#$&* course Phy 232 8/7 appx 9 015. `Query 15
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Given Solution: `qSTUDENT RESPONSE: The logitudinal waves had a higher velocity. That doesn't provide evidence that the high-pitched wave was longitudinal, since we didn't directly measure the velocity of those waves. The higher-pitches waves were damped out much more rapidly by touching the very end of the rod, along its central axis, than by touching the rod at the end but on the side. The frequency with which pulses arrive at the ear determines the pitch. The amplitude of the wave affects its intensity, or energy per unit area. For a given pitch the energy falling per unit area is proportional to the square of the amplitude. Intensity is also proportional to the square of the frequency. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery General College Physics and Principles of Physics 12.08: Compare the intensity of sound at 120 dB with that of a whisper at 20 dB. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: log(I / I_threshold) = dB / 10 I / I_threshold = 10^(120 / 10) = 12 I = I_threshold * 10^12. I_threshold = 10^-12 watts / m^2 dB = 120: I = 10^-12 watts / m^2 * 10^12 = 1 watt / m^2. dB = 20 watts, I = I_threshold * 10^(20 / 10) = 10^-12 watts / m^2 * 10^2 = 10^-10 watts / m^2. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe intensity at 120 dB is found by solving the equation dB = 10 log(I / I_threshold) for I. We get log(I / I_threshold) = dB / 10, so that I / I_threshold = 10^(120 / 10) = 12and I = I_threshold * 10^12. Since I_threshold = 10^-12 watts / m^2, we have for dB = 120: I = 10^-12 watts / m^2 * 10^12 = 1 watt / m^2. The same process tells us that for dB = 20 watts, I = I_threshold * 10^(20 / 10) = 10^-12 watts / m^2 * 10^2 = 10^-10 watts / m^2. Dividing 1 watt / m^2 by 10^-10 watts / m^2, we find that the 120 dB sound is 10^10 times as intense, or 10 billion times as intense. A more elegant solution uses the fact that dB_1 - dB_2 = 10 log(I_1 / I_threshold) - ( 10 log(I_2 / I_threshold) ) = 10 log(I_1 / I_threshold) - ( 10 log(I_2 / I_threshold) ) = 10 {log(I_1) - log( I_threshold) - [ ( log(I_2) - log(I_threshold) ]} = 10 { log(I_1) - log(I_2)} = 10 log(I_1 / I_2). So we have 120 - 20 = 100 = 10 log(I_1 / I_2) and log(I_1 / I_2) = 100 / 10 = 10 so that I_1 / I_2 = 10^10. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q**** Univ phy 16.79 11th edition 16.72 (10th edition 21.32): Crab nebula 1054 A.D.;, H gas, 4.568 * 10^14 Hz in lab, 4.586 from Crab streamers coming toward Earth. Velocity? Assuming const vel diameter? Ang diameter 5 arc minutes; how far is it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: fR = fS ( 1 - v/c) v = (fR / fS - 1) * c = 3 * 10^8 m/s * ( 4.586 * 10^14 Hz / (4.568 * 10^14 Hz) - 1) = 1.182 * 10^6 m/s 949 yrs * 365 days / yr * 24 hrs / day * 3600 seconds / hour * 1.182 * 10^6 m/s = 3.5 * 10^16 meters diameter is appx 7 * 10^16 meters. 5/60 * pi/180 radians = 1.4 * 10^-3 radians diameter = distance * above angle distance = diameter / angle = 7 * 10^16 m / (1.4 * 10^-3) = 2.4 * 10^19 m. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Since fR = fS ( 1 - v/c) we have v = (fR / fS - 1) * c = 3 * 10^8 m/s * ( 4.586 * 10^14 Hz / (4.568 * 10^14 Hz) - 1) = 1.182 * 10^6 m/s, approx. In the 949 years since the explosion the radius of the nebula would therefore be about 949 years * 365 days / year * 24 hours / day * 3600 seconds / hour * 1.182 * 10^6 m/s = 3.5 * 10^16 meters, the diameter about 7 * 10^16 meters. 5 minutes of arc is 5/60 degrees or 5/60 * pi/180 radians = 1.4 * 10^-3 radians. The diameter is equal to the product of the distance and this angle so the distance is distance = diameter / angle = 7 * 10^16 m / (1.4 * 10^-3) = 2.4 * 10^19 m. Dividing by the distance light travels in a year we get the distance in light years, about 6500 light years. CHECK AGAINST INSTRUCTOR SOLUTION: ** There are about 10^5 seconds in a day, about 3 * 10^7 seconds in a year and about 3 * 10^10 seconds in 1000 years. It's been about 1000 years. So those streamers have had time to move about 1.177 * 10^6 m/s * 3 * 10^10 sec = 3 * 10^16 meters. That would be the distance of the closest streamers from the center of the nebula. The other side of the nebula would be an equal distance on the other side of the center. So the diameter would be about 6 * 10^16 meters. A light year is about 300,000 km/sec * 3 * 10^7 sec/year = 9 * 10^12 km = 9 * 10^15 meters. So the nebula is about 3 * 10^16 meters / (9 * 10^15 m / light yr) = 3 light years in diameter, approx. 