YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A third order fringe is going to travel 3 times further for every one wavelength, so the path for the one slit will be 610nm(3) which is 1830nm. We then need to find how far apart the slits are based off of this information and the light being observed at 18 degrees. We do this using the formula Slit space * sin(18degrees)= distance difference (1830nm) solving for the slit space gives a space of 5922 nm.
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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: We know that we are working with 2d and that one light has a max at 460nm We also should know that when we are dealing with 2d, the max would be 2x the distance which would make it 920nm which we conclude to be our wavelength Once we figure this out the problem is fairly simple. The minimum value states that this wavelength would only be ½ of the actual wavelength, so the minimum should be 1840 nm. We could continue using other harmonics, like the minimum could be 3/2 of the wavelength or 4/3.
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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 didn’t think about the visible range when finishing my solution. Also some of my wording is not accurate. Also I stated that we could use 4/3 to calculate a minimum but I don’t think this works I think I went in the opposite direction when thinking of my harmonics and meant 4/2. This was what I did already. Self-critique Rating: 3 "