#$&* course phy 242 july 5 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.
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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 ********************************************* 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: If at the m=3, its at maximum, the red light is visible which means that the monochromatic light differs by a half wavelength. We can conclude that the one source of light travels 3 times further than the other. The possible wavelengths are 3/3.5*700 nm = 600 nm and 3.5/3*700 nm = 817 nm. We do not need to know the slit spacing to answer this question. 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): ok ------------------------------------------------ Self-critique Rating: ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!