on its both sies, the color soon faing off to a uniform white In the usual interference pattern with a monochromatic source, a large number of ientical interference fringes are obtaine an it is usually not possible to etermine the position of central maxima Interference with white light is use to etermine the position of central maxima in such cases A beam of light consisting of wavelengths 6000Å an 4500Å is use in a YDSE with D m an mm Fin the least istance from the central maxima, where bright fringes ue to the two wavelengths coincie D 6000 0 3 0 0 06 mm D β 045 mm Let n th maxima of an n th maxima of coincie at a position y then, y n P n P LCM of β an β y LCM of 06 cm an 045 mm y 8 mm Ans At this point 3r maxima for 6000 Å & 4th maxima for 4500 Å coincie X C X Teko Classes, Maths : Suhag R Kariya (S R K Sir, Bhopal Phone : 0 903 903 7779, 0 98930 5888 page 9 Successful People Replace the wors like; "wish", "try" & "shoul" with "I Will" Ineffective People on't
FREE Downloa Stuy Package from website: wwwtekoclassescom & wwwmathsbysuhagcom Get Solution of These Packages & Learn by Vieo Tutorials on wwwmathsbysuhagcom Ex 7 Sol White light is use in a YDSE with D m an 09 mm Light reaching the screen at position y mm is passe through a prism an its spectrum is obtaine Fin the missing lines in the visible region of this spectrum p D y for minima p (n / 9 0 4 0 3 m 900 nm P (n 800 (n 800 800 800 800,,, 3 5 7 of these 600 nm an 360 nm lie in the visible range Hence these will be missing lines in the visible spectrum 7 GEOMETRICAL PATH & OPTICAL PATH Actual istance travelle by light in a meium is calle geometrical path ( x Consier a light wave given by the equation E E 0 sin (ωt kx + φ If the light travels by x, its phase changes by k x v ω x, where ω, the frequency of light oes not epen on the meium, but v, the spee of light epens on the meium as v µ c Consequently, change in phase, φ k x c ω (µ x It is clear that a wave travelling a istance x in a meium of refractive inex µ suffers the same phase change as when it travels a istance µ x in vacuum ie a path length of x in meium of refractive inex µ is equivalent to a path length of µ x in vacuum The quantity µ x is calle the optical path length of light, x opt An in terms of optical path length, phase ifference woul be given by, φ c ω xopt π xopt (7 where 0 wavelength of light in vacuum However in terms of the geometrical path length x, ω π φ (µ x c 0 x (7 where wavelength of light in the meium ( µ 0 7 Displacement of fringe on introuction of a glass slab in the path of the light coming out of the slits On introuction of the thin glass-slab of thickness t an refractive inex µ, the optical path of the ray S P increases by t(µ Now the path ifference between waves coming form S an S at any point P is p S P (S P + t (µ (S P S P t(µ p sin θ t (µ if << D an p D y t(µ If y << D as well S S Figure : 7 D P O O' Teko Classes, Maths : Suhag R Kariya (S R K Sir, Bhopal Phone : 0 903 903 7779, 0 98930 5888 page 0 Successful People Replace the wors like; "wish", "try" & "shoul" with "I Will" Ineffective People on't
Get Solution of These Packages & Learn by Vieo Tutorials on wwwmathsbysuhagcom for central bright fringe, p 0 FREE Downloa Stuy Package from website: wwwtekoclassescom & wwwmathsbysuhagcom Ex 8 Sol y t(µ D y OO (µ t D (µ t β The whole fringe pattern gets shifte by the same istance (µ D (µ t B * Notice that this shift is in the irection of the slit before which the glass slab is place If the glass slab is place before the upper slit, the fringe pattern gets shifte upwars an if the glass slab is place before the lower slit the fringe pattern gets shifte ownwars In a YDSE with mm an D m, slabs of (t µm, µ 3 an (t 05 µm, µ are introuce in front of upper an lower slit respectively Fin the shift in the fringe pattern Optical path for light coming from upper slit S is S P + µm ( S P + 05 µm Similarly optical path for light coming from S is S P + 05 µm ( S P + 05 µm Path ifference : p (S P + 05 µm (S P + µm D y 5 µm for central bright fringe p 0 5µ m y mm m 5 mm The whole pattern is shifte by 5 mm upwars (S P S P 5 µm 8 YDSE WITH OBLIQUE INCIDENCE In YDSE, ray is incient on the slit at an inclination of θ 0 to the axis of symmetry of the experimental set-up for points above the central point on the screen, (say for P p sinθ 0 + (S P S P p sinθ 0 + sinθ (If << D an for points below O on the screen, (say for P p (sin θ 0 + S P S P sin θ 0 (S P S P p sin θ 0 sinθ (if << D We obtain central maxima at a point where, p 0 ( sin θ 0 sinθ 0 or θ θ 0 This correspons to the point O in the iagram Hence we have finally for path ifference (sin θ 0 + sinθ for points above O p (sin θ0 sinθ for points between O & O' (sinθ sinθ for points below O' 0 Ans sinθ 0 Figure : 8 (8 S S θ P θ 0 θ O' O P B 0 Teko Classes, Maths : Suhag R Kariya (S R K Sir, Bhopal Phone : 0 903 903 7779, 0 98930 5888 page Successful People Replace the wors like; "wish", "try" & "shoul" with "I Will" Ineffective People on't
