course Phy 202 020. `Query 18
.............................................
Given Solution: `aThe 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 1.028, or 102.8%. So the precent difference is about 2.8%. 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ############ This is something I should know, but sometimes lack simple math= where does the 2.8 percent “difference” come from?? Why wouldn’t the answer be 102.8%?? (I’m using the solution’s numbers)
.............................................
Given Solution: `aGOOD 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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Spectrum width?? I thought what I did would have been right, what is the difference??