Interference and Diffraction

Transcription

1 0_Interference.nb Chapter 0 Interference and Diffraction 0. Interference 0.. Two Sit Interference Pattern We have aready seen exampes of the interference between two waves. Given two counter propagating waves of the same frequency and waveength on a string, there exist fixed position that have zero ampitudes for a time. y x, t = A Sin kx-wt + A Sin kx+wt = A Cos w t Sin kx (0.) antinodes nodes The reative phase of the two waves at the node positions wi odd mutipes of p (80 out of phase) whereas the positions of the antinodes correspond to reative phases that are mutipes of p (0 phase shift). If we now consider two sources of waves embedded in or 3 dimensions, interference can ony occur if the two waves are coherent; the reative phase between the waves of the two source must be fixed for a time. An easy way to accompish this is to have two sits in a screen that is iuminated by a singe pane wave source.

2 0_Interference.nb 3 time d 7.93 Consider two rays that reach a point on a distant screen from the two sources

3 4 0_Interference.nb C r r d a J r b D To determine the reative phase between the two rays that reach the screen at point C, one can use the aw of Cosines to determine the distance that each ray has traveed. Noting that Cos a = Cos 90 -J = Sin J and Cos b = Cos 90 +J =-Sin J r = r + d 4 - d r Cos a = r + d 4 - d r Sin J (0.) r = r + d 4 - d r Cos b = d 4 + r + d r Sin J For D p d, the above expressions can be expanded in a Tayor series in d r = r - Sin J d + r - r Sin J d + O d 3 8 r r = r + Sin J d + r - r Sin J d + O d 3 8 r (0.3)

4 0_Interference.nb 5 To owest order in d, the difference between the paths is given by Eprint r, eq, HodForm r r,eq, d Sin Dr = r - r = d Sin J (0.4) If the two waves are in phase at the sources, then each wave wi accumuate a phase given by p times the number of waveengths contained within the corresponding path Df = p r Df = p r (0.5) so that the phase difference becomes Df = p r - r = p d Sin J (0.6) If Df = 0, p, 4 p.. = m p, the two waves wi interfere constructivey and if Df= p, 3 p.. = m + p, then the two waves wi interfere destructivey. Therefore, one obtains Constructive Interference whenever d Sin J = m Destructive Interference whenever d Sin J = + m = + m The intensity at positions with constructive interference shoud be arge (or bright ) and at position of destructive interference shoud be zero (or dark). To see such effects, it is necessary that d shoud be comparabe to the waveength. For exampe If =5000 Þ and d = mm, then the first non-centra peak (or maximum) corresponding to m = occurs at Sin J = d = 5000 Þ mm = 5μ0-7 m = μ0-6 m or J= p or 30 6 (0.7) If the screen is at a distance of meter, then the distance between the center of this peak and the centra peak is Dx = D Tan J = m Tan p 6 = m (0.8) If however, =5000 Þ and d = 0.5 mm, then the first non-centra peak (or maximum) corresponding to m = occurs at Sin J = d = 5000 Þ 0.5 mm = 5μ0-7 m = 5μ0-4 m 000 or J=0.00 (0.9) and the distance between the center of this peak and the centra peak is

5 6 0_Interference.nb Dx = D Tan J = m Tan 000 = 0.00 m (0.0) In this exampe of a arge separation one woud not be abe to resove the separate peaks and the resut woud appear to be a singe spot of ight. 0.. Fermat s Principe One can improve the anaysis a bit by introducing a ens between the screen and the sources. Consider a point source of ight paced at the focus, F, of a converging ens. A B F F The rays from the point source are refracted by the first ens into parae rays. The second ens then focuses these rays at its focus, F. Athough ray A traves a arger distance in air than the ray B, ray B traves arger distance in the two enses. Because the propagation veocity is sower within the gass of the enses, it is conceivabe that the two rays actuay take the same time in going from F to F. This is a statement of Fermat s Principe Light traveing from one point to another wi foow a path such that when compared to other paths, the time required to transverse the path wi either be a minimum, a maximum or remain the same. In the above figure, a rays take the same amount of time to transverse their separate paths. If the waves in each ray eave F in phase, they wi arrive at F in phase. This enabes one to simpify the anaysis of the -sit interference pattern. By pacing a ens in front of the sits with the screen at the foca pane of the ens, one need ony consider parae rays that eave the sits. These rays wi a be focused at the same point on the screen. Moreover, from our previous discussion, we see that the two rays wi accumuate the same phase from a point perpendicuar to the rays.