Get Solution of These Packages & Learn by Vieo Tutorials on wwwmathsbysuhagcom FREE Downloa Stuy Package from website: wwwtekoclassescom & wwwmathsbysuhagcom Ex 9 In YDSE with D m, mm, light of wavelength 500 nm is incient at an angle of 057º wrt the axis of symmetry of the experimental set up If centre of symmetry of screen is O as shown (i fin the position of central maxima (ii Intensity at point O in terms of intensity of central maxima 0 (iii Number of maxima lying between O an the central maxima Sol (i θ θ 0 057º 057 y D tan θ _~ Dθ meter ra 57 y cm (ii for point 0, θ 0 Hence, p sin θ 0 θ 0 mm (0 ra 0,000 nm 0 (500 nm p 0 Hence point O correspons to 0th maxima intensity at O 0 (iii 057º S S Figure : 8 9 maxima lie between central maxima an O, excluing maxima at O an central maxima 9 THIN-FILM INTERFERENCE In YDSE we obtaine two coherent source from a single (incoherent source by ivision of wave-front Here we o the same by ivision of Amplitue (into reflecte an refracte wave When a plane wave (parallel rays is incient normally on a thin film of uniform thickness then waves reflecte from the upper surface interfere with waves reflecte from the lower surface Clearly the wave reflecte from the lower surface travel an extra optical path of µ, where µ is refractive inex of the film Figure : 9 Air µ Air y P O Incient & Reflecte light Transmitte light Further if the film is place in air the wave reflecte from the upper surface (from a enser meium suffers a suen phase change of π, while the wave reflecte from the lower surface (from a rarer meium suffers no such phase change Consequently conition for constructive an estructive interference in the reflecte light is given by, µ n for estructive interference an µ (n + for constructive interference (9 where n 0,,, an wavelength in free space Interference will also occur in the transmitte light an here conition of constructive an estructive interference will be the reverse of (9 ie n µ (n + for constructive interference for estructive interference (9 This can easily be explaine by energy conservation (when intensity is maximum in reflecte light it has to be minimum in transmitte light However the amplitue of the irectly transmitte wave an the wave transmitte after one reflection iffer substantially an hence the fringe contrast in transmitte light is poor It is for this reason that thin film interference is generally viewe only in the reflecte light In eriving equation (9 we assume that the meium surrouning the thin film on both sies is rarer Teko Classes, Maths : Suhag R Kariya (S R K Sir, Bhopal Phone : 0 903 903 7779, 0 98930 5888 page Successful People Replace the wors like; "wish", "try" & "shoul" with "I Will" Ineffective People on't
FREE Downloa Stuy Package from website: wwwtekoclassescom & wwwmathsbysuhagcom Get Solution of These Packages & Learn by Vieo Tutorials on wwwmathsbysuhagcom compare to the meium of thin film If meium on both sies are enser, then there is no suen phase change in the wave reflecte from the upper surface, but there is a suen phase change of π in waves reflecte from the lower surface The conitions for constructive an estructive interference in reflecte light woul still be given by equation 8 However if meium on one sie of the film in enser an that on the other sie is rarer, then either there is no suen phase in any reflection, or there is a suen phase change of π in both reflection from upper an lower surface Now the conition for constructive an estructive interference in the reflecte light woul be given by equation 8 an not equation 8 Ex 0 White light, with a uniform intensity across the visible wavelength range 430 690 nm, is perpenicularly incient on a water film, of inex of refraction µ 33 an thickness 30 nm, that is suspene in air At what wavelength is the light reflecte by the film brightest to an observer? Sol This situation is like that of Figure (9, for which euqation (9 gives the interference maxima Solving for an inserting the given ata, we obtain µ ((33(30nm 85nm m + / m + / m + / for m 0, this give us 700 nm, which is in the infrare region For m, we fin l 567 nm, which is yellow-green light, near the mile of the visible spectrum For m, 340 nm, which is in the ultraviolet region So the wavelength at which the light seen by the observer is brightest is 567 nm Ans Ex A glass lens is coate on one sie with a thin film of magnesium fluorie (MgF to reuce reflection from the lens surface (figure The inex of refraction of MgF is 38; that of the glass is 50 What is the least coating thickness that eliminates (via interference the reflections at the mile of the visible specturm ( 550 nm? Assume the light is approximately perpenicular to the lens surface Sol The situation here iffers from figure (9 in that n 3 > n > n The reflection at point a still introuces a phase ifference of π but now the reflection at point b also oes the same ( see figure 9 Unwante reflections from glass can be, suppresse (at a chosen wavelength by coating the glass with a thin