- #1
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#1 A diffraction grating having 500lines/mm diffracts visible light at 30 degrees([tex]\pi/6[/tex]) What is its wavelength?
the distance between slits is .002mm(500l^-1)
[tex] d\sin\frac{\pi}{6} = \lambda = 1000nm[/tex], which is too big. The answer is 500nm. What am I doing wrong?
#2
The slit spacing in a diffration grating is .0002m. The screen is 2.0m behind the grating. The distance of the first bright fringe is .004m. What is the wavelength of light?
y = L[tex]\tan\theta[/tex]
[tex] \tan^{-1}.004/2 = \theta = .0019 [/tex]
[tex] d\sin\theta = \lambda [/tex]
[tex] .0002m\sin(.0019) = \lambda = 400nm [/tex]
If this is right then I read from a graph wrong. If this is wrong, then I would appreciate anybody's input. Thank you for your time.
the distance between slits is .002mm(500l^-1)
[tex] d\sin\frac{\pi}{6} = \lambda = 1000nm[/tex], which is too big. The answer is 500nm. What am I doing wrong?
#2
The slit spacing in a diffration grating is .0002m. The screen is 2.0m behind the grating. The distance of the first bright fringe is .004m. What is the wavelength of light?
y = L[tex]\tan\theta[/tex]
[tex] \tan^{-1}.004/2 = \theta = .0019 [/tex]
[tex] d\sin\theta = \lambda [/tex]
[tex] .0002m\sin(.0019) = \lambda = 400nm [/tex]
If this is right then I read from a graph wrong. If this is wrong, then I would appreciate anybody's input. Thank you for your time.