6 0_Interference.nb 7 C A d J B J The two rays accumuate the same phase starting from the points A and B to the focus at C. The overa phase difference between two rays is due to the difference in the path ength which is just d Sin J Intensity of the Doube Sit Pattern To determine the variation of the intensity on the screen as a function of J, the sum of the eectric fieds from the two rays. For convenience, assume that the eectric fied points perpendicuar to the paper (z-axis). The eectric fieds eave the sits in phase. As they trave to the screen, they accumuate a phase reated to the tota distance traveed.

7 8 0_Interference.nb C r E r d E The sum of the eectric fieds is E tota = E + E = E 0 Cos kr -wt z` + E 0 Cos kr -wt z` = E 0 Cos k r + r -wt Cos k r - r z` (0.) Corresponding to this eectric fied, the magnetic fied is B tota = E 0 c Cos k r + r -wt Cos k r - r ỳ (0.) (Note: because the spacing between the sits is sma and the screen is a arge distance away, we can assume that both magnetic fieds point in the same direction.) The intensity at the screen is determined by the time average of the Poynting vector E μ B I J = S = = 4 E 0 Cos m 0 c m 0 k r + r -wt Cos k r - r = E 0 c m 0 Cos k r - r (0.3) but r - r = d Sin J and k = p so that I J = E 0 c m 0 Cos p d Sin J (0.4) At J =0, the intensity is I J =0 = E 0. The genera form for the intensity as a function of J is m 0 c I J =I 0 Cos p d Sin J (0.5)

8 0_Interference.nb 9 Potting the intensity as a function of J and x = Tan J we obtain D d d =.3 I 0 d = p - p 4 J p 4 p I x D 3 0. Interference Due To Differences in Optica Path Length In some probems, the interfering rays trave through media of different indices of refraction. Consider the two rays shown. Over the physica ength L, the two rays trave through media of indices n and n.

9 0 0_Interference.nb n n Athough the two waves enter with the same phase, they accumuate different phases because now within the ength L, the two waves have different waveengths. The phase difference is given by Df = L p - L p (0.6) where = v n = v c c n = vac and = v n n = v c c n = vac, i.e. the waveengths inside the materia is n smaer. Using these definitions, the phase difference is Df = L p n - L p n = p Ln - Ln vac vac vac (0.7) The product of the physica ength, L, and the appropriate index of refraction is caed the optica path ength. As before If Df = m p m = 0 or Ln - Ln = m vac then the two waves wi interfere constructivey

10 0_Interference.nb If Df = + m p m = 0 or Ln - Ln = + m vac then the two waves wi interfere destructivey 0.. Soap Bubbes Consider the interference between two rays refecting from the front and back interfaces of a thin fim. A B st nd d The rays are drawn at an ange (away from norma incidence) for carity. At the first surface, the incident ray is both refected and refracted. The transmitted ray is aso partiay refected at the second surface. The ray refected from the second surface wi the be refracted through the first surface where it combines with the origina refected ray, A. (In principe, there wi be additiona refections as each ray encounters an interface.) Ray B wi no onger be in phase with ray A because it has traveed a different distance. The phase difference shoud be Df = p vac dn vac - 0 (0.8) because ray B traves a distance d in a medium with an index of refraction, n, whereas ray A does not. Based on this cacuation, one wod expect that as d -> 0, Df -> 0 and there woud be no phase shift between the two rays. Consequenty, the two rays interference constructivey and one woud expect to aways see a strong refection from very thin fims such as found in a soap bubbe. Experimentay, soap bubbe become transparent as the soap fim becomes thinner. To resove this probem, one must take into account the discrete phase shift associated with a refection. From the anaysis of a puse on a string, we found that the phase shift upon refection depended on the reative veocity of propagation.

