Physics E1bx: Assignment for April 28 May 5 Homework #9: Sound and Diffraction Due Tuesday, May 5, at 6:00PM This assignment must be turned in by 6:00PM on Tuesday, May 5. Late homework will not be accepted. Please write your answers to these questions on a separate sheet of paper with your name and your section TF s name written at the top. Turn in your homework to the mailbox marked with your section TF s name in the row of mailboxes outside of Sci Ctr 108. You are encouraged to work with your classmates on these assignments, but please write the names of all your study group members on your homework. After completing this homework assignment, you should be able to Single-Slit Diffraction Understand that light passing through a single slit produces a diffraction pattern consisting of a bright spot with alternating bright and dark bands on each side. Be able to calculate the angle at which the dark bands will appear due to destructive interference. Understand how changing one aspect of the apparatus will affect the diffraction pattern seen on the wall. Example: How does varying the width of the slit affect the diffraction pattern? Diffraction from a Circular Aperture / Rayleigh Criterion Understand that light passing through a circular aperture will produce an Airy pattern consisting of alternating bright and dark concentric circles. Know the position of the first dark ring in terms of angle, wavelength of the incident light, and width of the circular aperture. Understand what the angular position θ and the angular width are, and how they are related. Be able to use the Rayleigh criterion to determine whether it is possible to distinguish between two objects. Use the Rayleigh criterion to calculate such parameters such as the separation distance between two objects, the width of the circular aperture, the wavelength of incident light, the angular position, etc. Diffraction from a Grating or Crystal Understand how to find the areas of constructive interference when light passes through many narrow slits. Do not confuse the situation of light passing through many slits with that of light passing through a single slit, even though the equations look very similar. Standing Waves and Sound Explain how two identical waves traveling in opposite directions will create standing waves, and show mathematically how the standing wave equation emerges from the superposition of the two traveling waves. Describe the standing waves for a string held between two fixed ends, and calculate the allowed frequencies from the string length and the wave speed (or vice versa). Describe the standing waves in a pipe of air, depending on the boundary conditions (open or closed ends). Calculate the allowed wavelengths and frequencies. Define the decibel, and convert between sound intensity in decibels and sound intensity in W/m 2. Explain why there is a Doppler shift in the observed frequency of a sound wave when either the source or the observer is moving (or both). Calculate the Doppler shift for motion of the source, the observer, or both. Use the Doppler shift to calculate the speed of a source. 1
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1. Putting Everything Together (Exam-Type Question) (1 pt). a) The graph at right shows the intensity on a distant screen as a function of θ for a singleslit experiment. If λ = 580 nm, what is the width of the slit in mm? Show your work. The location of the minima are related to wavelength and slit width by sin θ =!"! where m = ±1, ±2, ±3, From the graph, we see that the first minima (m = 1) occurs when θ = 5. Rearranging the above equation so that we are solving for D and plugging in the appropriate values yields D = mλ sin θ = 1 580 nm sin 5 = 6.65 µμm For this particular setup, the width of the slit is 6.65 µm. 2. Putting Everything Together (Exam-Type Question) (2 pts). The human eye can be considered to be a circular aperture with a diameter of a few millimeters (depending on the contraction of your pupils). Assume that the index of refraction within your eye is n = 1.38, and that your retina is located 2.5 cm behind your pupils. a) When the pupil diameter is 2 mm, what is the angular diameter of the central diffraction maximum produced by a point source on your retina for light of wavelength 500 nm? What is the diameter of the central bright spot on your retina? The basic geometry for circular diffraction is shown in the diagram at left. Light diffracts through a circular aperture of diameter D and then spreads out. The angle from the central axis to the edge of the Airy disc (i.e., the angle to the first dark ring) on the screen is given by θ = 1.22λ D. Since we are interested in the angular diameter of the Airy disc, we take twice this angle. Since the diffraction occurs inside the eye, we need to use the wavelength inside the eye: 2θ = 2 1.22λ nd = 21.22(500 nm) = 4.4 10 4 rad or 0.025. 1.38(2 mm) 7
