Finding second furthest distance of const interference from 2 sources

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JoeyBob
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Homework Statement
See attached
Relevant Equations
d2-d1=n(wavelength)
So I can find the wavelength using v/f = 0.8995. for the distances, d=x and d=sqrt(9.7^2+x^2). So the full equation would be

sqrt(9.7^2+x^2)-x=n(0.8995)

Now I am going to take the derivative of the left side to see where the maxima is.

0= sqrt(x2+9409/100)−x

Now this doesn't have a solution so there is no maximum? I guess that makes sense because as x increases the overall expression decreases.

But now I am pretty sure my whole approach is inccorect. I am not really sure now how to find the highest x where const interference occurs. I could just say n=10000, but the x where n=1000000000 would be even higher ect. I don't think taking the limit will help either.

The answer is 25.25 btw
 

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JoeyBob said:
Now I am going to take the derivative of the left side to see where the maxima is.
Your quadratic already determines the values of x at which the waves produce a local maximum.
And how do you get 0= sqrt(x2+9409/100)−x by taking a derivative of it? Looks like you simply set n=0.

First, I strongly recommend working symbolically. Only plug in numbers at the end. So you have ##\sqrt{w^2+x^2}-x=n\lambda##.
See if you can simplify that.
Next, think about what happens to n as x increases. Does it increase or decrease?
 
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haruspex said:
Your quadratic already determines the values of x at which the waves produce a local maximum.
And how do you get 0= sqrt(x2+9409/100)−x by taking a derivative of it? Looks like you simply set n=0.

First, I strongly recommend working symbolically. Only plug in numbers at the end. So you have ##\sqrt{w^2+x^2}-x=n\lambda##.
See if you can simplify that.
Next, think about what happens to n as x increases. Does it increase or decrease?
I don't think you can simplify. It decreases?
 
You can do this without a quadratic or, at least, with an expression in which the x2 term cancels out.

Y
 
Without having to solve a quadratic

You have a right angled triangle where one side is 9.7, one side is the distance x you're looking for and the hypotenuse must be x + nλ

So start with n=1 and put

(x + λ)² = x² + 9.7² and solve for x

Then repeat for
(x +2λ)² = x² + 9.7² and (x + 3λ)² = x² + 9.7²

It gave me the answer you're looking for.
 
haruspex said:
Sure you can. How do you get rid of a square root?

You square it. Yeah I figured it out from that, thanks.