11 0_Interference.nb c > c c < c If a wave goes from a medium with a high propagation speed to a medium with a ow propagation speed, there wi be a p phase shift. If the wave goes from a medium with a ow propagation speed to one with a high propagation speed then the refected waves have no phase shift. fast -> sow sow -> fast p phase shift upon refection 0 phase shift upon refection In either case, the transmitted wave has no phase shift at the interface. Returning to the previous diagram, if the index of refraction changes from n = ton >, ray A picks up a phase shift of p upon refecting from the first interface.

12 0_Interference.nb 3 A B st p v = c d v = c nd n < c v = c Ray B does not pick up a phase shift upon refection because now the veocity of ight goes from sow to fast at the second interface. The overa phase shift between the two rays is then Df = p vac dn vac - 0 -p = 4 dnp vac -p (0.9) If 4 dnp -p = p m m = 0 or d = + m vac 4 n vac then the two waves wi interfere constructivey If 4 dnp -p = + m p m = 0 or d = + m vac n vac then the two waves wi interfere destructivey Therefore as d -> 0, the reative phase shift approaches -p and the two rays interfere destructivey and not refection is possibe. This expains the back specks that appear on a soap bubbe just before it bursts. Consider =530 Þ and n =.33 water. Constructive interference or bright bands wi occur at d = + m 4 n vac m = 0 d = 4 n vac = Þ m = d = 3 4 n vac = Þ m = d = 5 4 n vac = Þ (0.0) (0.)

13 4 0_Interference.nb Destructive interference or dark bands wi occur at d = + m n vac (0.) m = 0 d = n vac = Þ m = d = n vac = Þ (0.3) m = d = 3 n vac = 60.8 Þ As the fim thickens from the d = 0 band, the first coor shoud correspond to the smaest visibe waveength, i.e. bue. As the fim thickens, bue wi have a dark band whereas red wi have a bright band. When observed with white ight, each waveength wi have its own bright and dark bands at different thicknesses. Consequenty, the fim wi appear to have numerous bands of coor. The 93 CIE coour matching functions and standard iuminant data in the format: waveength nm,x bar,y bar,z bar, D65 standard iuminant ciedata 380, , , , , 385, 0.004, , , 5.380, 390, , 0.000, , , 395, , 0.000, 0.036, , 400, 0.043, , , , 405, 0.039, , 0.00, , 40, , 0.00, , , 45, , 0.008, , , 40, , , , , 45, 0.477, ,.03905, , 430, , 0.060,.38560, , 435, , ,.696, , 440, , ,.74706, , 445, , ,.7860, , 450, , ,.77, , 455, , ,.7440, , 460, , ,.6690, 7.800, 465, 0.50, ,.580, , 470, , ,.8764, , 475, 0.40, 0.60,.0490, ,

14 480, , 0.390, 0.895, , 485, , , 0.660,.36700, 490, 0.030, 0.080, , , 495, , , , , 500, , , 0.700, , 505, , , 0.30, , 50, , , 0.580, , 55, 0.090, , 0.70, , 50, , , , , 55, , , , , 530, , , 0.046, , 535, 0.575, , , , 540, , , , , 545, , , , , 550, , , , , 555, 0.505,.00000, , , 560, , , , , 565, , , , , 570, 0.760, , 0.000, , 575, , , , , 580, , , , , 585, , , , , 590,.0630, , 0.000, , 595,.05670, , , , 600,.060, , , , 605,.04560, , , , 60,.0060, , , , 65, , 0.440, , , 60, , , , , 65, , 0.300, , , 630, , , , , 635, , 0.700, , , 640, , , , , 645, , 0.380, , , 650, , , , , 655, 0.870, , , , 660, , , , , 665, 0.0, , , 8.460, 670, , , , , 675, , 0.030, , , 680, , , , , 0_Interference.nb 5

