course PHY 202 gЌFzassignment #029
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19:36:37 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 --> Since we are dealing with the speed of light, I will use the formula: Velocity= speed of light/ index of refraction for the material. I will find the speed of the red light and then the speed of the violet light and divde them to get the ratio or percent by wich the red light exceeds the violet light. Since wavelenths are given in the problem, I can use those in place of values in the table of index refractions because those values correspond to a wavelength of 569 nm. v=3.0E8/7.0E-7=4.28E14 for the red light v=3.0E8/4.0E-7=7.4E14 for the violet light v of red light/v of violet light = .579.
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19:37:49 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 --> Looks like my answer was an attempt at the second part of the solution. i just carried out my formula incorrectly. I understand now..
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19:49:02 **** 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 --> Given the formula for principal maxima grating we have sin omega=interger(wavelenth)/distance b/t slits. Assuming normal incidence, we have sin(90 degrees)=m(4ED-7)/75 m slit Solving for m, we get 3.0E-5 m. Although I am not sure if the question was asking us to solve for m because it is asking for the width of the spectrum.
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19:49:37 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
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f髳݃x assignment #029 029. `Query 18 Physics II 08-01-2007"