Chapter 28 Physical Optics: Interference and Diffraction 1
Overview of Chapter 28 Superposition and Interference Young s Two-Slit Experiment Interference in Reflected Waves Diffraction Resolution Diffraction Gratings 2
28-1 Superposition and Interference If two waves occupy the same space, their amplitudes add at each point. They may interfere either constructively or destructively. 3
28-1 Superposition and Interference Interference is only noticeable if: - Light sources are monochromatic (so all the light has the same wavelength) and coherent (different sources maintain the same phase relationship over space and time). Interference will be constructive where the two waves are in phase, and destructive where they are out of phase. 4
28-1 Superposition and Interference In this illustration: Interference will be constructive where the path lengths differ by an integral number of wavelengths. Destructive where they differ by a half-odd integral number of wavelengths. 5
28-1 Superposition and Interference Two path lengths will thus interfere constructively or destructively according to the following: 6
28-2 Young s Two-Slit Experiment In this experiment, the original light source need not be coherent; it becomes so after passing through the very narrow slits. 7
28-2 Young s Two-Slit Experiment If light consists of particles.. - The final screen should show two thin stripes, one corresponding to each slit. If light is a wave - Each slit serves as a new source of wavelets - Final screen will show the effects of interference. - This is called Huygens s principle. 8
28-2 Young s Two-Slit Experiment Pattern on the screen shows the light on the screen has alternating light and dark fringes, corresponding to constructive and destructive interference. The path difference is given by: Therefore, the condition for bright fringes (constructive interference) is: 9
28-2 Young s Two-Slit Experiment The dark fringes are between the bright fringes; the condition for dark fringes is: 10
28-2 Young s Two-Slit Experiment This diagram illustrates the numbering of the fringes. 11
28-3 Interference in Reflected Waves Reflected waves can interfere due to path length differences, but they can also interfere due to phase changes upon reflection. 12
28-3 Interference in Reflected Waves Constructive interference: Destructive interference: 13
28-3 Interference in Reflected Waves Interference can also occur when light refracts and reflects from both surfaces of a thin film Accounts for the colors we see in oil slicks and soap bubbles. Now, we have not only path differences and phase changes on reflection. We also must account for the change in wavelength as the light travels through the film. 14
28-3 Interference in Reflected Waves Wavelength of light in a medium of index of refraction n: Therefore, the condition for destructive interference, where t is the thickness of the film, is: 15
28-3 Interference in Reflected Waves The condition for constructive interference: The rainbow of colors we see is due to the different wavelengths of light. 16
28-4 Diffraction A wave passing through a small opening will diffract, as shown. Means that, after the opening, there are waves traveling in directions other than the direction of the original wave. 17
28-4 Diffraction Diffraction is why we can hear sound even though we are not in a straight line from the source. - Sound waves will diffract around doors, corners, and other barriers. Amount of diffraction depends on the wavelength, which is why we can hear around corners but not see around them. 18
28-4 Diffraction To investigate the diffraction of light, we consider what happens when light passes through a very narrow slit. As the figure indicates, what we see on the screen is a single-slit diffraction pattern. 19
28-4 Diffraction This pattern is due to the difference in path length from different parts of the opening. The first dark fringe occurs when: 20
28-4 Diffraction The second dark fringe occurs when: 21
28-4 Diffraction In general, then, we have for the dark fringes in a single-slit interference pattern: The positive and negative values of m account for the symmetry of the pattern around the center. Diffraction fringes can be observed by holding your finger and thumb very close together (it helps not to be too farsighted!) 22
28-5 Resolution Diffraction through a small circular aperture results in a circular pattern of fringes.. Limits our ability to distinguish one object from another when they are very close. The location of the first dark fringe determines the size of the central spot Diameter of aperture 23
28-5 Resolution Rayleigh s law (criterion) relates the size of the central spot to the limit at which two objects can be distinguished: If the first dark fringe of one circular diffraction pattern passes through the center of a second diffraction pattern, the two sources responsible for the patterns will appear to be a single source. Aperture with D 24
28-5 Resolution On the left, there appears to be a single source; on the right, two sources can be clearly resolved. First dark fringes overlap First dark fringes don t overlap 25
28-6 Diffraction Gratings A system with a large number of slits is called a diffraction grating. As the number of slits grows, the peaks become narrower and more intense. Here is the diffraction pattern for five slits: 26
28-6 Diffraction Gratings The positions of the peaks are different for different wavelengths of light. The condition for constructive interference in a diffraction grating: 27
28-6 Diffraction Gratings X-ray diffraction is used to determine crystal structure the spacing between crystal planes is close enough to the wavelength of the X- rays to allow diffraction patterns to be seen. A grating spectroscope allows precise determination of wavelength: 28
28-6 Diffraction Gratings Diffraction can also be observed upon reflection from narrowlyspaced reflective grooves; the most familiar example is the recorded side of a CD. Some insect wings also display reflective diffraction, especially butterfly wings. 29
30 Answer: Destructively
31 Answer : 0.18 khz
32 Answer: 623 nm
33 Answer: 1.8 µm
Answer: The resolution of an optical instrument is greater if the minimum angular separation θ is smaller. The minimum angle θ decreases if the wavelength decreases. Therefore, greater resolution is obtained with the blue light. 34
35 Answer: d = 0.12 nm