15 6 0_Interference.nb 685, , 0.09, , , 690, 0.070, 0.008, , , 695, , , , , 700, 0.036, , , , 705, 0.008, , , , 70, , , , , 75, 0.004, , , , 70, , , , , 75, , , , , 730, , , , , 735, , , , , 740, , , , , 745, , , , , 750, , 0.000, , , 755, , , , , 760, , , , , 765, 0.000, , , , 770, , , , , 775, , , , , 780, , , , ; Transmission refection function describing interference phenomena Transmission retardation_, _ : Cos Pi retardation ; noninear rotation matrix to convert CIE XYZ cooords to RGB inear M 3.406,.537, , ,.8758,.045,.0557,.04,.057 ; ca Tota ciedata A, ciedata A, 5, Tota ciedata A, 3 ciedata A, 5, Tota ciedata A, 4 ciedata A, 5 ; Gamma correction set at. standard computer monitor gc x_ : Cip If x ,.9 x, x ^.4, 0, ; definition for the vaue of a coour channe I channe_bar Iuminant d chan n_, ret_ : Tota Transmission ret, ciedata A, ciedata A, n ciedata A, 5 ca n ; XYZ ret_ : chan, ret, chan 3, ret, chan 4, ret Fprint Rotate Pot x 50, x 50, x, 0, 000, Frame True,

16 0_Interference.nb 7 RotateLabe True, PotRange 50, 000, 40, 50, FrameTicks None, Axes Fase, Fase, AspectRatio.30, Background Back, CoorFunction Function x, y, RGBCoor Map gc, M.XYZ 000 x, Fiing Axis, ImageSize 400, 0, Epiog White, Arrowheads 0.0, Arrow 0, 50, 0, 0, Stye Text d 0, 0, 30,,, 0,,, Arrow 500, 50, 500, 0, Stye Text d 000 fi, 50, 40,,, 0,,, Arrow 000, 50, 000, 0, Stye Text d 000 fi, 00, 40,,, 0,,, Arrow 500, 50, 500, 30, Stye Text d 3000 fi, 50, 40,,, 0,,, Pi

17 8 0_Interference.nb 0.. Non-Refecting Coatings On some optica eements, a coating is added to suppress refections. Consider a coating of MgF on gass. As before we can ook at the interference between the refections from the first and second surfaces. A B st Air n air = p nd d p MgF n =.38 Gass n gass =.5 Upon refection ray A has a p phase shift because the veocity of ight in air is faster than that in the MgF ayer. Simiary ray B wi aso obtain a p phase shift because the veocity of ight is faster in the MgF ayer than in the gass. The overa phase shift is then Df = Df B -Df A = p dn+p -p vac = 4 dnp vac (0.4) In order for there to be no refections, rays A and B must interfere destructivey. The phase difference is therefore Df = + m p (0.5) or 4 dnp = + m p vac (0.6) or d = + m vac 4 n (0.7) Therefore the thinnest coating wi correspond to m = 0 or d min = vac 4 n.

18 0_Interference.nb Mutipe Sit Interference Patterns 0.3. Intensity of Mutipe Sit Pattern Consider the interference for mutipe sits. As before we introduce a ens between the sits and the screen to simpify the anaysis. C d d d d E E E 3 E 4 E 5 J J Assuming that each wave starts at the same phase, the phase of each wave at the point C is determined by the tota ength that it propagates from the sits to the screen. Noting that each successive wave traves an additiona distance given by DL = d Sin J, the eectric fied from each sit can be expressed as E E E 3 E 4 E N = E 0 Cos kr-wt = Re E 0 Â k r-w t = Re E 0 Â k r+dl -w t = Re E 0 Â k r+ DL -w t = Re E 0 Â k r+3 DL -w t ª = Re E 0 Â k N- DL -w t (0.8) The tota eectric fied at the point C is then E tota = Re E 0 Â kr-w t + E 0 Â k r+dl -w t + E 0 Â k N- DL -w t = Re E 0 Â kr-w t + Â k DL + Â k N- DL (0.9)

19 0 0_Interference.nb This finite series can easiy be summed. Taking a finite geometric series S = + x + x + x 3 + x -+N S x = x + x + x 3 + x N (0.30) Subtracting the second series from the first S - x = - x N (0.3) or S = - xn - x (0.3) Appying this resut to our series E tota = Re E 0 Â kr-w t - Â kndl - Â k DL = Re E 0 Â kr-w t+ k DL N- - Â kndl - Â kndl - Â k DL - Â k DL Sin NkDL = Re E 0 Â kr-w t+ k DL N- Sin k DL = E 0 Cos kr-wt + Sin NkDL k DL N - Sin k DL The corresponding magnetic fied is (0.33) B tota = E 0 c Cos kr-wt + N - k DL The intensity is therefore Sin NkDL Sin k DL (0.34) E μb I J = S = = E 0 Cos kr-wt + Sin N - k DL m 0 c m 0 NkDL Sin k DL = E 0 c m 0 Sin NkDL Sin k DL (0.35)

