25 The vibration spiral

Transcription

1 25 The vibration spiral Contents 25.1 The vibration spiral Zone Plates Circular obstacle and Poisson spot Keywords: Fresnel Diffraction, vibration spiral. Ref: Jenkins and White: Optics; E. Hecht: Optics; M. Born and E. Wolf: Principles of Optics; I.E. Irodov: Problems in General Physics The vibration spiral The sum of amplitudes for infinite Fresnel zones discussed in the previous chapter is shown the following fig A 2 A 4 A 6 A 8 2A A A 1 A 3 A 5 A 7 A 9 Fig. 25.1: Sum of amplitudes from different Fresnel zones Now the division with respect to half period zones is actually too coarse. In

2 2 25 The vibration spiral reality the phase will change continuously as the path difference changes. To have a better gradation we now divide each half period zone in two quarter period zones, so that the path difference between the two consecutive zones is λ/4 or the phase difference is π/2. The situation is shown in the fig The blue and red arrows are the phasors (showing amplitudes as well Fig. 25.2: Sum of amplitudes from different Fresnel zones as the phase i.e shown by the direction of the arrow) from the first two Fresnel zones. The relative π phase is shown by inverting the direction of the arrow. Now each of these two zones are divided into two quarter period zones, the respective contributions from the first four quarter period zones are shown by the black arrows. Again note that the amplitude of each quarter period zone is almost the same and it is reduced by a factor of 2 from that of the individual half period zones and the relative phase difference between the two consecutive quarter period zones is now π/2. As we move to higher zones the same pattern is repeated and we get spiral of square

3 25.1 The vibration spiral 3 shape. Now each of the quarter period zones is further divided into two λ/8 45 o 45 o π/8 45 o π/8 π/8 Fig. 25.3: Sum of amplitudes from different Fresnel zones period zones. This situation is depicted in the left diagram of the fig. 25.3, where black phasors are contributions of consecutive quarter period zones and cyan phasors are correspond to successive λ/8 period zones. In this case the relative phase difference between two consecutive phasors is π/4. Next, each λ/8 period zones are divided into two λ/16 period zones and which results in the right diagram of the fig The λ/16 contributions are shown by the magenta phasors, which are turning by an angle of π/8 successively. When this processes repeated several times one reaches at a smooth curve known as vibration spiral, which is shown in the fig Now we describe how to read phasors from the vibration spiral. The lower starting point corresponds to the centre of the circular aperture around which the Fresnel half period zones are drawn concentrically. The radial progress in the aperture corresponds to

4 4 25 The vibration spiral st 1 FZ st (4/3) 1 FZ 2A st (1/2) 1 FZ A Fig. 25.4: The vibration spiral a progress along the spiral in this curve. Suppose we have an aperture which allows only the first Fresnel zone. Then we start from the bottom of the curve and stop at the uppermost side of the curve after covering the semicircle. We draw an arrow from the start to the end (shown by the blue arrow of length 2A) and this is the contribution of the first Fresnel zone. If there were no aperture then we start at the bottom and move along the spiral and finish at the centre of the spiral covering infinite Fresnel half period zones and get the contribution A, the amplitude of the original plane wave. Similarly if the aperture ends at half of the first Fresnel zone one has a phasor tilted at angle π/4 from the vertical with a length equal to 2A (shown by a brown arrow). An aperture allowing four thirds of the first Fresnel zone will contribute the phasor shown by the green arrow. We now illustrate the procedure with a few examples. Example 1: Find the intensity at P for the following two cases shown in

5 25.1 The vibration spiral 5 P (a) P (b) Fig. 25.5: Example 1 the figure 25.5, if a plane wave of intensity I is falling on these apertures normally from the other side of the observation point. For both the cases the shaded region extends upto the infinity. In the second case the circular part at the corner corresponds to the boundary of the first Fresnel zone for the observation point P. Solution: In the first case one fourth part is close with respect to P and three fourth part is open. So if the incident amplitude is A then the phasor contribution is 3A/4 and hence the intensity will be 9I/16. In the second case in addition to the previous situation one fourth of the first Fresnel zone is open. This would contribute an extra A/2 (since the full first Fresnel zone contribution is 2A) and this phasor is parallel to the previous one. So the total amplitude is 5A/4 and which make the intensity 25I/16 for this case. Example 2: FindtheintensityatOforwhichthezoneboundariesareshown

6 6 25 The vibration spiral Fig. 25.6: Example 2 in the fig The plane wave of intensity I is incident on this aperture normally from the other side. The area outside the boundary is opaque for the light. Fig. 25.7: Solution 2 Solution: We give the solution by couple of methods below. In the first method, fig 25.7, we divide the aperture in concentric circles covering different Fresnel zones and add the contribution from the individual zones. One sees

7 25.1 The vibration spiral 7 that the whole of the first Fresnel zone is present (shown by the blue circle) and that contributes to an amplitude 2A. Next we find only three fourths of the second Fresnel zone is avaliable, so it gives an amplitude (3/4) 2A = 3A/2 (negative sign shows a phase difference of π with respect to the first zone). Further, only half of the third zone and quarter of the fourth zones contribute amplitudes (1/2) 2A = A and (1/4) 2A = A/2 respectively. The total amplitude is equal to 2A (3A/2) + A (A/2) = A, giving an intensity at P equal to the original incident intensity, A 2 = I, of the plane wave. Fig. 25.8: Alternate solution 2 Alternate solution: In the second method we divide the whole aperture area in terms of four cake pieces as shown in the fig in fig In this case there are two quarter cake pieces which include odd number of Fresnel zones (the blue boundary has one zone and the green has first three zones). Since the whole contribution from odd number of zones is A for each of

