#$&* course phy201 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: First way: KE change is equal to the work done by the net force, which is net force * displacement, or Fnet * `ds. Second way: KE change is also equal to Kef - KE0 = .5 m vf^2 - .5 m v0^2. ** STUDENT QUESTION: I wasn’t sure what equation to use to find KE the second way. What does Kef stand for? INSTRUCTOR RESPONSE: In general f stands for 'final' and 0 for 'initial'. Just as v0 and vf stand for initial and final velocities, we'll use KEf and KE0 to stand for initial and final kinetic energies. STUDENT QUESTION: Ok I know the other equation now but I still don’t really understand it. How come you multiply each velocity by 0.5? I don’t really understand the second equation KE = 1/2 m v^2 INSTRUCTOR RESPONSE On one level, KE = 1/2 m v^2 is simply a formula you have to know. It isn't hard to derive that formula, as you'll see soon, and the 1/2 arises naturally enough. A synopsis of the derivation: If force F_net is applied to mass m through displacement `ds then: a = F_net / m, and vf^2 = v0^2 + 2 a `ds It's not difficult to rearrange the result of these two equations to get F_net * `ds = 1/2 m vf^2 - 1/2 m v0^2. You'll see the details soon, but that's where the formula KE = 1/2 m v^2 comes from; the 1/2 or 0.5 is part of the solution. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q (This question applies primarily to General College Physics students and University Physics students, though Principles of Physics students are encouraged, if they wish, to answer the question). In terms of the equations of motion why do we expect that a * `ds is proportional to the change in v^2, and why do we then expect that the change in v^2 is proportional to Fnet `ds? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Recall from the equation of motion Vf^2 = v0^2 + 2*a*ds If we subtract v0^2 from both side of the equation we end up with Vf^2 - v0^2 = 2*a*ds From this equation it is obvious that a*ds is twice the change in velocity squared. For the second part recall that f_net = m*a And recall that wd = f_net*ds= m*a*ds From this equation we can see that F_net*ds is proportional to a*ds(change in v^2) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ very confident
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Given Solution: In a nutshell: * since vf^2 = v0^2 + 2 a `ds, a `ds = 1/2 (vf^2 - v0^2), so a `ds is proportional to the change in v^2 * since F_net = m a, F_net * `ds = m a * `ds so F_net * `ds is proportional to a * `ds * Thus F_net `ds is proportional to a * `ds, which in turn is proportional to the change in v^2. * Thus F_net `ds is proportional to the change in v^2. More detail: It's very important in physics to be able to think in terms of proportionality. * To say that y is proportional to x is to say that for some k, y = k x. * That is, y is a constant multiple of x. To say that a * `ds is proportional to the change in v^2 is to say that * for some k, a * `ds = k * ( change in v^2)--i.e., that * a * `ds is a constant multiple of the change in v^2. In terms of the equations of motion, we know that vf^2 = v0^2 + 2 a `ds so a `ds = 1/2 (vf^2 - v0^2), which is 1/2 the change in v^2. So a `ds is a constant multiple (1/2) of the change in v^2. Formally we have a `ds = k ( change in v^2) for the specific k value k = 1/2. Now since Fnet = m a, we conclude that Fnet * `ds = m a * `ds and since a `ds = k * ( change in v^2) for the specific k value k = 1/2, we substitute for a * `ds to get Fnet `ds = m * k * (change in v^2), for k = 1/2. Now m and k are constants, so m * k is constant. We can therefore revise our value of k, so that it becomes m * 1/2 or m / 2 With this revised value of k we have Fnet * `ds = k * (change in v^2), where now k has the value m / 2. That is, we don't expect Fnet * `ds to be proportional to the change in velocity v, but to the change in the square v^2 of the velocity. STUDENT COMMENT: I am still a bit confused. Going through the entire process I see how these values correlate but on my own