Finding second furthest distance of const interference from 2 sources

AI Thread Summary
The discussion focuses on finding the second furthest distance of constant interference from two sources, utilizing the wavelength derived from the equation v/f = 0.8995. The participants explore the equation sqrt(9.7^2+x^2)-x=n(0.8995) and the implications of taking its derivative to identify maxima. There is confusion regarding the approach, particularly whether a maximum exists and how to handle the variable n as x increases. Recommendations are made to work symbolically rather than plugging in numbers early, emphasizing the need to simplify the expression effectively. Ultimately, the correct approach leads to the conclusion that the answer is 25.25.
JoeyBob
Messages
256
Reaction score
29
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
 

Attachments

  • question.PNG
    question.PNG
    11 KB · Views: 138
Physics news on Phys.org
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?
 
Last edited:
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?
 
JoeyBob said:
I don't think you can simplify.
Sure you can. How do you get rid of a square root?
JoeyBob said:
It decreases?
Yes, so for the largest possible x, what will n be?
 
  • Like
Likes JoeyBob
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.
 
Back
Top