Location of maximum desctructive intereference

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Gravitino22
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Homework Statement


Two radio sources in the xy plane are in phase and emitting the same wavelength, λ. Their locations are at coordinates (0,0) and (3λ,0). Show that the largest value of y at which fully destructive intereference may occur at coordinates (0, y) is given by
y=35λ/4



Homework Equations



Maximum destructive interference : dsinθ=(m+1/2)λ it’s derived from youngs two slit experiment
and ∆L = dsinθ <<< was used to derive the above, d is the source separation.


The Attempt at a Solution



My train of thought was that since we want the maximum difference in length the angle must be maximum but if the angle was 90 the waves wouldn’t even meet. So I constructed a right triangle from the problem and deduced that the difference in length would be
∆L =y-sqrt(y^2-9λ^2)=3λsinθ
And this is where i got stuck because I am confused on which angle we are looking at.


Any help is greatly appreciated ! thanks
 
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one bump because I still haven't solved this and my test is this week =/. Thanks for the help
 
Gravitino22 said:

The Attempt at a Solution



My train of thought was that since we want the maximum difference in length the angle must be maximum but if the angle was 90 the waves wouldn’t even meet. So I constructed a right triangle from the problem and deduced that the difference in length would be
∆L =y-sqrt(y^2-9λ^2)=3λsinθ
That's a reasonable start, but I see a couple of problems...

1. If the right triangle vertices are at (0,0), (3λ,0), and (0,y), then the legs of the right Δ are y and 3λ, and your sqrt expression should represent the hypotenuse (distance from (0,y) to (3λ,0). As you have written it, it does not represent the hypotenuse ... but a simple correction will fix that.

And this is where i got stuck because I am confused on which angle we are looking at.

2. I wouldn't worry about the angle. As you probably know, ∆L must be (m+1/2)λ here, where m is 0, ±1, ±2, etc. You can use this fact with the expression you wrote for ∆L.