20 0_Interference.nb = E 0 c m 0 Sin N p d Sin J Sin p d Sin J where k = p has been substituted. This expression can be simpified by ooking at the J =0 imit. In the imit as J->0, both the numerator and denominator vanish. To evauate the expression, one can use L Hospita s rue or the Tayor expansion. For sma J, p d Sin J wi be sma and Substituting dnpsin J dnpsin J Sin Ø and Sin d p Sin J Ø d p Sin J (0.36) I Jã0 Ø E 0 c m 0 N (0.37) and I J = I 0 N Sin N p d Sin J Sin p d Sin J = I 0 N Sin N a Sin a (0.38) where a =pd Sin J 0.3. Specia Case, N = Substituting N = I J Nã = I 0 4 = I 0 4 Sin a Sin a Sin a Cos a Sin a =I 0 Cos a (0.39) which is just our previous resut for the doube sit Exporing N = 4 To understand the dependence of the intensity as a function of J consider the case, N = 4. I J Nã4 = I 0 6 Sin 4 a Sin a (0.40) At J =0 a =0, the intensity is a maximum with I J =0 = I 0. As J increases so does a, but the argument of the Sine function in the numerator increases more rapidy. Consequenty the numer-

21 0_Interference.nb ator wi vanish at a =p 4 whereas the denominator wi have a non-zero vaue. The numerator wi aso vanish at p and 3 p 4. When a =p, both numerator and denominator wi vanish. To reduce the expression, use L Hospita s rue im a->4 Sin 4 a Sin a = im a->4 Sin 4 a a = im Sin a a->4 a 4 Cos 4 a Cos a = 4 (0.4) Therefore at a =p, I J =p = I 0 i.e. another maximum. As a function of a, one has zeros at a = p 4, p, 3 p 4, 5 p 4 and maxima at a =0, p, p, 3 p, 4 p Given that there are 3 minima between a =0 and p, there shoud be additiona maxima in between the minima as we. However, they wi not be as arge as the principe maxima because athough Sin 4 a = and Sin a <, there is the overa factor of /6. For exampe at a=3 p 8 I p 4 = I 0 Nã4 6 Sin 4 3 p 8 Sin 3 p 8 = I 0. (0.4) Potting the intensity as a function of J, Sin J and a=pd Sin J

22 0_Interference.nb 3 d d = 4. I 0 d = p - p 4 J p 4 p I Sin J Resoution Note that the condition for a principa maximum is the same as that for a doube sit interference pattern, d Sin J = m. In genera a N-sit interference patterns have their principe maxima in the same position. As N increase the principe maxima become sharper as more secondary maxima are introduced.

23 4 0_Interference.nb N N = I 0-4 p -3 p - p -p p p 3 p 4 p a=pdsin J N = I 0-4 p -3 p - p -p p p 3 p 4 p a=pdsin J The widths of the principe peaks is reated to the distance between the maximum position and the position of the subsequent zero. The ocation of the m th order peaks is a =m p or N a=nmp. The next minimum occurs when N a increments by p, i.e. a =NMp+p. Expressing a=pdsin J and soving for Sin J N a peak = dnpsin J peak = mnp or Sin J peak = m d (0.43) Simiary N a min = dnpsin J min =p+mnp or Sin J min = m d + For arge N, one woud expect that J min =J peak +DJ where DJ is sma. J min =J peak +DJ into the above expression and expanding dn Sin DJ+J peak = Sin J peak + Cos J peak DJ + O DJ = m d + dn (0.44) Substituting (0.45) Substituting Sin J peak = m d, and soving for DJ

