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When solving thin-film interference problems, construct an equation that is specific to the problem by using the following strategies 1 Make a sketch of the thin film and the two coherent waves For simplicity, draw the waves as rays 2 Read the problem Is the reflected light enhanced or reduced When it is enhanced, the two reflected waves undergo constructive interference When the reflected light is reduced, the waves undergo destructive interference 3 Are either or both waves inverted on reflection If the index of refraction changes from a lower to a higher value, then the wave is inverted If it changes from a higher to a lower value, there is no inversion 4 Find the extra distance that the second wave must travel through the thin film to create the needed interference a If you need constructive interference and one wave is inverted OR you need destructive interference Reflection from a Thin Film and either both waves or none are inverted, then the difference in distance is an odd number of half Ray 1 wavelengths: (m 1/2) film, where m 0, 1, 2, etc Ray 2 b If you need constructive interference and either both waves or none are inverted OR you need destructive Medium 1 n1 interference and one wave is inverted, then the difference is an integer number of wavelengths: m film, where m 1, 2, 3, etc n d Film 5 Set the extra distance traveled by the second ray to twice the film thickness, 2d 6 Recall from 18 that



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Oil and Water You observe colored rings on a puddle and conclude that there must be an oil slick on the water You look directly down on the puddle and see a yellow green ( 555 nm) region If the refractive index of oil is 145 and that of water is 133, what is the minimum thickness of oil that could cause this color

Sketch the thin film and layers above and below it Draw rays showing reflection off the top of the film as well as the bottom Known:

3 public class KaChung { 4 public static void main(String[] args) { 5 String s = ""; 6 if(IntegerparseInt("011") == IntegerparseInt("9")) s += 1; 7 if(021 == IntegervalueOf("17")) s += 2; 8 if(1024 == new Integer(1024)) s += 3; 9 Systemoutprintln(s); 10 } }

Unknown:





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The propagation of signals in a waveguide depends, in part, upon the operating frequency of the applied signal As covered earlier, the angle of incidence made by the plane wave to the waveguide wall is a function of frequency As the frequency drops, the angle of incidence increases toward 90 The propagation of waves depends on the angle of incidence and the associated reflection phenomena Indeed, both phase and group velocities are functions of the angle of incidence When the frequency drops to a point where the angle of incidence is 90 , then group velocity is meaningless We can define a general mode equation based on our system of notation: 1 ( c) where is the longest wavelength that will propagate c a, b are the waveguide dimensions (see Fig 19-2) m, n are integers that define the number of half-wavelengths that will fit in the a and b dimensions, respectively Evaluating Eq 197 reveals that the longest TE-mode signal that will propagate in the dominant mode (TE10) is given by

145 133

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Because noil nair, there is a phase inversion on the first reflection Because nwater noil, there is no phase inversion on the second reflection Thus, there is one wave inversion The wavelength in oil is less than it is in air Follow the problem-solving strategy to construct the equation 2d

(m )

What is the result A 2 B 3 C 13 D 23 E 123 F Compilation fails G An exception is thrown at runtime 3 Given the invocation "java GiveUp" and:

4noil 555 nm (4)(145)

Are the units correct The answer is in nm, which is correct for thickness Is the magnitude realistic The minimum thickness is smaller than one wavelength, which is what it should be

[197]

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