What Value of d/a Causes Diffraction to Eliminate the Third Bright Side Fringe?

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SUMMARY

The discussion focuses on determining the ratio of d/a in a double-slit system that results in the elimination of the third bright side fringe due to diffraction effects. The relevant equations include I(θ) = I_mcos²(β)(sin(α)/α)², β = (πd/λ)sin(θ), and α = (πa/λ)sin(θ). The critical conditions are d sin(θ) = 3λ and a sin(θ) = (3 + 1/2)λ, indicating that the angle θ corresponds to the third maximum in the interference pattern and the first minimum in the diffraction pattern. The correct ratio of d/a is essential for solving this problem accurately.

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  • Knowledge of the relationship between wavelength (λ), slit width (a), and slit separation (d)
  • Ability to manipulate trigonometric functions in the context of wave phenomena
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Zhalfirin88
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Homework Statement


(a) In a double-slit system, what value of d/a causes diffraction to eliminate the third bright side fringe?

Homework Equations


I(\theta) = I_mcos^2(\beta)(\frac{sin(\alpha)}{\alpha})^2

\beta =\frac{\pi d}{\lambda} sin \theta

\alpha = \frac{\pi a}{\lambda} sin \theta

d sin\theta = 3 \lambda

a sin\theta = (3+\frac{1}{2}) \lambda

The Attempt at a Solution


I solved for a and d in equations 4 and 5 above and got a ratio of 0.857, which isn't right. I'm not sure really how to go about this, I have a feeling that I need to use equations 2 and 3 but I'm not making the connections.
 
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Is everyone as lost as I am with this problem?
 
Zhalfirin88 said:
d sin\theta = 3 \lambda
Yes, the angle θ corresponds to the 3rd maximum (not counting central max) in the 2-slit interference term cos2β

a sin\theta = (3+\frac{1}{2}) \lambda
Actually, the idea here is that θ also corresponds to the first minimum of the diffraction term [(sinα)/α]2
 

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