course Phy 202
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12:24:25 **** query gen phy problem 24.36 width of 1st-order spectrum of white light (400 nm-750nm) at 2.3 m from a 7500 line/cm grating **** gen phy what is the width of the spectrum?
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RESPONSE --> IN the book it says a 8500 line/cm, this probably will make a difference from what you might have: sin (theta) = m(lambda)/d m = 1 d = 8500 lines/cm lambda= 410 through 700 I am not sure where to go from here. I know you would have to use this equation, but i am not sure what to use for lambda
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12:27:18 GOOD STUDENT SOLUTION We are given that the spectrum is from 400-750 nm. We are also given that the screen is 2.3 meters away and that the grating is 7500 lines/cm. To find this I will find where 400 nm wavelength falls on the screen and also where 750 nm wavelength falls onto the screen. Everything in between them will be the spectrum. I will use the formula... sin of theta = m * wavelength / d since these are first order angles m will be 1. since the grating is 7500 lines/cm, d will be 1/7500 cm or 1/750000 m. Sin of theta(400nm) = 1 * (4.0 * 10^-7)/1/750000 sin of theta (400nm) = 0.300 theta (400nm) = 17.46 degrees This is the angle that the 1st order 400nm ray will make. sin of theta (750nm) = 0.563 theta (750nm) = 34.24 degrees This is the angle that the 1st order 750 nm ray will make. We were given that the screen is 2.3 meters away. If we draw an imaginary ray from the grating to to the screen and this ray begins at the focal point for the rays of the spectrum and is perpendicular to the screen (I will call this point A), this ray will make two triangles, one with the screen and the 400nm angle ray and one with the screen and the 750 nm angle ray. Using the trigonomic function; tangent, we can solve for the sides of the triangles which the screen makes up. Tan of theta = opposite / adjacent tan of 34.24 degrees = opposite / 2.3 meters 0.6806 = opposite / 2.3 meters opposite = 1.57 meters tan of 17.46 degrees = opposite / 2.3 meters opposite = 0.72 meters So from point A to where the angle(400nm) hits the screen is 0.72 meters. And from point A to where the angle(750nm) hits the screen is 1.57 meters. If you subtract the one segment from the other one you will get the length of the spectrum on the screen. 1.57 m - 0.72 m = 0.85 meters is the width of the spectrum on the screen. CORRECTION ON LAST STEP: spectrum width = 2.3m * tan (31.33)) - 2.3m * tan (17.45) = 0.68m
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RESPONSE --> sin of theta (750nm) = 0.563 theta (750nm) = 34.24 degrees Can you explain this step. I am sure it is just a calculation done, I am just not sure how he got the degrees from .563 Other than that i follow the students solution. And see where I was lost
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12:27:22 **** query univ phy 36.59 phasor for 8 slits
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RESPONSE -->
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12:27:24 ** If you look at the phasor diagram for phi = 3 pi / 4 you will see that starting at any vector the fourth following vector is in the opposite direction. So every slit will interfere destructively with the fourth following slit. This is because 4 * 3 pi / 4 is an odd multiple of pi. The same spacing will give the same result for 5 pi / 4 and for 7 pi / 4; note how starting from any vector it takes 4 vectors to get to the antiparallel direction. For 6 pi / 4, where the phasor diagram is a square, every slit will interfere destructively with the second following slit. For phi = pi/4 you get an octagon. For phi = 3 pi / 4 the first vector will be at 135 deg, the second at 270 deg (straight down), the third at 415 deg (same as 45 deg, up and to the right). These vectors will not close to form a triangle. The fourth vector will be at 45 deg + 135 deg = 180 deg; i.e., horizontal to the left. The next two will be at 315 deg (down and toward the right) then 90 deg (straight up). The last two will be at 225 deg (down and to left) and 360 deg (horiz to the right). The resulting endpoint coordinates of the vectors, in order, will be -0.7071067811, .7071067811 -0.7071067811, -0.2928932188 0, 0.4142135623 -1, 0.4142135623 -0.2928932188, -0.2928932188 -0.2928932188, 0.7071067811 -1, 0 0, 0 For phi = 5 pi / 4 each vector will 'rotate' relative to the last at angle 5 pi / 4, or 225 deg. To check yourself the first few endpoints will be -0.7070747217, -0.7071290944; -0.7070747217, 0.2928709055; 0, -0.4142040038 and the final endpoint will again be (0, 0). For 6 pi / 4 you will get a square that repeats twice. For 7 pi / 4 you get an octagon. NEW PROBLEM: The longest wavelength is 700 nm and slit spacing is about 1250 nm. The path difference can't exceed the slit spacing, which is less than double the 700 nm spacine. So there are at most central max (path difference zero) and the first-order max (path difference one wavelength). Note that there will be a second-order max for wavelengths less than about 417 nm. **
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yqyN~ޘw assignment #019 ]ĕ圯N Physics II 10-14-2005
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14:25:27 **** 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 --> The distance of the bright spots from the V of the thread was approximately 5.5 meters. I ran out of room in my apartment to probably get a better answer but this is what i could get with the space i had.
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14:28:11 ** 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 --> This is close to what i got. Although my battery is not that dead, i did drop it from a good distance on the kitchen floor when using the bar to do a past experiment, the pointer still seems to be working ok.
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14:28:22 **** explain how you determined the approximate wavelength of light from your data
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RESPONSE --> I had trouble with this because I know you have to use the sin of theta = m * wavelength / d equation but i was having trouble figuring out what numbers were to go where, other than m= 1.
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14:32:36 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 i see what this student did. I am having trouble getting to the sin of theta = m * wavelength / d equation, or just getting the information needed to use this equation. After doing the problem in the book like this I am kind of getting the hold of what to do. Are there any pointers that would help me out in finding the wavelength or in using the sin of theta = m * wavelength / d equation?
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14:37:22 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 --> I see in this case you would want to use 2t = m(lambda) or t = .5m(lambda) lambda = 650 nm I am confused though, on what to use for m.
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14:38:50 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 i see the reasoning on the dark bands and why there is only 27 and not 28
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14:41:31 **** gen phy how many wavelengths comprise the thickness of the foil?
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RESPONSE --> You would use the same equation and solve for m, using the answer found in the previous question
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14:41:43 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 --> This is what i did, I just forgot to work it out, but i see how you would do it.
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