24 0_Interference.nb 5 DJ max-min = N d Cos J peak (0.46) Note: the true haf width of the principe maxima is determined by soving for DJ at which I DJ+J = I 0 N Sin N a Sin a = I 0 (0.47) Soving for DJ 3 6 DJ FWHM = Ø d - + N dnpcos J p Cos J (0.48) in the imit of arge N. The essentia dependence is the same. The width of the peaks decreases as N. For increasing N, the principa maxima become more we defined. Such arge N mutipe sits are ca gratings and are typicay used to disperse ight instead of prisms. To see how a grating can separate different coors, one can determine the anguar separation of two principa maxima generated by two waveengths. The anguar positions of the maxima are determined from d Sin J = m. Noting that DJ = J D, we have that and so d Sin J = d Cos J J = m (0.49) DJ max-max =D J = m d Cos J peak D (0.50) The dispersion of the grating increases with the order of the peak, m. Of course as we sha see, the intensity of the higher order maxima wi aso be smaer. One criterion for resoving two neighboring peaks associated with two different waveengths, is to require that the separation between the peaks be arger than the separation between a peak and the first minimum. DJ max-max DJ max-min

25 6 0_Interference.nb DJ max-max m D d Cos J peak > DJ max-min > N d Cos J peak (0.5) or D > mn (0.5) This factor of N m is often referred to as the resoving power of the grating. To see this in practice, one can pot the sum of the diffraction pattern for two different waveengths. Ony if the peaks from the two waveengths are separated at a distance greater than the distance from a principe peak and the first minimum, can one distinguish that there are in fact two peaks. N 8 N = I 0-4 p -3 p - p -p p p 3 p 4 p a=pdsin J N = 8 I 0-4 p -3 p - p -p p p 3 p 4 p a=pdsin J 0.4 Singe Sit To determine the intensity from a singe sit, one can artificiay subdivide the sit into N sits and sum the eectric fied from the individua sits.

26 0_Interference.nb 7 To determine the intensity from a singe sit, one can artificiay subdivide the sit into N sits and add the eectric fieds from the individua sits. sits Tabe Line 0, n 0.4, 0, n , n,,0 ; rays Fatten Tabe Line 0, n , 3, 0.5 n , Line 3.0, 0.5 n , 0, 5, n, 0,9 ; Fprint Graphics Thick, Line 0, 4, 0, 5, Line 0, 4, 0,, sits, Line 0, 0.0, 0, 5, Arrowheads 0.0, 0.0, Arrow.0,,.0,, Circe 3, 0, 7, Pi 6, Pi 6, Circe 9., 0, 7, Pi Pi 6, Pi Pi 6, Bue, Stye Text TDF a,.6, 0, 4, Dashed, Line 0,, 0,, Stye Text TDF,.978,.57, 4, Thin, Red, rays, Stye Text TDF C, 0.5, 5, 4

27 8 0_Interference.nb C a J From the anaysis of the N sit interference pattern, the intensity was found to be I J = E 0 c m 0 Sin p N DL Sin pdl (0.53) For the singe sit, we have that DL = a Sin J. The intensity from each section of the singe sit N shoud aso be reduced from E 0 to E 0 N to insure that the tota ight from the singe sit is fixed. E 0 I J = c N m 0 Sin p a Sin J Sin p a Sin J N (0.54) Taking the imit as N ->, the Sine function in the numerator can be approximated by its argument E 0 I J = c N m 0 Sin p a Sin J = p a Sin J N E 0 c m 0 Sin p a Sin J = p a Sin J E 0 Sin b c m 0 b (0.55) where b = a p Sin J (0.56) The intensity at J =0orb->0 is just

28 0_Interference.nb 9 I 0 = E 0 c m 0 (0.57) and so I J =I 0 Sin b b (0.58) Potting the intensity as a function of J and Sin J we obtain a a = 0.5 I 0 a =. - p - p 4 a = 0.5 I 0 J p 4 a =. p - - Sin J The minima on either side occurs at b = mp or a p Sin J = p or Sin J min = m a (0.59) Note the simiarity of this formua with the formua for the maxima of a mutisit, ony now the formua predicts the position of minima! As the sit size approaches, the first minimum approaches J =p and the centra maximum becomes broader and broader. Therefore for a <<, the centra maximum iuminates the entire screen with a uniform intensity. This was impicity assumed when we cacuated the intensity of a mutisit. The ight from each sit coud iuminate the screen at a possibe anges.

