course PHY 202 I was just letting you know I resubmitted labs on RC Current and Charging and Discharging a Capacitor. I first submitted them a few weeks ago. I didn't see them on the access portfolio page. Let me know if you had any problems receiving them.
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22:06:46 Principles of Physics and General Physics Problem 24.14: By what percent does the speed of red light exceed that of violet light in flint glass?
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RESPONSE --> First, we were to find the incices of refraction for each on the graph. It appears to be 1.650 for the violet light and 1.620 for the red light. We then were to find the percentage by which the red exceeded the violet. To do this one could simply divide the two together. 1.650/1.620=1.0185*100=101.85 To find how much it exceeds violet we just take the excess over 100% so it's 1.85%
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22:09:08 The respective indices of refraction for violet and red light in flint glass appear from the given graph to be about 1.665 and 1.620. The speed of light in a medium is inversely proportional to the index of refraction of that medium, so the ratio of the speed of red to violet light is the inverse 1.665 / 1.62 of the ratio of the indices of refraction (red to violet). This ratio is about 1.0028, or 100.28%. So the precent difference is about .28%. It would also be possible to figure out the actual speeds of light, which would be c / n_red and c / n_violet, then divide the two speeds; however since c is the same in both cases the ratio would end up being c / n_red / ( c / n_violet) = c / n_red * n_violet / c = n_violet / n_red, and the result would be the same as that given above.
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RESPONSE --> ok, I worked this one with your numbers and I got 2.78%. Am I missing something or was that a typo?
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22:13:59 **** query gen phy problem 24.34 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 --> First we use the formula sin'theta=m'lambda/distance In the above it mentions a 7500 line/cm grating. We have to find distance with the formula D=1/N meters So we solve for D and get 1/750000m=1.3*10^-6 Now we are able to substitute into this formula with 400nm and 750nm: 400*10-7/1.3*10^-6='sin theta 'sin theta = .2 theta =.2deg 750*10-7/1.3*10^-6='sin theta 'sin theta = .5625 theta =34.23deg With this we can solve to the distances of the width of the spectrum. WE kow that the screen is 2.3 meters away. 2.3* 'tan 34.23 =1.565 2.3'* tan 17.45 = .723 Now we simply subtract those two 1.565-.723f=.8519
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22:15:34 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 --> ok, This was a different problem than the one in the book. I just solved with the data above.
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22:15:37 **** query univ phy 36.59 phasor for 8 slits
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RESPONSE --> ok
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22:15:39 ** 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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RESPONSE --> k
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~¸Μ׃yn assignment #019 rYɐZ Physics II 05-04-2006
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22:52:41 Principles of Physics and General College Physics Problem 24.54: What is Brewster's angle for an air-glass interface (n = 1.52 for glass)?
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RESPONSE --> The angle would be tan(n1/n2)
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22:54:13 Brewster's angle is the smallest angle theta_p of incidence at which light is completely polarized. This occurs when tan(theta_p) = n2 / n1, where n2 is the index of refraction on the 'other side' of the interface. For an air-glass interface, n1 = 1 so tan( theta_p) = n2 / 1 = n2, the index of refraction of the glass. We get tan(theta_p) = 1.52 so that theta_p = arcTan(1.52). This is calculated as the inverse tangent of 1.52, using the 2d function-tan combination on a calculator. WE obtain theta_p = 56.7 degrees, approximately.
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RESPONSE --> ok
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22:59:02 gen phy problem 24.43 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 --> We use the formula 2t=m* 'lambda Since we have 28 lines we have only 27 intervals. Our wavelength is 670nm. 2t=27*670 2t=18090 t=9045nm
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23:01:38 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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23:02:42 **** gen phy how many wavelengths comprise the thickness of the foil?
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RESPONSE --> Use the same fomula and solve for m. As you stated before there are 27 half wavelengths. 2(9045)= m*670 18090=m670 27=m
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23:03:13 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 --> 2t=m'lambda 2(9.05 um)=m(6.70 *10^-7m) Convert all units to meters. m=27 wavelengths.
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