transparent film of magnesium fluorie of a properly chosen thickness which introuces a phase change of half a wavelength For this, the path length ifference L within the film must be equal to an o number of half wavelengths: L (m + / n, or, with n /n, n L (m + / Air n 00 r i r θ θ Figure : 9 MgF n 33 a c L Glass n 50 3 We want the least thickness for the coating, that is, the smallest L Thus we choose m 0, the smallest value of m Solving for L an inserting the given ata, we obtain L 4n 550nm (4(38 966 nm Ans 0 HUYGENS' CONSTRUCTION Huygens, the Dutch physicist an astronomer of the seventeenth century, gave a beautiful geometrical escription of wave propagation We can guess that he must have seen water waves many times in the canals of his native place Hollan A stick place in water an oscillate up an own becomes a source of waves Since the surface of water is two imensional, the resulting wavefronts woul be circles instea of spheres At each point on such a circle, the water level moves up an own Huygens' iea is that we can think of every such oscillating point on a wavefront as a new source of waves Accoring to Huygens' principle, what we observe is the result of aing up the waves from all these ifferent sources These are calle seconary waves or wavelets Huygens' principle is illustrate in (Figure : 0 in the simple case of a plane wave (i (ii At time t 0, we have a wavefront F, F separates those parts of the meium which are unisturbe from those where the wave has alreay reache Each point on F acts like a new source an sens out a spherical wave After a time t each of these will have raius vt These spheres are the seconary b F F A A B C D Figure : 0 B C D Teko Classes, Maths : Suhag R Kariya (S R K Sir, Bhopal Phone : 0 903 903 7779, 0 98930 5888 page 3 Successful People Replace the wors like; "wish", "try" & "shoul" with "I Will" Ineffective People on't
FREE Downloa Stuy Package from website: wwwtekoclassescom & wwwmathsbysuhagcom Get Solution of These Packages & Learn by Vieo Tutorials on wwwmathsbysuhagcom (iii (iv (v wavelets After a time t, the isturbance woul now have reache all points within the region covere by all these seconary waves The bounary of this region is the new wavefront F Notice that F is a surface tangent to all the spheres It is calle the forwar envelope of these seconary wavelets The seconary wavelet from the point A on F touches F at A Draw the line connecting any point A on F to the corresponing point A on F Accoring to Huygens, A A is a ray It is perpenicular to the wavefronts F an F an has length vt This implies that rays are perpenicular to wavefronts Further, the time taken for light to travel between two wavefronts Is the same along any ray In our example, the spee v of the wave has been taken to be the same at all points in the meium In this case, we can say that the istance between two wavefronts is the same measure along any ray This geometrical construction can be repeate starting with F to get the next wavefront F 3 a time t later, an so on This is known as Huygens' construction Huygens' construction can be unerstoo physically for waves in a material meium, like the surface of water Each oscillating particle can set its neighbours into oscillation, an therefore acts as a seconary source But what if there is no meium, as for light travelling in vacuum? The mathematical theory, which cannot be given here, shows that the same geometrical construction works in this case as well 0 REFLECTION AND REFRACTION We can use a moifie form of Huygens' construction to unerstan reflection an refraction of light Figure (0a shows an incient wavefront which makes an angle i with the surface separating two meia, for example, air an water The phase spees in the two meia are v an v We can see that when the point A on the incient wavefront strikes the surface, the point B still has to travel a istance BC AC sin i, an this takes a time t BC/v AC (sin i / v After a time t, a seconary wavefront of raius v t with A as centre woul have travelle into meium The seconary wavefront with C as centre woul have just starte, ie woul have zero raius We also show a seconary wavelet originating from a point D in between A an C Its raius is less than v t The wavefront in meium is thus a line passing through C an tangent to the circle centre on A We can see that the angle r' mae by this refracte wavefront with the surface is given by AE v t AC sin r' Hence, t AC (sin r'} / v Equating the two expressions for t gives us the law of refraction in the form sin i/sin r' v /v A similar picture is rawn in Fig (0 b for the reflecte wave which travels back into meium In this case, we enote the angle mae by the reflecte wavefront with the surface by r, an we fin that i r Notice that for both reflection an refraction, we use seconary wavelets starting at ifferent times Compare this with the earlier application (Fig0 where we start them at the same time The preceing argument gives a goo physical picture of how the refracte an reflecte waves are