4 A diffraction grating is designed to spread out the 1st order spectrum of visi

Question

4 A diffraction grating is designed to spread out the 1st order spectrum of visible light such that the end of the spectrum is an angular location of 7o.o degrees. Ca What is the angular location of the the end of the spectrum? Ob Calculate the number of lines per centimeter of diffraction grating needed to produce this spread? Oc the 2nd order Spectrum what range of wavelengths of light will be included for this diffraction grating? Gd) What will be the angular range for the 2nd order spectrum?

Explanation / Answer

part a:

for a diffraction grating, we know that d*sin(theta)=m*lambda

where d=spacing between two slits

theta=angular location

m=order of the spectrum

lambda=wavelength

given that for m=1, lambda=700 nm , theta=70 degrees

then d=1*700*10^(-9)/sin(70)=7.4492*10^(-7) m

then for lambda=400 nm, theta=arcsin(1*400*10^(-9)/(7.4492*10^(-7)))=32.4776 degrees

part b:

number of lines per cm=1 cm/d=0.01/(7.4492*10^(-7))=13424.26

so 13424 lines per cm is required.

part c:

for m=2, lambda=d*sin(theta)/m

maximum wavelength=7.4492*10^(-7)*sin(90)/2=3.7426*10^(-7) m=374.26 nm

so lights of wavelength upto 374.26 nm will be included.

part d:

angular range will be 0 degree to 90 degree to include light of wavelength upto 374.26 nm

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