WAVE OPTICS GENERAL. Fig.1a The electromagnetic spectrum

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1 WAVE OPTICS GENERAL - The ray optics cannot explain the results of the two following experimental situations: a) When passing by small openings or illuminating small obstacles, the light bends around borders and reaches inaccessible regions for optical rays. DIFFRACTION; b) When two or more light beams superpose in some special conditions, fringes of maxima and minima brightness (i.e. light intensity) are produced on a screen.interference; Diffraction and interference are typical wave phenomena; one can explain them by the wave model used for mechanical waves (TW and LW- sound waves). In the following, one starts by a review of basics principles for diffraction and interference in mechanical waves and sound in D; next, one applies them to light waves. - Multiple experiments and theoretical research showed that light waves are electro-magnetic (E.M) waves. Radio & TV waves, x & γ rays are part of E.M. spectrum, too ( see figures 1.a-b). Fig.1a The electromagnetic spectrum Fig.1.b Visible Light - The speed of all E.M. waves (light included) in vacuum is c = 3*10 8 m/s. When light propagates through a transparent material medium, its speed decreases to υ = c / n (1) where n is the refractive index of the medium (n >1). If the wavelength of light in vacuum is T * 0 c, when propagating in a medium its wavelength becomes n T *. The period T and the frequency (f =1/ T) of light wave remain constant but the wavelength decreases / / 1 n 0 c 1 n Remember that ; λ n < λ 0 and λ n = λ 0 /n () DIFFRACTION Fig.1.c The light frequency remains the same but the its wavelength decreases inside the matter. - In geometrical optics, the rays (directed straight lines) show the path followed by light particles or energy droplets of light in a transparent medium; i.e. the direction of light energy flow. In a wave model of light, the energy propagates perpendicular to the wave fronts. So, it comes out that the rays are perpendicular to the wave fronts. For a point source inside a homogenous medium, the wave fronts are spherical and rays follow the radius directions. As mentioned previously, at a big distance from the source, a portion of wave front can be fitted by a plane (see fig.). Note that the rays remain always perpendicular to wave fronts independently of their geometrical form. 1

2 Wave fronts Point Source Ray Figure. The rays show the direction of energy flow and they are perpendicular to the wave fronts. -When a plane wave hits a large opening (or obstacle), there is no visible wave motion out the space region covered by rays (fig.3.a). But, when the opening (or obstacle) dimension decreases beyond a certain limit (less than ~10λ), the wave motion appears inside zones that cannot be reached by rays(fig. 3.b). This phenomenon is known as diffraction and can be explained only by the concept of wave fronts and the Huygens Principle. Figure 3. a) Ray "normal situation" (d >>λ) b) Ray "abnormal situation" (d > λ or comparable to λ ) Huygens principle: Any point on the wave front is a source of secondary waves (wavelets). The new wave front is the envelope of wave fronts of these wavelets. This principle explains the effect of diffraction in all circumstances; Figure 4 a) Large openings (or obstacles). There are too many points on the wave front that contribute into building the new wave front by wavelets that they emit. The large central part of common envelope profile hides the effect of a few wavelets originated from the borders and the result is a new almost plane wave front (fig. 4) b) Small openings (or obstacles). There is a limited number of points on the wave front that contribute into building the new wave front by their wavelets. The central plane part of envelope cannot hide the curved parts due to the wavelets originated at borders; the result is a curved wave front (fig.5). Figure 5. a- many wavelets (d >λ) b- several wavelet (d λ) c- only a few wavelet

3 - All diffraction phenomena are explained by use of wave fronts and Huygens principle. Note that one may observe easy way the diffraction of sound waves because their wavelength is comparable to dimensions of many common opening (doors) or obstacles (objects). Exemple: for f =150Hz (low sound frequencies) the wavelength is λ =340m s -1 /150 s -1 =.m which is comparable to widths of corridors or door. - In the case of light, the diffraction effects appear only when light falls on very small openings or objects because λ -values are very small. But, the mechanism of diffraction is the same and the Huygens principle applies the same way because it does not depend on the physical nature of a wave. INTERFERENCE - The principle of linear superposition: If the paths of two (or more) waves that propagate in a medium cross to each other, then the total displacement at crossing point is equal to the sum of displacements "y i " due to each wave at this point of medium. y T n i1 yi ; i 1,,... (3) This principle is valid no matter what is the physical nature displacement. - Remember: When two waves pass by the same point of a medium, there is always a superposition but not always interference. There is interference only when the waves: a) Have the same frequency. This requirement is essential; To observe a stable interference, one must deal with a standing wave. b) Have displacements along the same space direction(fig 6.a,b). This condition is not always met always for TW waves. (Ex. Two crossed pulses on a rope; fig 6.c) Figure 6 (c) There is superposition but not interference because the displacements are perpendicular to each other - Assume that a wave (TW, LW, mechanical or not) propagates through a medium. The phase of a travelling wave, i.e. Φ = kx +/- ωt + φ, can be calculated at any location x and any time t, provided one has defined an axe Ox and knows k and ω. The wave front is defined (in D and 3D) as the locus of the points (locations) in medium at which the wave function has the same phase (Φ- value) at any moment; the displacement is the same at all points of a particular wave front, too. In general, one shows a D travelling wave by sketching a set of crests (wave fronts referred to maxima displacements) at a given moment of time. For an harmonic wave, at any time, there is a phase change by +/- π between any two consecutive wave fronts and there is the same displacement on any of them. One says that the points on two consecutive wave fronts oscillate in phase (or are in phase). One may easily show that the distance between two similar consecutive (Φ = Φ 1 +/- π) 3