built up from seconary wavelets We can also unerstan the laws of reflection an refraction using the concept that the time taken by light to travel along ifferent rays from one wavefront to another must be the same (Fig Shows the incient an reflecte wavefronts when a parallel beam of light falls on a plane surface One ray POQ is shown normal to both the reflecte an incient wavefronts The angle of incience i an the angle of reflection r are efine as the angles mae by the incient an reflecte rays with the normal As shown in Fig (c, these are also the angles between the wavefront an the surface (a (b (c Figure : 0 (Fig (a Huygens' construction for the (a refracte wave (b Reflecte wave (c Calculation of propagation time between wavefronts in (i reflection an (ii refraction We now calculate the total time to go from one wavefront to another along the rays From Fig (c, we have, we have Total time for light to reach from P to Q PO OQ AO sin i OB sin r OA sin i + (AB OAsinr AB sin r + OA(sini sinr + + v v v v v v Different rays normal to the incient wavefront strike the surface at ifferent points O an hence have ifferent Teko Classes, Maths : Suhag R Kariya (S R K Sir, Bhopal Phone : 0 903 903 7779, 0 98930 5888 page 4 Successful People Replace the wors like; "wish", "try" & "shoul" with "I Will" Ineffective People on't
FREE Downloa Stuy Package from website: wwwtekoclassescom & wwwmathsbysuhagcom Get Solution of These Packages & Learn by Vieo Tutorials on wwwmathsbysuhagcom values of OA Since the time shoul be the same for all the rays, the right sie of equation must actually be Inepenent of OA The conition, for this to happen is that the coefficient of OA in Eq (shoul be zero, ie, sin i sin r We, thus have the law of reflection, i r Figure also shows refraction at a plane surface separating meium (spee of light v from meium (spee of light v The incient an refracte wavefronts are shown, making angles i an r' with the bounary Angle r' is calle the angle of refraction Rays perpenicular to these are also rawn As before, let us calculate the time taken to travel between the two wavefronts along any ray Time taken from P to R PO v + v v OR OA sin i + (AC OAsin r' v AC sin r' sin i sin r' v + OA v v This time shoul again be inepenent of which ray we consier The coefficient of OA in Equation is, therefore, zero, That is, sin i sin r' v v n where n is the refractive inex of meium with respect to meium This is the Snell's law of, refraction that we have alreay ealt with from Eq n is the ratio of spee of light in the first meium (v to that in the secon meium (v Equation is, known as the Snell's law of refraction If the first meium is vacuum, we sin i c have n sin r' v where n is the refractive inex of meium with respect to vacuum, also calle the absolute refractive inex of the meium A similar equation efines absolute refractive inex n of the first meium From Eq we then get n v v c c n / n n n The absolute refractive inex of air is about 0003, quite close to Hence, for all practical purposes, absolute ref ractiv e inex of a meium may be taken with respect to air For water, c n 33, which means v, ie about 075 times the spee of light in vacuum The measurement of 33 the spee of light in water by Foucault (850 confirme this preiction of the wave theory Once we have the laws of reflection an refraction, the behaviour of prisms lenses, an mirrors can be unerstoo These topics are iscusse in etail in the previous Chapter Here we just escribe the behaviour of the wavefronts in these three cases (Fig (i (ii (iii (iv Consier a plane wave passing through a thin prism Clearly, the portion of the incoming wavefront which travels through the greatest thickness of glass has been elaye the most Since light travels more slowly in glass This explains the tilt in the emerging wavefront Similarly, the central part of an incient plane wave traverses the thickest portion of a convex lens an is elaye the most The emerging wavefront has a epression at the centre It is spherical an converges to a focus, A concave mirror prouces a similar effect The centre of the wavefront has to travel a greater istance before an after getting reflecte, when compare to the ege This again prouces a converging spherical wavefront Concave lenses an convex mirrors can be unerstoo from time elay arguments in a similar manner One interesting property which is obvious from the pictures of wavefronts is that the total time taken from a point on the object to the corresponing point on the image is the same measure along any ray (Fig For example, when a convex lens focuses light to form a real image, it may seem that rays going through the centre are shorter But because of the slower spee in glass, the time taken is the same as for rays travelling near the ege of the lens (a (b (c Figure : 03 Teko Classes, Maths : Suhag R Kariya (S R K Sir, Bhopal Phone : 0 903 903 7779, 0 98930 5888 page 5 Successful People Replace the wors like; "wish", "try" & "shoul" with "I Will" Ineffective People on't