Question #214918

A diffraction grating produces a third-order bright spot for wavelength 550 nm at 50o from the central maximum. What is/are the other angle(s) where the bright spot can occur? [angle is expressed in 2 significant figures]

A. 8o

B. 18o

C. 27o

D. 50o

1
Expert's answer
2021-07-19T09:35:44-0400

For a diffraction gratingdsinθm=mλ\text{For a diffraction grating}\newline d\sin\theta_m = m\lambda

λ=550nm\lambda= 550 nm

θ3=50°\theta_3 = 50\degree

d=mλsinθm=3550sin50°2154 nmd= \frac{m\lambda}{\sin\theta_m}=\frac{3*550}{\sin50\degree}\approx2154\ nm

m=dsinθmλm= \frac{d\sin\theta_m}{\lambda}

A.if θm=8°m=dsin8°λ=2154sin8°550=0.54mZA.\text{if }\theta_m= 8\degree\newline m= \frac{d\sin8\degree}{\lambda}= \frac{2154\sin8\degree}{550}=0.54\newline m\notin Z

m=dsin8°λ=2154sin8°550=0.54m= \frac{d\sin8\degree}{\lambda}= \frac{2154\sin8\degree}{550}=0.54

B.if θm=18°m=dsin18°λ=2154sin18°550=1.21mZB.\text{if }\theta_m=18\degree\newline m= \frac{d\sin18\degree}{\lambda}= \frac{2154\sin18\degree}{550}=1.21\newline m\notin Z

C.if θm=27°m=dsin27°λ=2154sin27°550=1.77mZC.\text{if }\theta_m= 27\degree\newline m= \frac{d\sin27\degree}{\lambda}= \frac{2154\sin27\degree}{550}=1.77\newline m\notin Z

D.if θm=50°m=dsin50°λ=2154sin50°550=3mZD.\text{if }\theta_m= 50\degree\newline m= \frac{d\sin50\degree}{\lambda}= \frac{2154\sin50\degree}{550}=3\newline m\in Z

Answer: D.50°\text{Answer: D}.50\degree


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