APPLIED OPTICS. Opaque obstacle
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1 Lecture Notes A. La Rosa APPLIED OPTIS I. Diffraction by an edge II. Diffraction of light by a screen Reference: Feynman, "Lectures on Physics," Vol I, Addison Wesley. Section 30-6 and Section I. Diffraction by an edge Figure 1 shows light travelling from left to right, obstructed by an opaque. We want to evaluate the light intensity profile on a screen located at a finite distance from the, particularly around the geometrical projection of the edge. Incident field Fig. 1 Monochromatic light obstructed by an opaque object. The edge will produce diffraction of the light, to be observed on the screen. Screen 1. Help from math theorems To evaluate the fields at the screen, we will use a theorem in electromagnetism theory that the field at a given point, let s call it (P' ), can be calculated (after finding a proper Green function G) in terms of the fields existent at an arbitrarily given boundary S enclosing the point.
2 S ( P' ) ( x' ) S G x' da Surface Field at Field at the surface S Fig. 2 The field at a given point is calculated based on the values of the field at an arbitrarily chosen boundary surface S enclosing the point. hoosing a proper surface S (one that fits to your problem) For the particular case illustrated in Fig.1, we can choose an infinitely extended syrface (as shown in Fig.3). For any practical situation the fields would decay at infinity. Accordingly, when applying the surface integral (the one indicated in Fig.2 ), only the contributions from the side T will effectively contribute. T To infinity Surface S To infinity To infinity Screen
3 T S Screen Fig. 3 The contributions to the field at from from the surface sections located at infinity would be zero. Hence, the field at will depend only on the values of the field on the surface that passes through and T. We will solve the problem illustrated in Fig.1 by assuming that we are dealing with a two-dimensional case (to simplify the math). The field at an arbitrary point will be calculated by assuming that we have a set of effective sources uniformly distributed over the infinite line that passes through T, as shown in Fig.4. Fields produced by the Incident field T E D Fig. 4 Experimental setup equivalent to the one presented in Fig. 1, based on the theoretical analysis presented in figures 2 and 3. The field at will have contributions from the multiple synchronous sources located along the vertical line along the T direction.
4 To find the field at (located at a finite distance from the ), we need to calculate the contribution from all the individual sources located along the line T (Fig. 4). Since the point is at a finite distance, it turns out (as we will see below) that the phase difference between the fields arriving to from two arbitrary sources (D and E for example) is not proportional to their separation distance (DE). (Such as proportionality between phase difference and the separation-distance between the sources would occur if the screen were at infinity). 2. alculation of the phase-difference between the sources As a reference point, let s take source D located oppsite to (see Fig. 4 and Fig. 5). Let s take another point E located a litle bit up (at a height h from D): Source E will produce a field at that has different phase and magnitude than the field produced by the source D. But, for sources close to D, the magnitude will not be affected as drastically as the phase. Hence, let s concentrate our effort for calculating only the phase difference between fields arriving from D and E. Such a phase difference is due to the difference in path length: =E-D We want to calculate in terms of h (Fig.5). E h D X Triangle EXD similar to h h EP' h 2 EP' triangle ED For a point E close to D: h << D then 2 2 h h (1) EP' 2 2 h DP' h 2 (2) DP' For a point E far away from D: h >> D then h (3) Fig. 5 Figure shows that the path length varies quadratically with h for points close to E; but linearly with h for point far away from D. Electric field at
5 To find the total field at the point on the screen (due to the incident fields coming from, D, E, T, etc.), i) we start with a phasor representing the field at due to the source located at ; subsequently, ii) we add phasors representing the contribution to the field at from the other sources (D, E, T, etc). Notice, according to expression (2) above, that the phase of the new phasors increases proportional to h 2 for points closer to. But for point far away from the phase changes linearly with h, according to expression (3). This analysis suggests the following graphical way to evaluate the total field phasor E at : Total field at point on the screen: This would be the path if the phase were to change linearly with the position distance of the sources E ( at ) T D This path correspond to the fact that the phase changes quadratically with h (i.e. for small h the phase practically does not change) screen Fig. 6 Total field phasor at point on the screen is indicated by the thick oriented segment in the figure. [Notice the figure is missing the small phasors that circle around in smaller and smaller circumferences. Please complete them on your own.] Total field at point B :
