#$&* course Phy 122 8:30 pm July 2 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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Given Solution: `aThe path difference for a 3d-order fringe is 3 wavelengths, so light from one slit travels 3 * 610 nm = 1830 nm further. The additional distance is equal to slit spacing * sin(18 deg), so using a for slit spacing we have a sin(18 deg) = 1830 nm. The slit spacing is therefore a = 1830 nm / sin(18 deg) = 5920 nm, or 5.92 * 10^-6 meters. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `q**** query 1 phy problem 35.54 11th edition 35.52 (37.46 10th edition) normal 477.0 nm light reflects from glass plate (n=1.52) and interferes constructively; next such wavelength is 540.6 nm. How thick is the plate? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It is twice the thickness to find the path difference. 477 nm / 1.52 540.6 nm / 1.52 find the first integer multiple of 477 nm / 1.52 (has to also an int. multiple of 540.6 nm / 1.52) 477 / 63.6 = 8.5, 2 * 477 / 63.6 = 17 wavelengths 17 wavelengths of 477 nm light 17 * 477 = 8109 8109 / 540.6 = 15 distance 17 * 477 nm / 1.52 = 5335 nm. thickness of the pane is 5335 nm / 2 = 2667 nm confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ok ------------------------------------------------ Self-critique Rating: ok" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!