#$&* course phy 122 2/8 3There are also questions in this document. 019. `Query 17
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Given Solution: I would like to understand this more. I will continue to read over the problem until it makes more sense to me. `aThe path difference for a 3d-order fringe is 3 wavelengths, so light from one slit travels 3 * 610 nm = 1830 nm further. The additional distance is equal to slit spacing * sin(18 deg), so using a for slit spacing we have a sin(18 deg) = 1830 nm. The slit spacing is therefore a = 1830 nm / sin(18 deg) = 5920 nm, or 5.92 * 10^-6 meters. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q**** query gen phy problem 24.7 460 nm light gives 2d-order max on screen; what wavelength would give a minimum? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Second order max occurs when the path difference is 2 wavelengths , and a minimum occurs when path difference is a whole number of wavelengths plus a half wavelength. Path difference is 2*460=920 nm First order minimum would occur for a path difference of ½ wavelength, ½ of the wavelength would be 920 nm and a wavelength would be 1860nm The minimum would also occur if 3/2 of the wavelength is 920 nm, in which case the wavelength would be 2/3 *920nm=613nm A minimum could also occur if 5/2 of the wavelength is 920nm, but this would give us a wavelength of about 370 nm confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Feeling confused.
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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): I’m going to continue reviewing until I give myself a headache or I understand this material. ------------------------------------------------ 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 for constructive interference is an integer multiple of the wavelength. The path difference here is twice the thickness. Wavelengths in glass are 477nm /1.52 and 540.6nm/1.52 So we know that double the thickness is an integer multiple of 477nm/1.52 and also an integer multiple of 477nm/1.52. We need to find the first integer multiple of 477nm/1.52, that is also an integer multiple of 540.6 nm/ 1.52 We first find an integer multiple of 477 that is also an integer multiple of 540.6. Integer multiples of 540.6 are 540.6, 1081.2, etc Dividing these numbers by 477 nm we obtain remainders 62.6, 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 477nm 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 follow that 17 wavelengths of 477nm/1.52 light span the same distance as 15 wavelengths of 540nm/1.52light. This distance is 17*477/1.52= 5335nm. The thickness is therefore 5335nm/2=2667nm If interference was destructive: n x 477nm/1.52=(n-1) x 540.6 nm/1.52 477nm=540.6 nm-540.6 N x (540.6-477) = 540.6 N x 63.6=540.6 N=540.6/63.6=8.5 This is an integer plus ½ integer of wavelengths, which would result in destructive interference for both waves. Multiplying 8.5 wavelengths by 477/1.52 we get a round trip distance 2667 nm, of thickness 1334nm confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Oh my…
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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): That was completely over my head. I’m hoping that I’m just having a “foggy” day and that this will make sense once I read over it once more. " 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 for constructive interference is an integer multiple of the wavelength. The path difference here is twice the thickness. Wavelengths in glass are 477nm /1.52 and 540.6nm/1.52 So we know that double the thickness is an integer multiple of 477nm/1.52 and also an integer multiple of 477nm/1.52. We need to find the first integer multiple of 477nm/1.52, that is also an integer multiple of 540.6 nm/ 1.52 We first find an integer multiple of 477 that is also an integer multiple of 540.6. Integer multiples of 540.6 are 540.6, 1081.2, etc Dividing these numbers by 477 nm we obtain remainders 62.6, 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 477nm 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 follow that 17 wavelengths of 477nm/1.52 light span the same distance as 15 wavelengths of 540nm/1.52light. This distance is 17*477/1.52= 5335nm. The thickness is therefore 5335nm/2=2667nm If interference was destructive: n x 477nm/1.52=(n-1) x 540.6 nm/1.52 477nm=540.6 nm-540.6 N x (540.6-477) = 540.6 N x 63.6=540.6 N=540.6/63.6=8.5 This is an integer plus ½ integer of wavelengths, which would result in destructive interference for both waves. Multiplying 8.5 wavelengths by 477/1.52 we get a round trip distance 2667 nm, of thickness 1334nm confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Oh my…
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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): That was completely over my head. I’m hoping that I’m just having a “foggy” day and that this will make sense once I read over it once more. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!