5 seconds of arc is 5/60 of a degree or 5 / (60 * 360) = 1 / 4300 of the circumference of a full circle, approx. If 1/4300 of the circumference is 6 * 10^16 meters then the circumference is about 4300 times this distance or about 2.6 * 10^20 meters. The circumference is 1 / (2 pi) times the radius. We're at the center of this circle since it is from here than the angular diameter is observed, so the distance is about 1 / (2 pi) * 2.6 * 10^20 meters = 4 * 10^19 meters. This is about 4 * 10^19 meters / (9 * 10^15 meters / light year) = 4400 light years distant. Check my arithmetic. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q **** query univ phy 16.74 / 16.66 (21.26 10th edition). 200 mHz refl from fetal heart wall moving toward sound; refl sound mixed with transmitted sound, 85 beats / sec. Speed of sound 1500 m/s. What is the speed of the fetal heart at the instant the measurement is made? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 200,000,085 Hz = 1500 m/s / (v - vHeart) * (1 + vHeart / v) * 2.00 v / (v - vHeart) * (1 + vHeart / v) = 200,000,085 Hz / 200,000,000 Hz = 1.000000475. (v + vHeart) / (v - vHeart) = 1.000000475 v + vHeart = 1.000000475 v - 1.000000475 vHeart 2.000000475 vHeart = .000000475 v vHeart = .000000475 v / (2.000000475) vHeart = .000000475 v / 2 = .000000475 * 1500 m/s / 2= .00032m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a. ** 200 MHz is 200 * 10^6 Hz = 2 * 10^8 Hz or 200,000,000 Hz. The frequency of the wave reflected from the heart will be greater, according to the Doppler shift. The number of beats is equal to the difference in the frequencies of the two sounds. So the frequency of the reflected sound is 200,000,085 Hz. The frequency of the sound as experienced by the heart (which is in effect a moving 'listener') is fL = (1 + vL / v) * fs = (1 + vHeart / v) * 2.00 MHz, where v is 1500 m/s. This sound is then 'bounced back', with the heart now in the role of the source emitting sounds at frequency fs = (1 + vHeart / v) * 2.00 MHz, the 'old' fL. The 'new' fL is fL = v / (v - vs) * fs = v / (v - vHeart) * (1 + vHeart / v) * 2.00 MHz. This fL is the 200,000,085 Hz frequency. So we have 200,000,085 Hz = 1500 m/s / (v - vHeart) * (1 + vHeart / v) * 2.00 MHz and v / (v - vHeart) * (1 + vHeart / v) = 200,000,085 Hz / (200,000,000 Hz) = 1.000000475. A slight rearrangement gives us (v + vHeart) / (v - vHeart) = 1.000000475 so that v + vHeart = 1.000000475 v - 1.000000475 vHeart and 2.000000475 vHeart = .000000475 v, with solution vHeart = .000000475 v / (2.000000475), very close to vHeart = .000000475 v / 2 = .000000475 * 1500 m/s / 2 = .00032 m/s, about .3 millimeters / sec. ** STUDENT COMMENT My final answer was twice the answer in the given solution. I thought that I used the Doppler effect equation correctly; however, I may have solved for the unknown incorrectly. INSTRUCTOR RESPONSE The equations tell you the frequency that would be perceived by a hypothetical detector on the heart. Suppose that each time the detector records a 'peak', it sends out a pulse. The pulses are sent out at the frequency of the detected wave. The source of these pulses is the detector, which is moving toward the 'listener', and as a result they are detected at an even higher frequency. Thus the doubled number of beats. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: Query 16 016. `Query 16 ********************************************* Question: `qquery Principles of Physics and General College Physics 12.40: Beat frequency at 262 and 277 Hz; beat frequency two octaves lower. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: beat frequency 277 Hz - 262 Hz = 15 Hz. Two octaves lower would be a fourth the frequency = 1/4 * 15 Hz = 3.75 Hz Confidence Rating: 3
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Given Solution: `aThe beat frequency is the difference in the frequencies, in this case 277 Hz - 262 Hz = 15 Hz. One ocatave reduces frequency by half, so two octaves lower would give frequencies 1/4 as great. The difference in the frequencies would therefore also be 1/4 as great, resulting in a beat frequency of 1/4 * 15 Hz = 3.75 Hz. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: `qquery gen phy problem 12.46 speakers 1.8 meters apart, listener three meters from one and 3.5 m from the other **** gen phy what is the lowest frequency that will permit destructive interference at the location of the listener? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: half-wavelength corresponds to .5 meters make a full wavelength equal to 1 The frequency of a sound with a 1-meter wavelength with v = 343 m/s is 343Hz. 1.5, 2.5 wavelengths will have dest interference corresponding to the .5 m path difference .5 m / (1.5) = .33 m f = 343/.33 = 1030Hz .5 m / (2.5) = .2 m, f = 34 / .2 = 1720 Hz confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSTUDENT SOLUTION: To solve this problem, I first realize that for destructive interference to occur, the path difference is an odd multiple of half of the wavelength(ex. 1/2, 3/2, 5/2). I used Fig.12-17 in the text to help visualize this problem. Part (A) of the problem asks to calculate the lowest frequency at which destructive interference will occur at the point where two loudspeakers are 2.5m apart and one person stands 3.0m from one speaker and 3.5m from the other. The text states that 'destructive interference occurs at any point whose distance from one speaker is greater than its distance from the other speaker by exactly one-half wavelength.' The path difference in the problem is fixed, therefore the lowest frequency at which destructive interference will occur is directly related to the longest wavelength. To calculate the lowest frequency, I first have to calculate the longest wavelength using the equation 'dL ='lambda/2, where `dL is the path difference. 'lambda=2*'dL =2(3.5m-3.0m)=1m Now I can calculate the frequency using f=v/'lambda. The velocity is 343m/s which is the speed of sound. f=343m/s/1m=343 Hz. Thus, the lowest frequency at which destructive interference can occur is at 343Hz. Keeping in mind that destructive interference occurs if the distance equals an odd multiple of the wavelength, I can calculate (B) part of the problem. To determine the next wavelength, I use the equation 'dL=3'lambda/2 wavelength=2/3(3.5m-3.0m) =0.33m Now I calculate the next highest frequency using the equation f=v/wavelength. f^2=343m/s/0.33m=1030Hz. I finally calculate the next highest frequency. 'del L=5/2 'lambda wavelength=0.20m f^3=343m/s/0.2m=1715 Hz. INSTRUCTOR EXPLANATION: The listener is .5 meters further from one speaker than from the other. If this .5 meter difference results in a half-wavelength lag in the sound from the further speaker, the peaks from the first speaker will meet the troughs from the second. If a half-wavelength corresponds to .5 meters, then the wavelength must be 1 meter. The frequency of a sound with a 1-meter wavelength moving at 343 m/s will be 343 cycles/sec, or 343 Hz. The next two wavelengths that would result in destructive interference would have 1.5 and 2.5 wavelengths corresponding to the .5 m path difference. The wavelengths would therefore be .5 m / (1.5) = .33 m and .5 m / (2.5) = .2 m, with corresponding frequencies 343 m/s / (.33 m) = 1030 Hz and 343 m/s / (.2 m) = 1720 Hz, approx. **** ********************************************* Question: `qgen phy what must happen in order for the sounds from the two speakers to interfere destructively, assuming that the sources are in phase? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Path difference must be an integer value of wavelengths plus a half wavelength Confidence Rating:
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Given Solution: `a** The path difference has to be and integer number of wavelengths plus a half wavelength. ** STUDENT QUESTION The book tells me that for these two speakers to interfere destructively, the distance from one speaker has to be greater than its distance from the other speaker by one-half wavelength. Destructive interference would occur if the distance would equal 1/2, 3/2, 5/2,… wavelengths. INSTRUCTOR RESPONSE That is correct. The given solution to the original problem says this as well. The book's explanation of course gives you a third option for the most appropriate way to think of the problem. In any case you need to understand why those path differences result in destructive interference. Once you're clear on that, a wide variety of interference problems become pretty straightforward. CRAB NEBULA PROBLEM? This Query will exit. Query 17 017. `Query 17 ********************************************* Question: `qPrinciples of Physics and General College Physics Problem 23.08. How far from a concave mirror of radius 23.0 cm must an object be placed to form an image at infinity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: focal distance = .5*radius of curvature 1/2 * 23.0 cm = 11.5 cm. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aRecall that the focal distance of this mirror is the distance at which the reflections of rays parallel to the axis of the mirror will converge, and that the focal distance is half the radius of curvature. In this case the focal distance is therefore 1/2 * 23.0 cm = 11.5 cm. The image will be at infinity if rays emerging from the object are reflected parallel to the mirror. These rays would follow the same path, but in reverse direction, of parallel rays striking the mirror and being reflected to the focal point. So the object would have to be placed at the focal point, 11.5 cm from the mirror. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery gen phy problem 23.11 radius of curvature of 4.5 x lens held 2.2 cm from tooth YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: object image distance = 4.5 * object distance = 9.9 cm 1 / i + 1 / o = 1 / f i = 9.9cm o = 2.20cm f = i * o / (i + o) = 21.8 cm^2 / (9.9 cm + 2.2 cm) = 1.8 cm confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The object distance is the 2.2 cm separation between tooth and mirror. The ratio between image size and object size is the same as the ratio between image distance and object distance, so object image distance = 4.5 * object distance = 9.9 cm. We use the equation 1 / i + 1 / o = 1 / f., where f stands for focal distance. Image distance is i = 9.9 cm and object distance is o = 2.20 cm so 1/f = 1/i + 1/o, which we solve for f to obtain • f = i * o / (i + o) = 21.8 cm^2 / (9.9 cm + 2.2 cm) = 1.8 cm or so. However i could also be negative. In that case image distance would be -9.9 cm and we would get • f = -21.8 cm^2 / (-9.9 cm + 2.2 cm) = 2.8 cm or so. MORE DETAILED SOLUTION: We have the two equations • 1 / image dist + 1 / obj dist = 1 / focal length and • | image dist / obj dist | = magnification = 4.5, so the image distance would have to be either 4.5 * object distance = 4.5 * 2.2 cm = 9.9 cm or -9.9 cm. If image dist is 9.9 cm then we have • 1 / 9.9 cm + 1 / 2.2 cm = 1/f. We solve this equation to obtain f = 1.8 cm. This solution would give us a radius of curvature of 2 * 1.8 cm = 3.6 cm, since the focal distance is half the radius of curvature. This positive focal distance implies a concave lens, and the image distance being greater than the object distance the tooth will lie at a distance greater than the focal distance from the lens. For this solution we can see from a ray diagram that the image will be real and inverted. The positive image distance also implies the real image. The magnification is • magnification = - image dist / obj dist = (-9.9 cm) / (2.2 cm) = - 4.5, with the negative implying the inverted image. There is also a solution for the -9.9 m image distance, which would correspond to a positive magnification (i.e., an upright image). The image in this case would be 'behind' the mirror and therefore virtual. For this case the equation is • 1 / (-9.9 cm) + 1 / (2.2cm) = 1 / f, which when solved give us • f= (-9.9 cm * 2.2 cm) / (-9.9 cm + 2.2 cm) = 2.9 cm, approx. This solution would give us a radius of curvature of 2 * 2.9 cm = 5.8 cm, since the focal distance is half the radius of curvature. This positive focal distance also implies a concave lens, but this time the object is closer to the lens than the focal length. For this solution we can see from a ray diagram that the image will be virtual and upright. The negative image distance also implies the virtual image. The magnification is • magnification = - image dist / obj dist = -(-9.9 cm) / (2.2 cm) = + 4.5. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Failed to assess I as negative ------------------------------------------------ Self-critique Rating: 2 ********************************************* Question: `q**** query univ phy problem 11.46 / 33.44 11th edition 33.38 (34.28 10th edition) 3 mm plate, n = 1.5, in 3 cm separation between 450 nm source and screen. How many wavelengths are there between the source and the screen? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Separation 1.5cm = 1.5*1^7nm wavelength in the glass is 450 nm / 1.5 = 300 nm 1.5 * 10^7 nm / (450 nm/wavelength) = 3.3 * 10^4 wavelengths_air 1.5 * 10^6 nm / (300 nm/wavelength) = 5.0 * 10^3 wavelengths in the glass confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The separation consists of 1.5 cm = 1.5 * 10^7 nm of air, index of refraction very close to 1, and 1.5 mm = 1.5 * 10^6 nm of glass, index of refraction 1.5. The wavelength in the glass is 450 nm / 1.5 = 300 nm, approx.. So there are 1.5 * 10^7 nm / (450 nm/wavelength) = 3.3 * 10^4 wavelengths in the air and 1.5 * 10^6 nm / (300 nm/wavelength) = 5.0 * 10^3 wavelengths in the glass. ** STUDENT QUESTION After reading the solution, I am unsure about why the path difference is equal to twice the plate thickness. INSTRUCTOR RESPONSE One ray is reflected from the 'top' surface of the plate. The other passes through the plate to the other surface, is reflected there. It has already passed through the thickness of the plate. It has to pass back through the plate, to the other surface, before it 'rejoins' the original ray. It has therefore traveled an extra distance equal to double that of the plate. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Had to look over solution but it became apparent. ------------------------------------------------ Self-critique Rating: 3 019. `Query 19 ********************************************* Question: `qGeneral College Physics and Principles of Physics Problem 24.2: The third-order fringe of 610 nm light created by two narrow slits is observed at 18 deg. How far apart are the slits? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since it’s a 3d fringe, path difference = 3 wavelengths 3*610 = 1830 nm further. All other distance is equal to slit spacing * sin(18) a sin(18) = 1830 nm a = 1830 nm / sin(18) = 5920 nm confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe path difference for a 3d-order fringe is 3 wavelengths, so light from one slit travels 3 * 610 nm = 1830 nm further. The additional distance is equal to slit spacing * sin(18 deg), so using a for slit spacing we have a sin(18 deg) = 1830 nm. The slit spacing is therefore a = 1830 nm / sin(18 deg) = 5920 nm, or 5.92 * 10^-6 meters. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q**** query gen phy problem 24.7 460 nm light gives 2d-order max on screen; what wavelength would give a minimum? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: path difference = 2 * 460 nm = 920 nm. minimum when 3/2 of the wavelength is 920 nm wavelength = 2/3 * 920 nm = 613 nm confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSTUDENT SOLUTION FOLLOWED BY INSTRUCTOR COMMENT AND SOLUTION: The problem states that in a double-slit experiment, it is found that bule light of wavelength 460 nm gives a second-order maximun at a certain location on the screen. I have to determine what wavelength of visible light would have a minimum at the same location. To solve this problem I fist have to calculate the constructive interference of the second order for the blue light. I use the equation dsin'thea=m'lambda. m=2 (second order) dsin'thea=(2)(460nm) =920nm Now, I can determine the destructive interference of the other light, using the equation dsin'thea=(m+1/2)'lambda=(m+1/2)'lambda m+(0,1,2...) Now that I have calculated dsin'thea=920nm, I used this value and plugged it in for dsin'thea in the destructive interference equation.(I assumed that the two angles are equal) because the problem asks for the wavelength at the same location. Thus, 920nm=(m+1/2)'lambda. m=(0,1,2,...) I calculated the first few values for 'lambda. For m=0 920nm=(0+1/2)'lambda =1.84*10^nm For m=1 920nm=(1+1/2)'lambda =613nm For m=2 920nm=(2+1/2)'lambda=368 nm From these first few values, the only one of thes wavelengths that falls in the visible light range is 613nm. Therefore, this would be the wavelength of visible light that would give a minimum. INSTRUCTOR COMMENT AND SOLUTION: good. More direct reasoning, and the fact that things like sines are never needed: ** The key ideas are that the second-order max occurs when the path difference is 2 wavelengths, and a minimum occurs when path difference is a whole number of wavelengths plus a half-wavelength (i.e., for path difference equal to 1/2, 3/2, 5/2, 7/2, ... of a wavelength). We first conclude that the path difference here is 2 * 460 nm = 920 nm. A first-order minimum (m=0) would occur for a path difference of 1/2 wavelength. If we had a first-order minimum then 1/2 of the wavelength would be 920 nm and the wavelength would be 1860 nm. This isn't in the visible range. A minimum would also occur If 3/2 of the wavelength is 920 nm, in which case the wavelength would be 2/3 * 920 nm = 613 nm, approx.. This is in the visible range. A niminum could also occur if 5/2 of the wavelength is 920 nm, but this would give us a wavelength of about 370 nm, which is outside the visible range. The same would be the case for any other possible wavelength. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q**** query 1 phy problem 35.54 11th edition 35.52 (37.46 10th edition) normal 477.0 nm light reflects from glass plate (n=1.52) and interferes constructively; next such wavelength is 540.6 nm. How thick is the plate? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The path difference here is twice the thickness. Wavelength = 477 nm / 1.52 and 540.6 nm / 1.52 477 / 63.6 = 8.5 2 * 477 / 63.6 = 17 17 * 477 = 8109 8109 / 540.6 = 15 distance = 17 * 477 nm / 1.52 = 5335 nm = twice the pane thickness = 5335 nm / 2 = 2667 nm confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The path difference for constructive interference is an integer multiple of the wavelength. The path difference here is twice the thickness. Wavelengths in glass are 477 nm / 1.52 and 540.6 nm / 1.52. So we know that double the thickness is an integer multiple of 477 nm / 1.52, and also an integer multiple of 540.6 nm / 1.52. We need to find the first integer multiple of 477 nm / 1.52 that is also an integer multiple of 540.6 nm / 1.52. We first find an integer multiply of 477 that is also an integer multiply of 540.6. Integer multiples of 540.6 are 540.6, 1081.2, 1621.8, etc. Dividing these numbers by 477 we obtain remainders 63.6, 127.2, etc. When the remainder is a multiple of 477 then we have an integer multiple of 477 which is also an integer multiple of 540.6. SInce 477 / 63.6 = 8.5, we see that 2 * 477 / 63.6 = 17. So 17 wavelengths of 477 nm light is the first multiple that is equivalent to an integer number of wavelengths of 540.6 nm light. 