6 E ( at B ) B B B screen Fig. 7 Total field phasor at point B on the screen is indicated by the thick oriented segment in the figure. [Notice the figure is missing the small phasors that circle around in smaller and smaller circumferences. Please complete them on your own.] Total field at point A : E ( at A ) B A A Fig. 8 Total field phasor at point A on the screen is indicated by the thick oriented segment in the figure. [Notice the figure is missing the small phasors that circle around in smaller and smaller circumferences. Please complete them on your own.] A
7 This analysis leads to the following intensity profile. Here I o is the intensity of the light at one point on the screen when no is present. The intensity is proportional to 2 E. D A D R Fig. 9 Profile of the light intensity of the screen (see also Fig. 4 above). II. Diffraction of light by a screen Field at the other side of an opaque screen that has an aperture Metallic wall P E =E Source E =E Source + E wall Fig. 10 Light incident on a metallic wall that has an aperture at the center. Objective: In this section, we illustrate that the settings displayed in Fig. 10 and Fig. 13 are equivalent when the observation point P is away from the rims of the aperture Light incident on the metal place the charges in oscillation, which re-emit light also reaching the point P. The total field at P has contributions from the source and from the radiators on the wall.
8 E (at P) =E source + E wall (1) Hence, the setting in Fig. 10 could be depicted a bit more accurate by the one in Fig. 11. I general, a peculiar accumulation of charges occur around the edges of the aperture. Hence, the exact field at points very near those borders may be a bit complicated (actually it is a lot complicated). E wall is a bit complicated to calculate So let s consider a point P away from the aperture. For points P located away from those borders these peculiar charge distributions around the rim of the aperture may have no net significant effect. Metallic wall P E =E Source E (at P) =E Source + E wall Fig. 11 Light incident on a metallic wall that has an aperture at the center. The incident like drives fee electrons on the metal, which produce an additional field at point P. An opaque screen Now it comes a trick to evaluate E wall is in an approximate way. Let s block the aperture with a metal plug (similar material to that of the wall). The total field at P will then be equal to zero. But, still charges on the metal are being driving by the incident light from the source. It must occur, then, that the field from the wall E wall plus the field from the plug E plug cancel out; E Source + E wall + E plug = 0 (2) Wall Plug P E =E Source E =E Source + E wall + E plug = 0
9 Fig. 12 Light incident on a metallic wall whose aperture has been covered with a metallic plug. Notice, by placing a perfect plug, the peculiar charge distribution around the wall aperture may have disappeared (a new charge distribution, a more uniform one, will occur). It is for this reason that we write E wall instead E wall. For points P located away from those borders these peculiar distributions of charges around the rim of the aperture may have no net significant effect. We will make such an approximation at the end. Subtracting (2) from (1), where E wall E (at P) = - E wall - E plug + E wall = (E wall - E wall ) - E plug (3) correspond to the original field that takes into account the complicated charge distribution around the aperture rim E wall correspond to a uniform charge distribution on the rim (because of the presence of the metallic plug) E wall correspond to a uniform charge distribution on the plug. As pointed out above, in general E wall E wall. But for a point P located away from those borders these peculiar distributions of charges around the rim of the aperture may have no net significant effect.) For such a point, E wall ~ E wall. Expression (3) then becomes, E (at P) = - E plug for points P away from the borders (4) of the aperture Plug P E (at P) =- E plug Fig. 13 Experimental setting equivalent to the one depicted in Fig. 11 when evaluating the field at points P away from the borders of the aperture.
10 I short, the setting in Fig. 11 shows that the field at P is the superposition of fields produced by the light source plus the field from the charges in the metallic screen. We have shown that, an equivalent way to calculate the filed at P is to consider fictitious sources located inside the aperture (except for a minus sign). uriously, we place sources right where Fig 11 indicate that in fact there is none. References: Feynman, "Lectures on Physics," Vol I, Addison Wesley. - Section 30-6 Diffraction by opaque screens. - Section 31-6 Diffraction of light by a screen.
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