29 30 0_Interference.nb If however, a >, then the singe sit patter wi moduate the N sit pattern. Moreover, some of the principe maxima of the N sit pattern may be missing because they occur at anges at which the singe sit pattern has a minimum. Consider 4 sits with a spacing d = 4 in which the width of the sits is a =. a = I 0 a = a = - 4 I 0 Sin J 4 = a a = - 4 I 0 Sin J 4 = a 3 4 missing principe maximum missing principe maximum Sin J In the above exampe the principe maxima of the mutisit pattern occur at Sin J max = m d = 0, 4,, 3 4, (0.60) There are 9 maxima (one coud argue that there are ony 7 maxima because at Sin J max = the maxima are not perfecty deveoped. The minima of the singe sit occur at Sin J min = m a =, (0.6) Consequenty, the principe maxima at Sin J min = and are eiminated because no ight

30 0_Interference.nb 3 from the sits are directed in these directions. For arbitrary vaues of N, a and d, the overa pattern wi be more compex. a d N a =.35 I 0 a = d =.96-4 I 0 Sin J 4 d = d = I 0 Sin J 4 a = Sin J Taking the width of the centra maximum of the singe sit pattern to be the width at haf maximum, one needs to find the vaue of b such that Sin b b = (0.6)

31 3 0_Interference.nb The vaue of b must be soved numericay. To get an estimate, the above equation can be reduced to Sin b =b. Ceary one soution of this equation is when b is sma. Taking the Tayor expansion, Sin b ºb- b3 6 + O b 4 º b or b- b - b3 6 = 0 (0.63) Soving for b - 6 b b = 0 (0.64) or b Ø 0 = 0. b Ø- 3 - = (0.65) b Ø 3 - =.3565 The exact numerica resut is b = a p Sin J maximum is therefore DJ width = ArcSin a If the sit size is arge compared to, then a =.3956 or Sin J = The width at haf a 0.5 Circuar Hoe For a circuar hoe, the first minimum next to the centra maximum is given by Sin J min =. D (0.66) where D is the diameter of the aperture. If one uses the same criterion as before then the image of two point sources can be resoved if their images have an anguar separation that is greater than. D.

32 0_Interference.nb 33 Dx point sources Dx Dy D 0 D i For distance objects the anguar separation of the images is equa to the anguar separation of the two objects DJ sep = Dx =DJ image = Dy D 0 D i (0.67) Therefore to resove the two images as separate images, Dx D 0 >. D. Note the dependence on the waveength. In order to improve the resoution of a microscope, in addition to making the enses arger, one coud aso use shorter waveengths. For teescopes, one can not change the waveengths of the incoming ight so one must make the enses or mirrors arger. The same criterion appies to radio teescopes. In New Mexico, there is the VLA (Very Large Array) and now it is possibe to ink teescopes across the word to form a baseine (or aperture) of 5000 km. One can estimate, the resoving power of a human eye. The pupi is about D~0.5 cm. Because the diffraction is occurring inside the eye, one needs to incude the index of refraction of the vitreous humor (assume the index is that of water, n~.33). Choosing =5000 Þ Dx > Þ = D 0 Dn.33 μ 0.5 cm = 9.7μ0-5 (0.68) The headights on a car are about.4 meters apart and therefore the distance at which one coud sti resove the two headights becomes Dx D 0 < = m = ft μ0 (0.69) So in principe, one shoud be abe to resove the two headights at the norma cruising atitude of an airpane ~ ft. Unfortunatey, the size of a singe rod is ony about mm and has an anguar size of DJ = mm cm = 5 μ 0-5. Consequenty, the eye may sti be unabe to resove the two headights because there are not enough ces to image the separate signas.

