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140 Mercury emits light with wavelengths 1 = 404.66 nm and 2 = 435.83 nm. This radiation is incident on a 1.00 cm wide diffraction
140 Mercury emits light with wavelengths 1 = 404.66 nm and 2 = 435.83 nm. This radiation is incident on a 1.00 cm wide diffraction grating and the diffraction pattern is formed on a screen beyond the grating. You would like the angular separation between the first two lateral light fringes due to the two wavelengths to be greater than 0.200.Does the angular separation between the two lines of light depend on the distance from the screen?A thousand slits in the lattice are sufficient to have the required angular separation
140 .. I1 mercurio emette luce con lunghezze d'onda 21 = 404,66 nm e A2 = 435,83 nm. Questa radiazione viene fatta incidere su un reticolo di diffrazione largo 1,00 cm e la figura di diffrazione si forma su uno schermo posto oltre il reticolo. Vorresti fare in modo che la separazione angolare tra le prime due frange luminose laterali dovute alle due lunghezze d'onda fosse maggiore di 0,200. . La separazione angolare tra le due righe luminose dipende dalla distanza dallo schermo? . Mille fenditure nel reticolo sono sufficienti per avere la separazione angolare richiesta?N = 1000 W= 1 cm 7 = 404, 66 am 9:90 08 2 = 435- 83 nm slit separation d = w 1X 10 - 2 N = 10 - 5 m 10 00 = > AD = 8 in " 435 x 10 9 J - Bin 1 404.66 *18 ) 10- 5 = > 101 = 2. 4 9 315 3109 - 2. 31916 42 45 10 = 0 174 This is less than the required or 2" ", we need more than 1000 8/its to achieve an angular separation greater than 0. 20Step by Step Solution
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