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Secure | https/s1.lite.msu.edu/res/brookscole/serway/College Phy... a Principles of Biochen 2 Physics 131 Homewc C Apps Other bookmarks Kyla Walton v (Student) PHY131S18 Messages Courses Help Logout nu Contents Grades ? Light of wavelength 570 nm illuminates a single 0.72 mm wide slit. At what distance from the slit should a screen be placed if the first minimum in the diffraction pattern is to be 1.24 mm from the central maximum? 000791m Submit Answer Incorrect. Tries 2/20 Previous Tries What is the width of the central maximum? mit Answer Tries 0/20 NEW Anonymous 1 Reply (Wed Apr 4 06:26:50 pm 2018 (EDT)) Has anyone figured out how to solve the width of the central maximum in part 2? NEW Re: Anonymous 2 Reply. (Thu Apr 5.11:36:20 am 2018 (EDT)) The distance from the central maximum to the first minimum is given, that is only half of the width. Therefore you have to multiply that distance by 2, for the width of the central maximum.

Explanation / Answer

Given,

lambda = 570 nm ; d = 0.72 mm ; y = 1.24 mm

We know that

d sin(theta) = m lambda

for a single slit, m = 1

sin(theta) = lambda/d

sin(theta) = 570 x 10^-9/0.72 x 10^-3 = 7.92 x 10^-4

Also, sin(theta) = y/L

L = y/sin(theta)

L = 1.24 x 10^-3/(7.92 x 10^-4) = 1.57 m

Hence, L = 1.57 m

b)w = 2 x y

w = 2 x 1.24 = 2.48 mm

Hence, w = 2.48 mm

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