Double Slit Diffraction d/a

[Zhalfirin88]

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.

 

[Zhalfirin88]

Is everyone as lost as I am with this problem?

 

[Redbelly98]

$$d sin\theta = 3 \lambda$$

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