course PHY202 滜ϴassignment #019
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14:42:06 **** Query experiment 29 **** how far from the V of the thread did you measure the separation of bright spots, how far apart were the spots, and what the you estimate was the average separation of threads at the point where you shined the light through
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RESPONSE --> didn't do experiment
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14:42:09 ** STUDENT RESPONSE: The distance from my V of thread to the wall that was my 'screen' was 6 meters, I tried to move it farther back but If I go much farther the dots are harder to see. I played with the cat too much with my laser pointer and have nearly used up the battery. ** It's hard not to do that with a cat. One battery per cat is about right. Beyond that they tend to get psychotic, though how you separate pyschosis from normal behavior in a cat I'm not sure. ** I measured the spots on the wall to be about 0.29 centimeters apart. The threads on my plastic were about 13 threads per centimeter where I tried to keep the lazer pointer which would make them have 7.7 * 10^-4 meters in between them.
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RESPONSE --> ok
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14:42:13 **** explain how you determined the approximate wavelength of light from your data
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RESPONSE --> ok
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14:42:15 STUDENT RESPONSE: Well the first thing I did was to measure the distance between the dots which came out to be about 0.29 cm. Then since I knew the distance to my screen was right at 600 cm I used the trigonomic function for tangent to solve for the angle theta which in this case was my 1st order angle for my dot. tan of theta = opposite / adjacent opposite = 0.29cm adjacent = 600 cm so angle theta = 0.02768 Then I used the formula... sin of theta = m * wavelength / d to solve for wavelength m = 1 d = 7.7 * 10^-4 For the wavelength I got 3.72 * 10^-7 m or 372 nm which can not be right because it is not even on the visible spectrum. From this experiment I would expect to get around 700 nm because that is about what red should be and the lazer pointer has red light. I honestly expected to get something that was not even close due to crudeness of the apparati that I have rigged up. None the less, providing that I did everything correctly calculation wise and didn't miss anything, this technique should work if I tweeked it a bit more and refined my tools some more. INSTRUCTOR COMMENT: ** It's hard to be much more accurate than that with the crude apparatus, but it shows you what's going on here. **
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RESPONSE --> ok
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14:45:45 gen phy problem 24.44 foil separates one end of two stacked glass plates; 28 lines observed for normal 650 nm light gen phy what is the thickness of the foil?
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RESPONSE --> t=0.5*2*670nm =9.05*10^3nm=9.05um
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14:45:49 STUDENT SOLUTION: To solve this problem, I refer to fig. 24-31 in the text book as the problem stated. To determine the thickness of the foil, I considered the foil to be an air gap. I am not sure that this is correct. Therefore, I used the equation 2t=m'lambda, m=(0,1,2,...). THis is where the dark bands occur . lambda is given in the problem as 670nm and m=27, because between 28 dark lines, there are 27 intervals. Solve for t(thickness): t=1/2(2)(670nm) =9.05 *10^3nm=9.05 um INSTRUCTOR RESPONSE WITH DIRECT-REASONING SOLUTION:** Your solution looks good. Direct reasoning: ** each half-wavelength of separation causes a dark band so there are 27 such intervals, therefore 27 half-wavelengths and the thickness is 27 * 1/2 * 670 nm = 9000 nm (approx) **
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RESPONSE --> ok
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14:47:29 **** gen phy how many wavelengths comprise the thickness of the foil?
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RESPONSE --> 2-90.5um=m(6.70*10^-7m) m=27wavelengths
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14:47:35 GOOD STUDENT SOLUTION: To calculate the number of wavelengths that comprise the thickness of the foil, I use the same equation as above 2t=m'lambda and solve for m. 2(9.05 um)=m(6.70 *10^-7m) Convert all units to meters. m=27 wavelengths.
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RESPONSE --> ok
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