JamesD: please don\'t cut and paste your WRONG answer AGAIN here,unless you come

Question

JamesD: please don't cut and paste your WRONG answer AGAIN here,unless you come up with something different. Thanks.
A diffraction grating is used to analyze the wavelengths of lightpresent in the spectrum emitted by s distant star. To analyze the light properly, a separation of1mm between the center of the first order bright fringes is required when the spectrum isdisplayed on a screen located 1m away. a) How far apart should the slits be spaced in order to allow 477nmand 475nm wavelength maxima to be resolved? b) In order for these lines not to overlap, they must be narrowerthan their separation. What is the minimum number of lines the grating must have in order toachieve this result?
Please show ALL your work (with numbers and units), and I promiseto rate lifesaver! JamesD: please don't cut and paste your WRONG answer AGAIN here,unless you come up with something different. Thanks.
A diffraction grating is used to analyze the wavelengths of lightpresent in the spectrum emitted by s distant star. To analyze the light properly, a separation of1mm between the center of the first order bright fringes is required when the spectrum isdisplayed on a screen located 1m away. a) How far apart should the slits be spaced in order to allow 477nmand 475nm wavelength maxima to be resolved? b) In order for these lines not to overlap, they must be narrowerthan their separation. What is the minimum number of lines the grating must have in order toachieve this result?
Please show ALL your work (with numbers and units), and I promiseto rate lifesaver! JamesD: please don't cut and paste your WRONG answer AGAIN here,unless you come up with something different. Thanks.
A diffraction grating is used to analyze the wavelengths of lightpresent in the spectrum emitted by s distant star. To analyze the light properly, a separation of1mm between the center of the first order bright fringes is required when the spectrum isdisplayed on a screen located 1m away. a) How far apart should the slits be spaced in order to allow 477nmand 475nm wavelength maxima to be resolved? b) In order for these lines not to overlap, they must be narrowerthan their separation. What is the minimum number of lines the grating must have in order toachieve this result?
Please show ALL your work (with numbers and units), and I promiseto rate lifesaver!

Explanation / Answer

the positions of the first maxima from the central axisis y = L * sinbright or y1 = L *sin1 ----------(1) the position of the second maxima from the central axisis y2 = L * sin2 -----------(2) the separation between the first order bright fringes is y2 - y1 = 1 * 10-3 or L * sin2 - L *sin1 = 1 * 10-3 or L * (sin2 - sin1) = 1 *10-3 or (sin2 - sin1) = (1 *10-3/L) L = 1 m or (sin2 - sin1) = (1 *10-3/1) = 1 * 10-3 -------------(3) the condition for maxima in the interference pattern at theangle bright is d * sinbright = m * m =0,±1,±2,±3,................ when m = 1 then d * sin1 = 1 * or sin1 = (/d) -------------(4) when m = 2 then d * sin2 = 2 * or sin2 = (2/d) ------------(5) subtracting (4) from (5) we get sin2 - sin1 = ((2/d) -(/d)) = (/d) or d = (/sin2 - sin1)-------------(6) from (3) and (6) we get d = (/1 * 10-3) when 1 = 477 nm = 477 * 10-9 mthen d1 = (477 * 10-9/1 * 10-3) =477 * 10-6 m = 477 m when 2 = 475 nm = 475 * 10-9 mthen d2 = (475 * 10-9/1 * 10-3) =475 * 10-6 = 475 m the distance between the slits so that the maxima is resolvedis d = d1 - d2 = 477 - 475 = 2m b)Resolvance or "chromatic resolving power" for a device usedto separate the wavelengths of light is defined as R = (/) = 477 nm and = (477 - 475) nm = 2 nm Since the space between maxima for N slits is broken up intoN-2 subsidiary maxima, the distance to the first mimimum isessentially 1/N times the separation of the main maxima. This leadsto a resolvance for a grating of R = (/) = m * N or N = (1/m) * (/) m = 1 or N = (1/1) * (477 * 10-9/2 * 10-9) =238.5 the condition for maxima in the interference pattern at theangle bright is d * sinbright = m * m =0,±1,±2,±3,................ when m = 1 then d * sin1 = 1 * or sin1 = (/d) -------------(4) when m = 2 then d * sin2 = 2 * or sin2 = (2/d) ------------(5) subtracting (4) from (5) we get sin2 - sin1 = ((2/d) -(/d)) = (/d) or d = (/sin2 - sin1)-------------(6) from (3) and (6) we get d = (/1 * 10-3) when 1 = 477 nm = 477 * 10-9 mthen d1 = (477 * 10-9/1 * 10-3) =477 * 10-6 m = 477 m when 2 = 475 nm = 475 * 10-9 mthen d2 = (475 * 10-9/1 * 10-3) =475 * 10-6 = 475 m the distance between the slits so that the maxima is resolvedis d = d1 - d2 = 477 - 475 = 2m b)Resolvance or "chromatic resolving power" for a device usedto separate the wavelengths of light is defined as R = (/) = 477 nm and = (477 - 475) nm = 2 nm Since the space between maxima for N slits is broken up intoN-2 subsidiary maxima, the distance to the first mimimum isessentially 1/N times the separation of the main maxima. This leadsto a resolvance for a grating of R = (/) = m * N or N = (1/m) * (/) m = 1 or N = (1/1) * (477 * 10-9/2 * 10-9) =238.5

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