course Phy 242 ??????????????yassignment #018
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18:21:28 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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18:21:30 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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18:21:33 **** 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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18:21:35 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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18:45:43 **** query univ phy 36.59 phasor for 8 slits
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RESPONSE --> For the slits: 3'pi/4, 5'pi/4, 7'pi/4 the fourth slit of each of the following will have a vector in the opposite direction. For the slits: 6'pi/4 the second following slit with have antiparallel direction.
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18:46:03 ** 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 --> I understand that there is a lot more needed to understand this problem.
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??????O??|?? assignment #019 ?????????? Physics II 11-27-2006
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18:46:45 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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18:46:47 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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18:46:50 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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18:46:51 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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18:46:53 **** gen phy how many wavelengths comprise the thickness of the foil?
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18:46:55 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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€Y???z^?????x??assignment #020 ?????????? Physics II 11-27-2006
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18:51:04 Query introductory set #1, 1-8 Explain how we calculate the magnitude and direction of the electrostatic force on a given charge at a given point of the x-y plane point due to a given point charge at the origin.
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RESPONSE --> F = k q1 * Q / r^2 direction of the electrostatic force find arctan(y / x), adding 180 deg if x is negative. If q1 and Q are either both positive or both negative, that is the direction. If q1 and Q are unlike the direction of the field is opposite the direction.
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18:51:09 ** The magnitude of the force on a test charge Q is F = k q1 * Q / r^2, where q1 is the charge at the origin. The force is one of repulsion if q1 and Q are of like sign, attraction if the signs are unlike. The force is therefore directly away from the origin (if q1 and Q are of like sign) or directly toward the origin (if q1 and Q are of unlike sign). To find the direction of this displacement vector we find arctan(y / x), adding 180 deg if x is negative. If the q1 and Q are like charges then this is the direction of the field. If q1 and Q are unlike then the direction of the field is opposite this direction. The angle of the field would therefore be 180 degrees greater or less than this angle.**
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RESPONSE --> ok
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18:51:42 Explain how we calculate the magnitude and direction of the electric field at a given point of the x-y plane point due to a given point charge at the origin.
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RESPONSE --> Same as the other problem.
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18:51:45 ** The magnitude of the force on a test charge Q is F = k q1 * Q / r^2, where q1 is the charge at the origin. The electric field is therefore F / Q = k q1 / r^2. The direction is the direction of the force experienced by a positive test charge. The electric field is therefore directly away from the origin (if q1 is positive) or directly toward the origin (if q1 is negative). The direction of the electric field is in the direction of the displacement vector from the origin to the point if q1 is positive, and opposite to this direction if q1 is negative. To find the direction of this displacement vector we find arctan(y / x), adding 180 deg if x is negative. If q1 is positive then this is the direction of the field. If q1 is negative then the direction of the field is opposite this direction, 180 degrees more or less than the calculated angle. **
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RESPONSE --> ok
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