17 * 477 = 8109. Since 8109 / 540.6 = 15, we see that 17 wavelengths of 477 nm light span the same distance as 15 wavelengths of 540.6 nm light. It easily follows that that 17 wavelengths of (477 nm / 1.52) light span the same distance as 15 wavelengths of (540.6 nm / 1.52) light. This distance is 17 * 477 nm / 1.52 = 5335 nm. This is double the thickness of the pane. The thickness is therefore pane thickness = 5335 nm / 2 = 2667 nm. IF INTERFERENCE WAS DESTRUCTIVE: n * 477 nm / 1.52 = (n-1) * 540.6 nm / 1.52, which we solve: Multiplying by 1.52 / nm we get 477 n = 540.6 n - 540.6 n * (540.6 - 477 ) = 540.6 n * 63.6 = 540.6 n = 540.6 / 63.6 = 8.5. This is a integer plus a half integer of wavelengths, which would result in destructive interference for both waves. Multiplying 8.5 wavelengths by 477 nm / 1.52 we get round-trip distance 2667 nm, or thickness 1334 nm. ** STUDENT QUESTION I understand that we need an integer multiple to find the plate thickness, but I don’t understand how you find that integer multiple. INSTRUCTOR RESPONSE It's obvious, for example, that 20 is an integer multiple of both 4 and 5. There are easier ways to find this, but we could have reasoned it out as follows: 2 * 4 = 8, and 8 / 5 is not an integer.3 * 4 = 12, and 12 / 5 is not an integer.4 * 4 = 16, and 16 / 5 is not an integer.4 * 5 = 20 and 20 / 5 is an integer. So 20 is the smallest number which is an integer multiple of both 4 and 5. We used a similar brute-force process here. Alternatively we might have found the result as follows: We find the least common multiple of 4770 and 5406. The prime factorization of 4770 is 2 * 3^2 * 5 * 53. The prime factorization of 5406 is 2 * 3^2 * 17 * 53. The least common multiple is therefore 2 * 3^2 * 5 * 17 * 53 = 81090, which is 17 * 4770. In actual practice, with experimental uncertainties, we might not get exact equality, which would limit the usefulness of the least-common-multiple procedure. We would likely instead look for a multiple of one wavelength whose remainder on division by the other was a sufficiently small fraction of that wavelength. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q**** query univ phy prob 35.58 / 35.52 11th edition 35.50 (10th edition 37.44): 700 nm red light thru 2 slits; monochromatic visible ligth unknown wavelength. Center of m = 3 fringe pure red. Possible wavelengths? Need to know slit spacing to answer? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When m=3, max for red light_far travels 3 wavelengths more than the light from the other. The light of the unknown color travels 3.5 wavelengths further. unknown wavelength = 3/3.5 * red = 600 nm Confidence Rating:3
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Given Solution: `aSTUDENT SOLUTION: The pure red band at m = 3 suggests that there exists interference between the wavelength of the red light and that of the other light. Since only the red light is present at m = 3 it stands to reason that the wavelength of the other light is a half of a wavelength behind the red wavelength so that when the wavelength of the red light is at its peak, the wavelength of the other light is at its valley. In this way the amplitude of the red light is at its maximum and the amplitude of the other light is at it minimum - this explains why only the red light is exhibited in m = 3. INSTRUCTOR COMMENT At this point you've got it. At the position of the m=3 maximum for the red light the red light from the further slit travels 3 wavelengths further than the light from the closer. The light of the unknown color travels 3.5 wavelengths further. So the unknown wavelength is 3/3.5 times that of the red, or 600 nm. You don't need to know slit separation or distance (we're assuming that the distance is very large compared with the wavelength, a reasonable assumption for any distance we can actually see. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: 020. `Query 20-22 ********************************************* Question: `q**** query univ phy 36.61 11th edition 36.59 phasor for 8 slits YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: phi = pi/4 octagon phi = 3 pi / 4 first vector @ 135 deg second @ 270 deg, the third at 415 deg The fourth vector at 45 deg + 135 deg = 180 deg 315 deg 90 deg 225deg 360 deg The resulting endpoint coordinates of the vectors, in order, will be -0.707, .707 -0.707, -0.292 0, 0.414 -1, 0.414 -0.292, -0.292 -0.292, 0.707 -1, 0 0, 0 phi = 5 pi / 4 each vector will 'rotate' relative to the last at angle 5 pi / 4, or 225 deg. To check yourself the first few endpoints will be For 6 pi / 4 ? For 7 pi / 4: octagon confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** If you look at the phasor diagram for phi = 3 pi / 4 you will see that starting at any vector the fourth following vector is in the opposite direction. So every slit will interfere destructively with the fourth following slit. This is because 4 * 3 pi / 4 is an odd multiple of pi. The same spacing will give the same result for 5 pi / 4 and for 7 pi / 4; note how starting from any vector it takes 4 vectors to get to the antiparallel direction. For 6 pi / 4, where the phasor diagram is a square, every slit will interfere destructively with the second following slit. For phi = pi/4 you get an octagon. For phi = 3 pi / 4 the first vector will be at 135 deg, the second at 270 deg (straight down), the third at 415 deg (same as 45 deg, up and to the right). These vectors will not close to form a triangle. The fourth vector will be at 45 deg + 135 deg = 180 deg; i.e., horizontal to the left. The next two will be at 315 deg (down and toward the right) then 90 deg (straight up). The last two will be at 225 deg (down and to left) and 360 deg (horiz to the right). The resulting endpoint coordinates of the vectors, in order, will be -0.7071067811, .7071067811 -0.7071067811, -0.2928932188 0, 0.4142135623 -1, 0.4142135623 -0.2928932188, -0.2928932188 -0.2928932188, 0.7071067811 -1, 0 0, 0 For phi = 5 pi / 4 each vector will 'rotate' relative to the last at angle 5 pi / 4, or 225 deg. To check yourself the first few endpoints will be -0.7070747217, -0.7071290944; -0.7070747217, 0.2928709055; 0, -0.4142040038 and the final endpoint will again be (0, 0). For 6 pi / 4 you will get a square that repeats twice. For 7 pi / 4 you get an octagon. NEW PROBLEM: The longest wavelength is 700 nm and slit spacing is about 1250 nm. The path difference can't exceed the slit spacing, which is less than double the 700 nm spacine. So there are at most central max (path difference zero) and the first-order max (path difference one wavelength). Note that there will be a second-order max for wavelengths less than about 417 nm. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Query 23 ********************************************* Question: `qIn your own words explain how the introductory experience with scotch tape illustrates the existence of two types of charge. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: When the first two pieces of tape were torn apart they were attracted to each other. When the second two pieces were torn apart, one piece was attracted to a separate piece and repelled the other. Second piece from second pair was attracted to the piece from the first pair. This illustrates the existence of two types of charge. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `qIn your own words explain how the introductory experience with scotch tape supports the idea that the force between two charged particles acts along a straight line through those particles, either attracting the forces along this line or repelling along this line. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: When two pieces with opposing charges were held together, the ends were attracted into a line. The bottom of the tape didn’t pull towards the middle of the other piece of tape; it was attracted to the portion of the tape directly in front of it. The opposite behavior was observed for pieces of tape that had the same charge. The bottom of one piece repelled the bottom of the other, and they each moved back in a very straight manner confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `qIn your own words explain why this experience doesn't really prove anything about actual point charges, since neither piece of tape is confined to a point. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The tape had point charges all over it. The piece was entirely too big to be a point, so it can’t prove anything about charges. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `qIf one piece of tape is centered at point A and the other at point B, then let AB_v stand for the vector whose initial point is A and whose terminal point is B, and let AB_u stand for a vector of magnitude 1 whose direction is the same as that of AB_u. Let BA_v and BA_u stand for the analogous vectors from B to A. Vectors of length 1 are called unit vectors. • If the pieces attract, then in the direction of which of the two unit vectors is the tape at point A pushed or pulled? • If the pieces repel, then in the direction of which of the two unit vectors is the tape at B pushed or pulled? Explain. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: If the pieces attract, then point A will be pulled in the AB_v direction. If the pieces repel, then B will be pushed in the AB_v direction. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If the pieces attract, then the tape at point A is pulled toward point B. • The vectors AB_v and AB_u point from A to B. • Of these the vector AB_u is the unit vector. • So the tape at A experiences a force in the direction of the vector AB_u. If the pieces repel, then the tape at point B is pushed away from point A. • The direction of the force is therefore from A towards B. • The direction is therefore that of the vector AB_u. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qUsing the notation of the preceding question, which you should have noted on paper (keep brief running notes as you do qa's and queries so you can answer 'followup questions' like this), how does the magnitude of the vector AB_v depend on the magnitude of BA_v, and how does the magnitude of each vector compare with the distance between A and B? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: distance from A to B = distance from B to A. AB_v and BA_v have the same length Each vector’s magnitude = distance from A to B confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The distance from A to B is the same as the distance from B to A. So the vectors AB_v and BA_v have the same length. Each vector therefore has magnitude equal to the distance between A and B. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qUsing the notation of the preceding question, how is the force experienced by the two pieces of tape influenced by the magnitude of AB_v or BA_v? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: force exerted by two charges on one another is inversely proportional to the square of the distance between them confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The expected answer is that the force exerted by two charges on one another is inversely proportional to the square of the distance between them. So as the magnitudes of the vectors, which are equal to the distance between A and B (i.e., to the separation between the two pieces of tape), increases the force decreases with the square of the distance (similarly if the distance decreases the force increases in the same proportion). This answer isn't exactly correct, since the two pieces of tape are not point charges. Since some parts of tape A are closer to B than other parts of tape A, and vice versa, the inverse square relationship applies only in approximation for actual pieces of tape. STUDENT COMMENT: Still not sure how to answer this question. I dont know what sort of difference there is between the magnitude vector and the unit vector. INSTRUCTOR RESPONSE: As noted in the preceding solution, AB_v stands for the vector whose initial point is A and whose terminal point is B, and AB_u stands for a vector of magnitude 1 whose direction is the same as that of AB_u. To get a unit vector in the direction of a given vector, you divide that vector by its magnitude. So AB_u = AB_v / | AB_v |. The purpose of a unit vector is to represent a direction. If you multiply a unit vector by a number, you get a vector in the same direction as the unit vector, whose magnitude is the number by which you multiplied it. The problem does not at this point ask you to actually calculate these vectors. However, as an example: Suppose point A is (4, 5) and point B is (7, 3). Then AB_v is the vector whose initial point is A and whose terminal point is B, so that AB_v = <3, -2>, the vector with x component 3 and y component -2. The magnitude of this vector is sqrt( 3^2 + (-2)^2 ) = sqrt(11). Therefore AB_u = <3, -2> / sqrt(11) = (3 / sqrt(11), -2 / sqrt(11) ). That is, AB_u is the vector with x component 3 / sqrt(11) and y component -2 / sqrt(11). (Note that these components should be written with rationalized denominators as 3 sqrt(11) / 11 and -2 sqrt(11) / 11, so that AB_u = <3 sqrt(11) / 11, -2 sqrt(11) / 11 > ). The vector AB_u is a unit vector: it has magnitude 1. The unit vector has the same direction as AB_v. If we want a vector of magnitude, say, 20 in the direction of this vector, we simply multiply the unit vector AB_u by 20 (we would obtain 20 * <3 sqrt(11) / 11, -2 sqrt(11) / 11 > = <60 sqrt(11) / 11, -40 sqrt(11) / 11 >). You don't need to understand this example at this point, but if you wish to understand it you should probably sketch this situation and identify all these quantities in your sketch. STUDENT RESPONSE AB_v/(AB_v)absolute valueSo AB_u would either be 1 or -1. INSTRUCTOR COMMENT AB_u and BA_v are vectors, so they have both magnitude and direction. Your answer is correct if everything is in one dimension (e.g., if all charges are on the x axis and all forces directed along the x axis). In one dimension direction can be specified by + and - signs. In two dimensions + and - signs are not sufficient. In two dimensions the direction of a vector is generally specified by its angle as measured counterclockwise from the positive x axis. In two dimensions, the magnitude of AB_u would be the same as that of BA_u, but the vectors would be in the opposite direction. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Recognizing A and B aren’t point charges ------------------------------------------------ Self-critique Rating: 3 "