Find the Optimum Spot to Minimize Wave Action in Large Water Tank Experiment

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SUMMARY

The discussion centers on determining the optimal position to minimize wave action in a large water tank experiment involving two openings spaced 5.00 m apart. The relevant equations include the interference condition for wave fronts, specifically [P(n)*S(1)] - [P(n)*S(2)] = (n - 1/2)*wavelength and sin ANGLE(n) = (n - 1/2)*wavelength/(distance between sources). The key insight is that standing at points where the distance from the openings differs by half a wavelength will result in minimal wave action. The wavelength in this experiment is established as 3.00 m.

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Homework Statement


In a large water tank experiment, water waves are generated with straight, parallel wave fronts, 3.00 m apart. The wave fronts pass through two openings 5.00 m apart in a long board. The end of the tank is 3.00 m beyond the board. Where would you stand, relative to the perpendicular bisector of the line between the openings, if you want to receive little or no wave action?

2. Homework Equations

[P(n)*S(1)] - [P(n)*S(2)] = (n - 1/2)*wavelength

sin ANGLE(n) = (n - 1/2)*wavelength/(distance between sources)

x(n) / L = (n - 1/2) * wavelength / (distance between sources)


The Attempt at a Solution


I tried using the third formula, allowing the n be 1, the wavelength 3.00 m and the distance between the two sources 5.00 m, but i don't know what value i should assign x(n) and L. I don't know if I'm doing the question right. Can anyone help?
 
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This sounds like 2 slit interference. The points of no wave motion would be the places where the distances from the slits differ by 1/2 wavelength..
 

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