8 8 25 The vibration spiral these quarter cakes we have an amplitude A/2 (refer to fig 24.2). On the other two quarter cake pieces compromise of even number of zones (the red boundary has two zones whereas the magenta has four). We know that even number of zones contribute almost zero amplitude so these two give negligible contribution and the total amplitude is coming form the other two parts, i.e. A/2+A/2 = A and we arrive at the same result as before. The whole idea of tackling these problems is to divide the total aperture area in convenient non overlapping regions for which the individual contributions could be found efficiently. The final result is obtained by summing these individual phasors vectorially. One more example may make the understanding better. Example 3: FindtheintensityatPforwhichthezoneboundariesareshown in the fig The plane wave of intensity I is incident on this aperture normally from the other side. The area outside the boundary is opaque for the light. Solutions are in the following figures which are are self explanatory. Only thing to note is that if you have a θ portion of some particular zone then you chop the the phasor by θ/2π factor. The resultant amplitude in this case is again the same as the incident amplitude. Or alternatively one can do it in the following way as depicted in fig There is no third phasor in this case since the contribution from area covering the first two zones is negligibly small.

9 25.1 The vibration spiral 9 P Fig. 25.9: Example 3 (2/3) 3 A (1/3) 3 A A 3A Fig : Solution 3 In the above problems we were mainly concerned with the point on the axis of the aperture. Since the aperture is circular the diffraction pattern on the screen consists of concentric bright and dark regions. Finding intensities on off centre points exactly is difficult but can be found approximately by the above method. We illustrate the procedure with the fig below. The portion of the each Fresnel zone captured by the aperture should be estimated and phasors should be drawn accordingly and summed.

10 10 25 The vibration spiral (1/3) 3A A (1/3) 3 A Fig : Alternate solution 3 O 2 Fig : Zones in the off centre points Zone Plates An application of Fresnel zones is the construction of zone plates. Zone plates are devices which work like a lens but they are flat. They can focus parallel light. They can be made on thin plates so they are light compared to lenses. They have multiple foci. A zone plate is made by hiding the alternate Fresnel zones as shown in the fig In this way we can add up the phasors to

11 25.1 The vibration spiral 11 6 a 2 a 5a a 2a 3 a I=36 I 0 Fig : Alternate solution 3 make a big amplitude. Suppose we draw twenty Fresnel zones and hide ten of them alternatively. We would have an amplitude 10 2A = 20A and we which corresponds to an intensity 400I, where I is the incident amplitude. So this focuses light at a point P along the axis of the zone plate for which the boundaries correspond to consecutive Fresnel zones. In the usual construction the the first radius is taken as a and consecutive concentric circles are drawn with the radii, 2a, 3a, 4a, 5a, 6a, etc.. The n th radius will be na. For visible light a is in sub-millimeters and for microwaves it is in tens of cms. Next the alternative zones are made opaque. The fig shows two zone plates with 6 zones in each. Now we would like to find the primary focal length f of a zone plate whose first zone has a radius equal to a. From equation (24.2), we have a = λf and hence f = a 2 /λ (25.1)

12 12 25 The vibration spiral where λ is the wavelength of the light used. Next consider a point on the axis of the zone plate for which the first radius corresponds to the boundary of the third Fresnel zone. If the distance of this point from the centre of the zone plate is f 2, then a = 3λf 2 and f 2 = a 2 /3λ = f/3. (25.2) In this case the second circle of radius, 2a = 6λf 2, will correspond to the sixth Fresnel zone for the point. The consecutive circles will correspond to 9 th, 12 th, etc. zones. Since in this either three consecutive zones are open or closed the contribution from each open or opaque is again 2A. Effectively it will act like the former case for the phasors from the alternate open areas will add up to give a large amplitude at this point which is at a distance of f/3 from the centre of the zone plate. Since at a time a region closes three consecutive zones in this case, many zones are covered compared to the previous one and the intensity is weaker at this focal point and hence it is called a secondary focus. Similarly at a point on the axis of the zone plate for which the radius a corresponds to the fifth Fresnel zone, we have a tertiary focus, f 3, which has a focal length of f/5. Similarly there are more foci with focal lengths f/7, f/9,...etc..

13 25.1 The vibration spiral Circular obstacle and Poisson spot In case of a circular obstacle one finds there is a bright spot behind the obstacle on the axis if a plane wave falls normally on the obstacle from one side. The intensity of this spot is equal to that of the incident light and it is known as the Poisson spot. This apparently looks counter intuitive but a careful look at the vibration spiral explains it easily. We refer to the fig Since the radius of the obstacle will correspond to arbitrary Fresnel zone for a point on the axis. One would start the phasor one that zone and complete it at the centre covering the infante Fresnel zones. In any case the the resultant phasor has a length A corresponding to the incident amplitude as shown in the fig Fig : Phasors for the Poisson spot Problem 1: A plane wave of intensity I is incident on the aperture as shown in fig The boundaries of the aperture for the observation point P are 1/2 of I st Fresnel zone and I st Fresnel zone. If the intensity at P is 8I/3, find

14 14 25 The vibration spiral the angle θ. Only the white part is open and the rest is opaque. I st FZ 1/2 P θ I st FZ P Fig. 3 Fig : Problem 1 and 2 Problem 2: A plane monochromatic light wave of intensity I and wavelength 0.6µm along the +z axis is normally incident on an aperture placed at z = 0 plane with its centre, P, at origin. The aperture, as shown in the second diagram of fig 25.15, has an opaque semicircle with radius (0.8/ 2) mm. The outer circle has radius 0.8 mm. Only the white area inside the circle is open and the rest outside is opaque. Find the intensities on the z axis at z = (1.6/3) m and z = 0.8 m from the origin.