4 wave fronts is equal to the wavelength. Just choose an axe Ox (as shown in figure 7.a), use the expression of phase Φ = kx- ωt + φ (+Ox direction of propagation) and find out that, at any given moment of time "t " : 1 kx1 t ;_ kx t ; (4) 1 k( x x1) So, any two similar ( say crests ) consecutive wave fronts are at a distance λ to each other (fig.6a). O λ x Φ 1 Φ Φ1 Φ= Φ1+π Figure 7.a Figure 7.b Phase shift π between phasors of two consecutive wave fronts THE SATUS OF INTERFERENCE PRODUCED BY TWO WAVES AT GIVEN SPACE LOCATION -The interference status at P-point is defined by phase shift between the waves that interfere at this location. S 1 S 1 Figure 8 r 1 r P x x 1 If this point P is at distance r 1 from source S 1 and r from source S (fig.8), the phases of the first and second wave at P location are 1 kr 1 t 1 _ and _ kr t. So, the phase difference between the two waves at P point is 1 k( r r1 ) ( 1) p s (5) ΔΦ p = k(r -r 1 ) = kδ=(π/λ)δ phase shift due to path lengths difference ΔΦ s = φ - φ 1 the phase shift between the two sources As k =π/λ is a constant, the phase shift ΔΦ p depends only on the paths length difference δ between the two waves at P point. r r1 (6) r 1 is the distance from the first source; r is the distance from the second source; If reflections happen in the path of these two waves, there is a phase shift due to reflections ΔΦ r and - There is a constructive interference if ΔΦ tot = ± π*m (M=0,1,,3,...) In a constructive interference the two phasors are parallel, all time. There is a destructive interference if ΔΦ tot = ± π*(m+1) (m=0,1,,3,...) In a constructive interference the two phasors are inverse parallel, all time. ΔΦ tot = ΔΦ s + ΔΦ r + ΔΦ p (7) Two phasors in constructive interference Two phasors in destructive interference 4

5 -Assume that sources S 1,S oscillate in phase (ΔΦ s = 0) and there is no wave reflections (ΔΦ r = 0) till P point. In this case ΔΦ tot = ΔΦ p and one finds out that there is a constructive interference if a) ΔΦ p = 0 which happens for δ = 0, i.e. r 1 = r ; b) or ΔΦ p = ± π*m which happens for δ = r - r 1 = ± Mλ (8.a) There is destructive interference if : ΔΦ p = ± π*(m+1) which happens for δ = r - r 1 = ± *(m+1) λ/ (8.b) Remember: At locations of constructive interference there is oscillations with a maximum amplitude of displacement while at locations of destructive interference there is zero amplitude of displacement. -IMPORTANT : The derivation of conditions (7-8) is based only on phase difference requirements. So, they are valid for the interference of any type of waves, sound waves, light waves, radio waves, TV waves... These same rules explain bad or good sound hearing at different positions inside the same concert hall and also the low or high quality reception of radio signal and TV signals for the different positions of an antenna. OBSERVATION OF INTERFERENCE ON A TWO DIMENSIONNAL SPACE (plane surface) -The standing waves along a string are an example of 1D space interference. A D space interference can be built by using two plungers that vibrate at the same frequency and phase on the still surface of a water tank. Figure 9 Interference observed on water surface. Figure 10 - The wave model explains perfectly the picture fig.9 produced on water surface by the interference of two mechanical waves originated at two point sources (S 1, S ) that oscillate with same frequency, same amplitude and the same space direction of displacements. Actually, one can easily figure out that; There is CONSTRUCTIVE (double amplitude) interference at any point where ΔΦ p = 0 or ±π*m (a crest wave front meets another crest or a trough wave front meets another trough wave front). There is DESTRUCTIVE interference (zero amplitude) at any point where ΔΦ p = ± π*(m+1) (a crest wave front meets trough wave fronts or vice versa). As ΔΦ p = k *δ = k*(r -r 1 ) the interference status at a given point depends only on its location. From the figure 10, one may see that there is a constructive interference with the phase difference : a) ΔΦ p = 0 for each point located on central line(r 1 = r and δ = 0). b) ΔΦ p = ± M*π (at red points) also at all points (red dots) where δ = r - r 1 = ± M*λ. It comes out that all those points are located on hyperbola wings. 5

