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12 Nov 2019
Thin-film interference. A thin flat sheet of diamond (N = 2.42) is surrounded by air. A beam of monochromatic light (lambda = 588nm) is normally incident (theta = 0degree)b on the upper surface of the sheet, as shown in Figure 8.4. The light ray is partially reflected At the upper surface (that reflected ray not shown in the Figure). and there Are also multiple reflections in the sheet. (Figure 8 4 is schematic in that rays reflected off the bottom surface are drawn slanted to help in visualization. These rays are actually reflected perpendicular to the interface.) As are result, a series of transmitted rays emerges from the bottom surface of the sheet. Only the first two such rays (rays 1 and 2) are shown in the diagram. If the sheet is thin, these transmitted rays can interfere with each other. The Fresnel relations give the refection coefficient (for intensity) at normal incidence (theta = 0): where n = n2/n. For normal optical materials (that do not exhibit birefringence; the result for R| e=0 is independent of the direction of travel of the light through the interface, and also independent of the state of polarization of the incident ray The intensity transmission coefficient at the interface is given by Use (4.1) and (4.2) to evaluate R and T for the air/diamond interfaces in this problem (we will assume that diamond is not a birefringent material.) Use your results of (a) to find the intensities I 1 and I 2of rays 1 and 2, expressed as fractions of the incident intensity I 0 (assume no absorption of light in the sheet.) Find the minimum non-zero thickness d of the plate for which the electric field oscillations of rays 1 and 2 will be exactly in phase with each other. Find the resultant intensity (in terms of I0) resulting from the interference of rays 1 and 2, assuming the electric field oscillations of the rays are in phase with each other (constructive interference.)
Thin-film interference. A thin flat sheet of diamond (N = 2.42) is surrounded by air. A beam of monochromatic light (lambda = 588nm) is normally incident (theta = 0degree)b on the upper surface of the sheet, as shown in Figure 8.4. The light ray is partially reflected At the upper surface (that reflected ray not shown in the Figure). and there Are also multiple reflections in the sheet. (Figure 8 4 is schematic in that rays reflected off the bottom surface are drawn slanted to help in visualization. These rays are actually reflected perpendicular to the interface.) As are result, a series of transmitted rays emerges from the bottom surface of the sheet. Only the first two such rays (rays 1 and 2) are shown in the diagram. If the sheet is thin, these transmitted rays can interfere with each other. The Fresnel relations give the refection coefficient (for intensity) at normal incidence (theta = 0): where n = n2/n. For normal optical materials (that do not exhibit birefringence; the result for R| e=0 is independent of the direction of travel of the light through the interface, and also independent of the state of polarization of the incident ray The intensity transmission coefficient at the interface is given by Use (4.1) and (4.2) to evaluate R and T for the air/diamond interfaces in this problem (we will assume that diamond is not a birefringent material.) Use your results of (a) to find the intensities I 1 and I 2of rays 1 and 2, expressed as fractions of the incident intensity I 0 (assume no absorption of light in the sheet.) Find the minimum non-zero thickness d of the plate for which the electric field oscillations of rays 1 and 2 will be exactly in phase with each other. Find the resultant intensity (in terms of I0) resulting from the interference of rays 1 and 2, assuming the electric field oscillations of the rays are in phase with each other (constructive interference.)