Chapter 28 Physical Optics: Interference and Diffraction

Transcription

1 Chapter 28 Physical Optics: Interference and Diffraction

2 Units of Chapter 28 Superposition and Interference Young s Two-Slit Experiment Interference in Reflected Waves Diffraction Resolution Diffraction Gratings

3 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.

4 28-1 Superposition and Interference Interference is only noticeable if the 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). If this is true, interference will be constructive where the two waves are in phase, and destructive where they are out of phase.

5 28-1 Superposition and Interference In this illustration, interference will be constructive where the path lengths differ by an integral number of wavelengths, and destructive where they differ by a half-odd integral number of wavelengths.

6 28-1 Superposition and Interference To summarize, the two path lengths and will interfere constructively or destructively according to the following:

7 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.

8 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. However, if light is a wave, each slit serves as a new source of wavelets, as shown, and the final screen will show the effects of interference. This is called Huygens s principle.

9 28-2 Young s Two-Slit Experiment As the 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:

10 28-2 Young s Two-Slit Experiment The dark fringes are between the bright fringes; the condition for dark fringes is:

11 28-2 Young s Two-Slit Experiment This diagram illustrates the numbering of the fringes.

12 28-4 Diffraction A wave passing through a small opening will diffract, as shown. This means that, after the opening, there are waves traveling in directions other than the direction of the original wave.

13 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. The amount of diffraction depends on the wavelength, which is why we can hear around corners but not see around them.

14 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.

15 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:

16 28-4 Diffraction The second dark fringe occurs when:

17 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!)

18 28-5 Resolution Diffraction through a small circular aperture results in a circular pattern of fringes. This 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:

19 28-5 Resolution Rayleigh s 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. The size of the spot increases with wavelength, and decreases with the size of the aperture.

20 28-5 Resolution On the left, there appears to be a single source; on the right, two sources can be clearly resolved.

21 21

22 450 m 150 m 300 m m vt m vt set m vt 150 m vt 150 m m vt 150 m 150 m vt 450 m vt 150 m 2300 m 150 m vt 150 m vt 450 m 2150 m m vt 300 m m vt m t 17 m/s m m 30 s 22

23 23

24 Solution: 1. Find Moe s distance to the near speaker: 2. Find Moe s distance to the far speaker: 3. Set the difference equal to a half wavelength: 4. Divide the speed by the wavelength: 3 5. Set to find : 2 6. Divide the speed by the wavelength: m 0.40 m 3.00 m m m 0.40 m 3.00 m m m m m v 343 m/s f 0.68 khz m m m m f v 343 m/s m 2.0 khz 24

25 25

26 Solution: 1. (a) Calculate the angles for minima of order m and (m + 4): eq 28-3 y=l tan theta 1 sinm m m 2 d d 1 7 m 4 m m y y Ltan Ltan L m 4 m m+4 m m+4 m d set 7 1 Lm 2 m 2 4L 23.0 mm d d d mm d m m 4L m 647 nm 26

27 27

28 28

29 Calculate the angle to the second-order minima using equation 28-3, into equation and solve for the slit width. where y is half of 15.2 cm. y Solve for the second-order angle: y m L 1.50 m Ltan 2 2 tan tan 2.90 Solve equation for the slit width: m nm sin 2 m W 25.0 m W sin sin

30 30

31 (a) On a cloudy day your pupil diameter will increase in order to allow more light to hit the retina. This will decrease the angular size of the diffraction pattern created by the pupil and increase the resolution as indicated by equation We conclude that your eyes have greater resolution on a than they do on a bright sunny day. 2. (b) The best explanation is your eyes have better resolution on a cloudy day because your pupils will expand to allow more light to enter the eye. 31

32 32

33 m 6 sinmin min rad D 0.12 m 6 5 2min rad rad 2y 2Ltan 2L min min m rad m 7.0 m 33