33 34 0_Interference.nb Names "Goba` " nb EvauationNotebook FrontEndTokenExecute nb, "SeectA" FrontEndTokenExecute nb, "SeectionCoseAGroups" SeectionMove nb, Next, Ce a, A, air, A, antinodes, arc, arc$, as, as$385, as$385$$, as$46, as$46$$, as$6357, as$6357$$, as$6434, as$6434$$, as$839$$, as$949$$, as$$, a$, b, B, B, B, B$534, B$6449, B3, B30, B30$406, B30$893, B3, B3$406, B3$893, Beff, BitDepth, BS0, BS0$406, BS0$893, BS, BS$406, BS$3979, BS$893, BS$9064, BS, BS$049, BS$534, BS$5344, BS$6449, BS3, BS3$335, BS3$840, c, cm, d, dd, dd$, dia, dr, dr$, ds, ds$345, ds$345$$, ds$540, ds$540$$, ds$6398, ds$6398$$, ds$6593, ds$6593$$, ds$386, ds$386$$, ds$463, ds$463$$, ds$54, ds$54$$, ds$4687$$, ds$579$$, ds$6358, ds$6358$$, ds$6435, ds$6435$$, ds$654, ds$654$$, ds$830$$, ds$950$$, ds$$, dth, d$5, d$5$$, d$433$$, d$5956, d$5956$$, d$6309, d$6309$$, d$$, eff, Energy, eps, f, F, ff, Fux, form, form$5384, form$5384$$, form$3099, form$3099$$, form$$, ft, FuScreenArea, FWHM, g, GAFont, GFDInt, GF3DInt, GFCInt, GFCross, GFCur, GFDiv, GFLim, h, Hz, Goba`I, ie, ii, inc, inches, inc$60, inc$60$$, inc$5385, inc$5385$$, inc$756, inc$756$$, inc$30300, inc$30300$$, inc$$, Int, Int$, Goba`I$, k, kg, L, eft, ens, iq, m, max, min, mix, mm, m$, n, n, n$, n$398, n$398$$, n$6450, n$6450$$, n$6778$$, n$$, n, n$, n$399, n$399$$, n$645, n$645$$, n$6779$$, n$$, nb, nbs, nbstye, nn, nn$5493, nn$5493$$, nn$5534, nn$5534$$, nn$5588, nn$5588$$, nn$59, nn$59$$, nn$6644, nn$6644$$, nn$30$$, nn$464, nn$464$$, nn$30400, nn$30400$$, nn$3044, nn$3044$$, nn$3049, nn$3049$$, nn$6436, nn$6436$$, nn$83$$, nn$$, nodes, notes, Nt, n$, n$496, n$496$$, n$3978, n$3978$$, n$494, n$494$$, n$64, n$64$$, n$8893, n$8893$$, n$9379, n$9379$$, n$$, obs, P, Pascas, peak, ph, ph, phi, pabe, pabe$, pane, pane$644, pane$644$$, pane$660, pane$660$$, pane$$, pt, pts, pyg, pyg$, pos, r, R, range, range$,

34 0_Interference.nb 35 rays, Rc, Rc$, ref, Resoution, right, r$, s, S, ScreenArea, sec, sits, stye, t, T, th, th, th, time, tota, trans, tt, tt$, t$, t$5, t$5$$, t$60, t$60$$, t$397, t$397$$, t$43$$, t$5386, t$5386$$, t$5437, t$5437$$, t$568, t$568$$, t$5667, t$5667$$, t$5708, t$5708$$, t$5750, t$5750$$, t$5793, t$5793$$, t$5955, t$5955$$, t$6308, t$6308$$, t$6449, t$6449$$, t$6777$$, t$598, t$598$$, t$643, t$643$$, t$3059, t$3059$$, t$3097, t$3097$$, t$373, t$373$$, t$757, t$757$$, t$3030, t$3030$$, t$3035, t$3035$$, t$3053, t$3053$$, t$30570, t$30570$$, t$306, t$306$$, t$30653, t$30653$$, t$30696, t$30696$$, t$484$$, t$63$$, t$6570, t$6570$$, t$669, t$669$$, t$85, t$85$$, t$854, t$854$$, t$89, t$89$$, t$$, v, V, vac, vec, v$3096, v$3096$$, v$37, v$37$$, v$853, v$853$$, v$88, v$88$$, v$$, water, wind, x, xnode, xnode$, x$, y, Y, ynode, ynode$, y$, z, z$, E, P, B, E, k, r, S,,, L, P, P, r, t, V, x,,,,,,,,,, m,,,,,,, t, fi NotebookObject 0_Interference.nb $Aborted