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Light of wavelength 587.5 nm illuminates a slit of width 0.75 mm. (a) At what distance from the slit should a screen be placed if the first minimum in the diffraction pattern is to be 0.85 mm from the central maximum? (b) Calculate the width of the central maximum.

a) 1.09 $\mathrm{m}$

b) 1.7 $\mathrm{mm}$

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Cornell University

Rutgers, The State University of New Jersey

Numerade Educator

Hope College

for this problem on the topic of wave optics were given the wavelength of light, and we're told that it illuminates a split of a given with. We want to know the distance from the slip that the screen should be placed to have a first minimum in the diffraction pattern at 0.85 millimeters from the central maximum. We then want to calculate the width of the central maximum. Now we know that dark bands or minima, occur when science theater is equal to times lambda over a. Now for the first minimum, we know em is equal to one, and the distance from the center of the central maximum y one is equal two l 10. The to which we can approximate for the first minimum to be our signed data or simply l times the wavelength lambda over the width of the slit A and so the distance needed to the screen Oh is equal to why one times the with the slit a divided by the Waveland lambda. So if we substitute our values in here, this is zero 0.85 times 10 to the minus 3 m times 0.75 times 10 to the minus. The meters divided by Lambda Waveland is 587 0.5 times 10 to the minus nine meters. So calculating we get the distance of the screen. How to be one 0.1 meters now in part B, we want to find the central the width of the central maximum. The winner of the central maximum is to buy one, and so this is to time zero 0.85 millimeters, which gives US one 0.7 millimeters.

University of Kwazulu-Natal