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Transcript
2/27/2015 Wave Optics Physics 116 Eyres Topics • Double-slit interference • Diffraction gratings • Thin-film interference • • Single-slit diffraction Why do we see colors in a soap bubble? Circular-aperture diffraction A hummingbird’s feather colors have are unlike that of ordinary pigments. Why do they change subtly depending on the angle at which they’re viewed? 1 2/27/2015 Terms • • • • • • • • • antireflection double-slit interference refraction diffraction single-slit diffraction thin-film interference sharp-ended shadows double-slit interference More? • Waves: Through a Slit Opening 2 2/27/2015 Young's double-slit experiment © 2014 Pearson Education, Inc. Mathematical location of the bright & dark bands • Using trigonometry, we find: Under what conditions can one say: = Remember that mλ is simply the extra path length to point P. It is labelled Delta in the picture. You can use the same equation for dark bands. However, there you have half wavelengths. © 2014 Pearson Education, Inc. 3 2/27/2015 Same equations as written in your text: © 2014 Pearson Education, Inc. Example Two narrow slits 0.04 mm apart are illuminated by light from a HeNe laser (λ = 633 nm). A. What is the angle of the first (m = 1) bright fringe? B. What is the angle of the thirtieth bright fringe? 4 2/27/2015 Young's interference with white light • When white light is used in the double-slit experiment, we see the following: • Because the angular deflection of red light appears greater than that of blue light, we can conclude that red light must have a longer wavelength than blue light. © 2014 Pearson Education, Inc. Relating the refractive index and the speed of light in a substance • The wave model of light not only explains why light bends at the boundary of two media, but also explains Snell's law by connecting the medium's index of refraction to the speed of light in that medium. © 2014 Pearson Education, Inc. 5 2/27/2015 Refractive index: A review from chapter 21 © 2014 Pearson Education, Inc. Does the refractive index depend on the color of the light? • The different indexes of refraction for different colors mean that light of different colors and light waves of different frequencies travel at different speeds in the same medium. © 2014 Pearson Education, Inc. 6 2/27/2015 Chromatic aberration in lenses: A practical problem in optical instruments • The image locations for each wavelength of light are slightly different, leading to distortions: • Remember that the index of refraction is different for different colors of light. © 2014 Pearson Education, Inc. White light incident on grating • A spectrum produced by a grating is a result of the light of different wavelengths interfering constructively at different locations. © 2014 Pearson Education, Inc. 7 2/27/2015 Gratings: An application of interference • A typical grating has hundreds of slits per millimeter. • The bright bands are very intense and narrow, with almost complete darkness between them. Notice that this equation is the same as that for 2-slit interference! © 2014 Pearson Education, Inc. Quantitative analysis of single-slit diffraction • The width of the central diffraction maximum (the central bright band on the screen) increases as the width of the slit decreases. © 2014 Pearson Education, Inc. 8 2/27/2015 Quantitative analysis of single-slit diffraction • In the single-slit situation, the slit is not infinitely narrow. • Notice that the result is an interference pattern similar to that of multiple slit interference. © 2014 Pearson Education, Inc. Single-slit diffraction Compare this to the 2-slit equation: = © 2014 Pearson Education, Inc. 9 2/27/2015 Resolving power: Putting it all together © 2014 Pearson Education, Inc. Resolving ability of a lens Rayleigh Criterion 1.22 = When the lens is small (small D), then a large angle of separation is required to resolve the objects. © 2014 Pearson Education, Inc. 10 2/27/2015 Rayleigh criterion © 2014 Pearson Education, Inc. Example 23.7 A. What is the limit of resolution of the human eye? B. The rectangular box shown in the figure has vertical lines separated by 2 mm (as seen in your text). At what maximum distance (according to Rayleigh's criterion) will you be able to resolve the lines in picture? D for eye is 0.5 cm. Use wavelength of 550 nm = 1.22 Solve for θ: = 1.34 10 Now solve for s, given that y is 2 mm. S=15 m. Now try it with the distance as seen on the screen. You will have to measure y on the screen. Distance between lines: y Distance from you to the lines: s 11