The diameter of the actual Airy disc is just this angle times the distance to the screen L (2.5 cm, in this case) or 11 µm. b) In the fovea (the central region of your retina), the light receptor cells (cones) are separated by a distance of about 2 µm. Would you have sharper central vision if the cones were more closely spaced than this? Explain. Due to diffraction at the pupil, even a point source of light outside the eye will be spread out into an Airy disc about 11 microns in diameter on the fovea; larger objects will form correspondingly larger Airy discs. This size sets a limit on optical resolution of the human eye. If all of the Airy discs are on the order of 10 microns or more, there is no real benefit to having fovea resolution at the level of 2 microns (or smaller) because linear resolution in the fovea is not the limiting constraint for your overall optical resolution. c) An eagle has larger pupils than a human, around 6 mm in diameter during daylight, but a smaller eyeball, so that the retina is only 1 cm behind the pupil. If an eagle is flying 200 m high and spots two field mice on the ground, how far apart must the field mice be in order for the eagle to be able to resolve them as separate objects? (Assume the same wavelength of light for eagles as humans.) For the eagle, the angular radius θ of the Airy disc is 1.22(500 nm) θ = = 7.4 10 5 rad or 0.0042. 1.38(6 mm) By the Rayleigh criterion, this angle is also the minimum angular separation of two barelyresolvable point sources through this aperture. If the field mice are separated by a distance d, then the angular separation (in radians) from the perspective of the eagle at height h is just d/h. The minimum separation of the field mice is d = hθ = 200 m ( ) = 1.5 cm. ( ) 7.4 10 5 The size of the eagle s eyeball doesn t affect the result, although it does affect how closely spaced the eagle s receptor cells (cones) need to be on its fovea. 3. Flute hero (2 pts). Unlike the clarinet, the flute (and its relative, the piccolo) acts like a tube that is open at both ends. Instead of a reed, a flute s mouthpiece has a hole that the flautist blows across in order to excite standing sound waves in the flute s body. a) The lowest note on a flute is middle C, with a frequency of f = 262 Hz. How long is a flute? For the lowest note (the fundamental), corresponding to the n = 1 mode, the length is just L = λ 2 = v 2 f 343 m/s = = 65.5 cm. 2( 262 Hz) b) What are the three lowest harmonics supported by a flute? How does this contrast with the harmonics supported by a clarinet? 8
In an open-ended tube, both ends are nodes for pressure variation, so the allowed modes of standing waves are just like those on a fixed-end string. The length must be an integer multiple of a halfwavelength, so and the corresponding frequencies are λ n = 2L n f n = nv 2L. The three lowest harmonics are n = 1, 2, and 3, with frequencies of 262 Hz, 524 Hz, and 786 Hz. In contrast, a clarinet s lowest harmonics are the first, third, and fifth. 4. Holy echolocation, Batman! (6 pts) Bats are not blind, but they can hunt in complete darkness, using ultrasonic sound waves to detect both the position and the motion of their prey. ( Ultrasonic means at a higher frequency than humans can hear. Supersonic means faster than the speed of sound.) Highintensity, high-frequency sound pulses are emitted from their snouts, and they hear echoes from sound waves reflected from the prey. This process is known as echolocation. a) Suppose the friendly and graceful bat pictured at right tries to locate a moth by emitting sound pulses and detecting the echo. The bat sounds are emitted at f 0 = 100 khz. If the echo returns a time t = 0.25 s after the bat emits the sound, how far away is the moth? The echo returns after the sound has had a chance to