6 Between the central line and closest wing of constructive interference there is a line of destructive interference at points (empty circles) where ΔΦ p = +/-π or δ = r - r 1 = ± λ/. Those points are located on lines that have a hyperbola shape, too. So, based on the phase shift during interference of the two waves, one may explain the whole picture of fringes observed on the surface of the water tank. YOUNG S EXPERIMENT - Thomas Young showed the wave nature of light by proving that it does interfere like any other wave phenomena. He used an experimental set up similar to that presented in figure 11. In this scheme: Two sources of in phase oscillations are two secondary narrow slits S 1,S equidistant from the primary source (pinhole receiving sun light). The considerable distance from the pinhole gives plane waves at slits input. The narrow slits S 1,S produce cylindrical wave fronts. The model of a screen far from the slits simplifies expression for path length difference. Using a coloured glass one can get monochromatic light (one λ). Nowadays, one gets easily the monochromatic light by using a laser source. Coloured glass Figure 11 Figure 1 -The geometrical path length difference for a given point P on the screen (fig 1) is δ = r - r 1 As L >> d (screen far away from sources), the two paths are almost parallel and from fig.13, one gets d sin (9) d- distance between the middle points of the two narrow slits θ- the angle that correspond to the location of point P on screen Figure 13 - After noting that in this case ΔΦ S=0, ΔΦ R = 0 and ΔΦ tot = ΔΦ p, one might remember that there is: constructive interference for p k M * M * destructive interference for p k ( m 1) (m 1) /. 6

7 So, one can derive the corresponding directions (angles) for fringes of a maximum intensity by the expression (10) and a minimum intensity by (11) d sin M M M = 0, ±1, ±.. (10) d sinm (m 1) m = 0, ±1, ±.. (11) Notes: a) For M = 0 one gets the central maximum ( located in the middle of interference pattern). b) The first minimum upside ( θ > 0) the central maximum corresponds to m = 0. -Those interference fringes are parallel to slits. The position "y" (see fig 1) of low order maxima fringes (located close to the center of interference pattern where is small and tan sin ) can be found as follows: y L M y L M y d M L d M tan M sinm M (1) - The term coherence is widely used in interference studies. If the wave is a pure harmonic function (i.e. one frequency and infinite long) one says that it is a coherent wave emitted by a coherent source. Two waves (or sources) that oscillate with the same frequency (ω 1 = ω ) are said to be coherent between them in the sense that there is a constant phase shift between their phasors all the time ΔΦ 1, =[(ω t+φ ) - (ω 1 t+φ 1 )] = φ - φ 1. (13) Two coherent waves or sources are said to oscillate in phase if ΔΦ 1, = M*π where M=0,±1,±,.. S 1, S in the Young s experiment scheme are two coherent sources and in phase because ΔΦ 1, = 0. If the phase shift is constant but ΔΦ 1, M*π, the two sources are considered to be out of phase. Very often this nomination if used for two sources in opposite phase ; ΔΦ 1, = (m+1)π where m = 0,± 1,±,.. - Remember: The phase shift due to difference in path lengths is. If the two interfering waves propagate in vacuum (or air), the refraction coefficient is n =1 and k=k 0 =π/λ 0. When the two waves propagate in another medium n >1 and k k n = π/λ n = π/(λ 0 /n) = n*(π/λ 0 )= n* k 0 So, inside a medium P kn nko kon kon (14) When both waves propagate in the same medium the optical path length difference δ n is δ n = n*δ (15) When they propagate in two different mediums, δ n = n *r - n 1 *r 1 (16) The basic expression for the phase shift due to path length difference is k (17) path 0 n THE INTENSITY OF INTERFERENCE PATTERNS - In Young s experiment, one uses a narrow pinhole at big distance to slits to fit to the plane wave fronts at the slits input. Actually, this means equal frequency, equal phase and equal amplitude for the two waves at sources (slits). As the path lengths r 1, r have comparable values and the emitted powers are equal, these waves produce similar intensity ( I~P 0 /r ) and similar amplitude (I~A ) at each point of the screen. 7