I am not coming up with the correct solution. I am getting lost after we discover the a `ds is a constant multiple of (1/2) the change in v^2. Is it that I should simply substitute the k into the equation? Or am I missing something else? INSTRUCTOR RESPONSE: The short answer is that by the fourth equation of uniformly accelerated motion, a `ds = 1/2 (vf^2 - v0^2), which is half the change in v^2, so that a `ds is proportional to the change in v^2. (The proportionality constant between a `ds and change in v^2 is the constant number 1/2). F_net = m a, where m is the mass of the object. So F_net is proportional to a. (The proportionality constant between F_net and a is the constant mass m). Thus F_net `ds is proportional to a `ds, which we have seen is proportional to the change in v^2. The conclusion is the F_net `ds is proportional to the change in v^2. (The proportionality constant between F_net `ds and change in v^2 is 1/2 m.) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ------------------------------------------------ Self-critique Rating: ********************************************* Question: How do our experimental results confirm or cause us to reject this hypothesis? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The experiments carried out thus far confirm the two equations above. We got results that indicates that a*ds is approximately twice the change in velocity squared. And figures also indicate that the F_net*ds is approximately equal to a* m*ds confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The explanation for this result: On a ramp with fixed slope the acceleration is constant so * a `ds is simply proportional to `ds * specifically a `ds = k * `ds for k = a. In the preceding question we saw why * a * `ds = k * (change in v^2), with k = 1/2. In our experiment the object always accelerated from rest. So the change in v^2 for each trial would be from 0 to vf^2. the change would therefore be just * change in v^2 = vf^2 - 0^2 = vf^2. Thus if a `ds is proportional to the change in vf^2, our graph of vf^2 vs. a `ds should be linear. * The slope of this graph would just be our value of k in the proportionality a * `ds = k * (change in v^2), where as we have seen k = 1/2 We wouldn't even need to determine the actual value of the acceleration a. To confirm the hypothesis all we need is a linear graph of vf^2 vs. `ds. (we could of course use that slope with our proportionality to determine a, if desired) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qGeneral College Physics and Principles of Physics: convert 35 mi/hr to km/hr, m/s and ft/s. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Recall that 1 miles = 1.609344 kilometers (35 mi)/hr×(1.609344 km)/(1 mi) ? (56.327 km)/hr Recall that 1 kilometer = 1000meter and 1 hr = 3600s this implies that to convert km/hr to m/s first we must multiply the km/hr by 1000m/km then multiply the result by 1 hr/3600s. (56.32704 km)/hr×1000m/km×(1 hr)/(3600 s) = (56327.04 m)/(3600 s) = (15.646 m)/s Recall that 1 foot = 0.3048 meters this implies that to convert m/s to ft/s we must multiply the m/s by 1ft/0.3048 meter (15.6464 m)/s ×(1 ft)/(0.3048 m) = (51.333 ft)/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ very confident
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Given Solution: `aWe need a conversions between miles and meters, km and ft, and we also need conversions between hours and seconds. We know that 1 mile is 5280 ft, and 1 hour is 3600 seconds. We also know that 1 inch is 2.54 cm, and of course 1 foot is 12 inches. 1 mile is therefore 1 mile * 5280 ft / mile = 5280 ft, 5280 ft = 5280 ft * 12 in/ft * 2.54 cm / in = 160934 cm, which is the same as 160934 cm * 1 m / (100 cm) = 1609.34 m, which in turn is the same as 1609.34 m * 1 km / (1000 m) = 1.60934 km. Thus 35 mi / hr = 35 mi / hr * (1.60934 km / 1 mi) = 56 (mi * km / (mi * hr) ) = 56 (mi / mi) * (km / hr) = 56 km / hr. We can in turn convert this result to m / s: 56 km/hr * (1000 m / km) * (1 hr / 3600 sec) = 15.6 (km * m * hr) / (hr * km * sec) = 15.6 (km / km) * (hr / hr) * (m / s) = 15.6 m/s. The original 35 mi/hr can be converted directly to ft / sec: 35 mi/hr * ( 5280 ft / mi) * ( 1 hr / 3600 sec) = 53.33 ft/sec. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qGen phy and prin phy prob 2.16: sports car rest to 95 km/h in 6.2 s; find acceleration YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: v0 = 0 km/hr, vf = 95 km/hr, dt = 6.2 s For the next steep we can convert the time to hours or convert the velocity to distance per second (km/s or m/s), I am going to convert the km/hr to m/s by multiplying the km/hr by ((1 hr/3600s) ×(1000m /km)) vf = (95 km/hr) ×(1 hr/3600s)×(1000m /km) =26.389m/s recall from the equation of motion vf = v0 +a*dt this implies that a = (vf - v0)/dt a = (vf - v0)/dt a = (26.389m/s - 0 m/s)/6.2 s a = 4.26 m/s^2 To change the acceleration to km/hr^2 we just have to multiply the m.s^2 by ((1 km/ 1000m)×(3600s/hr)^2) a = 4.26 m/s^2 × ((1 km/ 1000m)×(3600s/hr)^2) = 55209.6 km/hr^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ very confident
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Given Solution: `a** 95 km/hr = 95,000 m / (3600 sec) = 26.3 m/s. So change in velocity is `dv = 26.3 m/s = 0 m/s = 26.3 m/s. Average acceleration is aAve = `dv / `dt = 26.3 m/s / (6.2 s) = 4.2 m/s^2. Extension: One 'g' is the acceleration of gravity, 9.8 m/s^2. So the given acceleration is -4.2m/s^2 / [ (9.8 m/s^2) / 'g' ] = -.43 'g'. STUDENT QUESTION: How did we know that the final velocity was 0? INSTRUCTOR RESPONSE: The final velocity was 0 because the car came to rest. Summary of what we were given: * Initial velocity is 95 km/hr, or 26.3 m/s. * Final velocity is 0, since the car came to rest. * The velocity makes this change in a time interval of 6.2 seconds. We can easily reason out the result using the definition of acceleration: The acceleration is the rate at which velocity changes with respect to clock time, which by the definition of rate is (change in velocity) / (change in clock time) The change in velocity from the initial 0 m/s to the final 26.3 m/s is 26.3 m/s, so acceleration = change in velocity / change in clock time = 26.3 m/s / (6.2 s) = 4.2 m/s^2. We could also have used the equations of uniformly accelerated motion, with vf = 26.3 m/s, v0 = 0 and `dt = 6.2 seconds. However in this case it is important to understand that the definition of acceleration can be applied directly, with no need of the equations. (solution using equations: 2d equation is vf = v0 + a `dt, which includes our three known quantities; solving for a we get a = (vf - v0) / `dt = (26.3 m/s - 0 m/s) / (6.2 s) = 4.2 m/s^2.) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): in the given solution vf is 0m/s and v0 = 26.3 m/s, I am not sure I understand why, the question state that car rest to 95 km/h in 6.2 s this question Is not clear because it did not sate if the car start from rest then reach a velocity of 95 km/h or if the car came to rest from the 95 km/h. For my calculation I assumed that the car started from rest then if reaches a velocity of 95 km/hr the 6.2 second ------------------------------------------------ Self-critique Rating: ********************************************* Question: univ phy 2.66 train 25m/s 200 m behind 15 m/s train, accel at -.1 m/s^2. Will the trains collide and if so where? Describe your graph. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To solve this problem let’s call the train that is ahead train1 and the train behind the first we will call it train 2. If the two trains are to collide then their position has to be equal at some point in time. Let’s make the initial position of train2 be in the reference point so train2 is at position 0 while train one is in a position of 200m Train1 The velocity is a constant 15 m/s so the a = 0 m/s^2 The distance train 1 achieve with time is given by the equation ds1 = v0 `dt + .5 a `dt^2 +200m recall the ‘200m’is in the equation because train 1 started 200m ahead of the reference point ds1 = 15 m/s *dt + .5*0m/s^2*dt^2+200m ds1 = 15 m/s*dt+200m Train2 The velocity is a constant 25 m/s so the a = -0.1 m/s^2 The distance train 2 achieve with time is given by the equation ds2 = v0 `dt + .5 a `dt^2 (from the equation of motion) ds2 = 25 m/s *dt + .5*-0.1m/s^2*dt^2 ds2 = 25 m/s*dt-0.05 m/s^2 *dt^2 Recall that in order for the train to collide their distance has to be the same at some point in time so I will equate ds1=ds2 then solve for the time that will satisfy the equation ds1 = ds2 15 m/s*dt+200m = 25 m/s*dt-0.05 m/s^2 *dt^2 25 m/s*dt - 15 m/s*dt-0.05 m/s^2 *dt^2-200m=0 10 m/s*dt - 0.05m/s^2*dt - 2000m = 0 -0.05m/s^2*dt + 10 m/s*dt - 2000m = 0 From the quadratic equation dt = (-b +sqrt(b^2 - 4*a*c))/2*a and /or dt = (-b - sqrt(b^2 - 4*a*c))/2*a dt = (-10 +sqrt(10^2 - 4*-0.5*-200))/2*-0.05 dt = (-10 +sqrt(60))/-0.1 dt = 22.54 s or dt = (-10 - sqrt(10^2 - 4*-0.5*-200))/2*-0.05 dt = (-10 -sqrt(60))/-0.1 dt = 177.46 s Since the second train is heading straight towards the first a collision will take place after 22.54 seconds of travelling the second answer of 177.46 can be discarded. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ This was a good one but I think I got it correct
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Given Solution: If we assume the passenger train is at position x = 0 at clock time t = 0 we conclude that the position function is x(t) = x0 + v0 t + .5 a t^2; in this case a = -.1 m/s&2 and x0 was chosen to be 0 so we have x(t) = 25 m/s * t + .5 * (-.1m/s^2) * t^2 = 25 m/s * t - .05 m/s^2 * t^2. To distinguish the two trains we'll rename this function x1(t) so that x1(t) = 25 m/s * t - .05 m/s^2 * t^2. At t = 0 the freight train, which does not change speed so has acceleration 0 and constant velocity 15 m/s, is 200 m ahead of the passenger train, so the position function for the freight train is x2(t) = 200 m + 15 m/s * t . The positions will be equal if x1 = x2, which will occur at any clock time t which solves the equation 25 t - .05 t^2 = 200 + 15 t(units are suppressed here but we see from the units of the original functions that solutions t will be in seconds). Rearranging the equation we have -.05 t^2 + 10 t - 200 = 0. The quadratic formula tells us that solutions are t = [ - 10 +- sqrt( 10^2 - 4 * (-.05) * (-200) ) ] / ( 2 * .05 ) Simplifying we get solutions t = 22.54 and t = 177.46. At t = 22.54 seconds the trains will collide. Had the trains been traveling on parallel tracks this would be the instant at which the first train overtakes the second. t = 177.46 sec would be the instant at which the second train again pulled ahead of the slowing first train. However since the trains are on the same track, the accelerations of both trains will presumably change at the instant of collision and the t = 177.46 sec solution will not apply. GOOD STUDENT SOLUTION: for the two trains to colide, the 25 m/s train must have a greater velocity than the 15 m/s train. So I can use Vf = V0 + a('dt). 15 = 25 + (-.1)('dt) -10 = -.('dt) 'dt = 100 so unless the displacement of the 25 m/s train is greater than the 15 m/s train in 100 s, their will be no colision. 'ds = 15 m/s(100) + 200 m 'ds = 1700 m 'ds = 25 m/s(100) + .5(-.1)(100^2) = 2000 m. The trains collide. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe path difference for a 3d-order fringe is 3 wavelengths, so light from one slit travels 3 * 610 nm = 1830 nm further. The additional distance is equal to slit spacing * sin(18 deg), so using a for slit spacing we have a sin(18 deg) = 1830 nm. The slit spacing is therefore a = 1830 nm / sin(18 deg) = 5920 nm, or 5.92 * 10^-6 meters. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `q**** query gen phy problem 24.7 460 nm light gives 2d-order max on screen; what wavelength would give a minimum? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 univ 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: 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q**** query univ phy prob 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: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: