Solve Slit Grating Problem: Compute Wavelength of Light

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SUMMARY

The slit grating problem involves calculating the wavelength of light from a gaseous discharge tube incident on a grating with a slit separation of 2.23 μm, producing sharp maxima at angles θ = -17.1° and +17.1°. The correct formula to use is nλ = d sin θ, where n represents the order of the maxima. In this scenario, the absence of a central maximum indicates destructive interference at θ = 0, suggesting the use of n(λ + 1/2) = d sin θ for calculations. The discussion highlights the complexity of the problem and the need for a clear understanding of interference patterns.

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  • Understanding of wave interference principles
  • Familiarity with the diffraction grating equation nλ = d sin θ
  • Knowledge of coherent light sources and their properties
  • Basic trigonometry for angle calculations
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  • Explore the concept of destructive interference and its implications in optics
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1. With light from a gaseous discharge tube incident normally on a grating with slit separation 2.23 μm, sharp maxima for a certain wavelength of light are produced at angles θ = -17.1° and +17.1°. Compute the wavelength of the light in nanometers.



2. I thought this would work... (wavelength) = d * sin θ, but it turns out its a bit more complicated, i can't find anything in my books to describe a situation like this or how to solve it.
 
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I would think the equation you mentioned n \lambda = d sin\theta would work. The only think I can think of is that the problem seems to imply that there is no central maximum, meaning that for \theta =0 the superimposed waves destrucively interfere. In that case, I think you use n(\lambda + 1/2) = d sin\theta , but I'm not sure about that.
 
It seems to me that there are not enough information to solve the question. A maxima is formed from a double slit when the difference between the distance from each slit to a point is a multiple of the wavelength. If the light is coherent and basically the same, there should be a zeroth order maxima between the slits.
 

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