Light of wavelength 500 nm is incident normally on a diffraction grating. If the third-order maximum of the diffraction pattern is observed at 32.0°, (a) what is the number of rulings per centimeter for the grating? (b) Determine the total number of primary maxima that can be observed in this situation.
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or any interference maximum for this light going through this grating,
sinθ=m(
d
λ
)
=
2.83×10
−6
m
m(5.00×10
−7
m)
=m(0.177)
For sinθ≤1, we require that m(0.177)≤1 or m≤5.66.
Because m must be an integer, its maximum value is really 5.
Therefore, the total number of maxima is 2m+1=11
Explanation:
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