8 - Therefore, at a given point on the screen, the light wave displacement E produced by the source S 1 is E1 ( t) E0 sint (18) produced by the source S is E ( t) E0 sin( t ) (19) E 0 - same amplitude of light wave displacement on the screen ω - same frequency (first requirement for coherence) φ - (Δφ p-1, φ) is constant in time. -The phase shift φ in (19) is defined by the position of considered point and does not change in time because the two waves are coherent. Actually, as φ =(π/λ)* δ, this phase shift is fixed by the path length difference δ from two slits to that point. -The principle of linear superposition gives the total displacement E T at this point as E T E1 E E E1 E0 sin( t ) E0 sin( t) E0 cos sin( t ) (0) A T E 0 cos (1) The amplitude of total displacement is -Here we remind another time that the intensity I = Power/Area of light oscillations at a point P on screen is proportional to (amplitude of light wave displacement) and the evolution of light intensity over different points on the screen is the same as the evolution of A. To make things easier, one may note I =A so that, if a single source sends a light wave at a point on the screen, the light intensity at this point is I0 A E. 0 If the light waves from two coherent sources illuminate simultaneously this point, the light intensity becomes I A 4E0 cos 4I 0 cos T () Figure 14 presents the graph of this function for different locations on screen ( different phase shifts φ). Actually, its maxima and minima are placed along the space directions that fit to expressions (10,11). -If the two waves had different frequency (ω ω 1 ) they would simply superpose on the screen but they would not interfere. In this case the intensity of light at considered point on the screen would be simple sum Io + Io = I 0. Because of interference, at the same point on screen, the light intensity: Figure 14 a) increases twice (4*I 0 = *( I 0 ) ) if the considered point is at a maximum location; b) is reduced to zero if the considered point is at a minimum location. Important Notes: a) The interference produces only energy redistribution but the average (see fig.14) (*I 0 ) intensity and the total energy on the screen remain unchanged. 8

9 b) If the amplitudes of two interfering waves are not equal, the interference fringes are not well seen. Even, the fringes may become invisible if E 01 >> E 0 or vice versa. This happens because the intensity of minima is not zero and it may become comparable to that of maxima. The figure 15 explains what happens in these situation. sum of phasors(maxima locations) difference of phasors (minima locations) Figure 15 The amplitudes of resultant phasors are comparable; the intensities become comparable, too. Exemple: A plane monochromatic wave with λ 0 = 600nm falls on two small slits and produces a set of maxima(bright) and minima(dark) fringes on a screen located on the other side of slits. Assume that one covers the upper slit ( as shown in figure ) by a plastic transparent strip with n =1.5. a) Show that the interference pattern shifts upward. b) Find its thickness L of the strip such that the pattern shifts up exactly by one inter fringe distance. Figure 16 The highest n-value of plastic increases the optical length of upper path n*r 1 and this affects the phase difference at any point on the screen. This effect appears as shift of fringes upside. r r 1 M=+1 M=0 M= -1 r 1 r M=0 M= -1 a) Without the plastic strip, the maximum of "order 0" on screen is located over a line passing through the midway of two slits. So, initially, δ op = 0 and Δφ = 0 at this location; i.e. the maximum of order M=0 is located at center (fig.16.a). In presence of the plastic strip on upper slit, the optical path length of the first wave to the center increases. As the length of optical path for second wave "r " does not change, it comes out that the introduction of plastic strip makes δ op = r - r 1 negative at center. This makes Δφ < 0 at center of patter and the location with Δφ = 0 is shifted upside the center (together with all the set of fringes) b)to shift the whole pattern up by one inter fringe distance, the thickness "L" of plastic strip should be such that the phase shift between wave_ and wave_1 at center becomes exactly "-π" (or M= -1). In presence of the strip with thickness L, the path length of the first wave changes from r 1 to r' 1 = (r 1 - L) + nl and the difference of optical lengths for the two waves at center (where r 1 = r ) becomes δ 1 = (r - r' 1 ) =[r -((r 1 - L) + nl)] = [r -r 1 + L - nl)]= [(r -r 1 ) - ( nl - L)] As (r - r 1 ) = 0, δ 1 = (r - r' 1 ) = - ( nl - L) = -L(n-1) This means that the phase shift between waves at center of screen becomes Δφ 1 = (π/λ) δ 1 = - (π/λ)* L(n 1) As one looks to have Δφ 1 = -π at center, it comes out that -π = - (π/λ)* L(n 1) or (L/λ)*(n 1) = 1 So, and one gets L = λ / (n-1) = 600 / (1.5-1) = 600 / 0.5 = 100nm L = 1.*10-6 m = 1.μm 9