travel from the bat to the moth and back. So the distance to the moth is d = 1 2 vt = 1 ( 343 m/s ) 2 0.25 s ( ) = 43 m. b) The frequency of the returning echo pulse is 101.0 khz. What is the speed v m of the moth? Is it moving toward or away from the bat? Note: the echo from the moth is essentially a sound absorption (in which the moth acts as a moving observer) plus an immediate re-emission (in which the moth then acts as a moving source of sound at the frequency that it just observed). The frequency has been shifted up, which means that moth is moving toward the bat. When the ultrasound waves hit the moth, the moth acts as the observer. The Doppler shift for a moving observer is given by:! v + v f ' = f m $ 0 " # v % &, where f ' is the frequency at which the sound reaches the moth, i.e. the number of wavefronts that reach the moth per second. For the echo (reflection), the moth immediately re-emits a sound back towards the bat at this same frequency. But this time the moth is a moving source, so we have to use the Doppler shift for a moving source, with f ' for the original frequency: v v + v f '' = f ' = f m 0. v v m v v m Now we can solve for v m with a little algebra: 9
v m = v f '' 1 f 0 f '' + 1 f 0 f '' " # $ f 0 v v m v % & ' = v + v m v f '' v f '' v m = v + v m f 0 f 0 " f '' % 1 # $ f 0 & ' v = " f '' % + 1 # $ f 0 & ' v m = ( 343 m/s) 1.010 1 = 1.7 m/s. 1.010 + 1 c) Now the bat itself begins to move toward the moth at a speed of v b = 3 m/s. What is the frequency of the sound pulses as heard by the moth? What is the frequency of the echo pulses received by the bat? The moth is a moving observer who hears sound emitted by a moving source. Both observer and source are moving towards each other. The frequency heard by the moth is: v + v f ' = f m 343 + 1.7 0 = ( 100 khz) = 101.4 khz. v v b 343 3 Once again, the moth acts as a moving source for the echo (with an original frequency of f '), and finally the bat acts as a moving observer. So the final frequency heard by the bat is: f '' = f ' v + v b v + v = f m v + v b 0 = ( 100 khz) ( 343 + 1.7 )( 343 + 3) = 102.8 khz. v v m v v b v v m ( 343 3) ( 343 1.7) Note that this problem involves essentially four Doppler shifts, each of which shifts the frequency up slightly. d) Bats require extremely sensitive hearing to detect small prey using echolocation. Suppose the bat emits a 100.0-kHz sound whose intensity is measured 1 cm from the bat s mouth to be 0.1 W/m 2. If the bat s threshold of hearing is 10 12 W/m 2 at this frequency, what is the maximum distance from the bat at which it could detect a moth of area 10 cm 2 You can assume that the bat emits sound isotropically (i.e., uniformly in all directions) with power P, that 100% of the intensity is reflected, and that the reflected sound is emitted isotropically from the moth. The intensity of the sound once it reaches the moth is I = P 4πd. 2 The power incident on the moth is the intensity times the moth s area: P moth = PA 4πd. 2 We can treat the echo as a new source of sound with this much power. (This is an approximation; in reality, somewhat less than 100% of the sound energy received by the moth would be echoed. Also, it might not be reflected isotropically; chances are, the intensity in the direction back towards the bat would be greater than in other directions. Both of these effects are hard to quantify, but they would at least serve to cancel each other out somewhat, so it s not ridiculous to neglect them both.) The intensity that reaches the bat is then 10
I = P moth 4πd 2 = PA 16π 2 d 4. Notice how the intensity of the echo decreases as the fourth power of the distance. This fact severely limits the maximum range of echolocation. The power originally emitted by the bat is related to the measured intensity by: P = 4π 1 cm ( ) = 1.26 10 4 W ( ) 2 0.1 W/m 2 Now we can use the expression for the intensity of the echo that reaches the bat, plug in the threshold intensity for I, and solve for the distance d: d = PA 4 = 16π 2 I ( 1.26 10 4 W) ( 10 3 m 2 ) 4 = 5.3 m. 16π 2 10 12 W/m 2 ( ) That s not very far at all! In order to detect prey at a significant distance, a bat would have to have much better sensitivity to ultrasound than this. For example, in order to detect this butterfly at 53 m, the bat would have to be able to hear signals 0.0001 times as intense (40 db quieter). 11