631 CH 17 Hw Problem 17.20 - Enhanced-with Feedback C) 12 of 20 PartA Constants

Question

631 CH 17 Hw Problem 17.20 - Enhanced-with Feedback C) 12 of 20 PartA Constants A 500 ines per mm diffraction grating is luminated by light of wavelength 540 nm What is the maximum diffraction order seen? Express your answer as an integer You may war toreview (P92 M.547 Correct Here we leam how to find the seen maximum difraction order Part B starting from zero dffraction order to themaximum vsible iffraction ode? What is the angle of each difraction order Enter your wers using two signMeant ngures in ascending order Mparad bycoms X Incorrect, Try Again; 4 attempts remaining denhe

Explanation / Answer

Given

wavelength is lambda = 540 nm

the number of lines per mm is N = 500

the space between two lines is d = l/N = 10^-3/500 m = 2*10^-6 m or 2 micro meters

Part A

we know that for maximum order the angle is 90 degrees so

condition is d sin theta = m*lambda

m = d sin theta / lambda

m = 2*10^-6sin90 /(540*10^-9)

m = 3.70

so the maximum order i s 3

Part B

the angles are for first order m=1  

d sin theta = m*lambda

sin theta = 1*540*10^-9/(2*10^-6) = 0.27

theta = arc sin (0.27) = 15.66 degrees

for second order m=2  

d sin theta = m*lambda

sin theta = 2*540*10^-9/(2*10^-6) = 0.54

theta = arc sin (0.27) = 32.68 degrees

for second order m = 3  

d sin theta = m*lambda

sin theta = 3*540*10^-9/(2*10^-6) = 0.81

theta = arc sin (0.27) = 54.1 degrees

for second order m = 4

d sin theta = m*lambda

sin theta = 4*540*10^-9/(2*10^-6) = 1.08

which is not possible because sin value is not greater than 1

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