Interference - Q done, is it right?

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    Interference
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Homework Help Overview

The discussion revolves around a problem related to wave interference, specifically focusing on the calculation of wavelength and frequency based on a given distance. The original poster seeks validation of their calculations and reasoning after revisiting the material.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster presents their calculations for wavelength and frequency, expressing uncertainty about their correctness. Some participants point out errors and suggest additional considerations regarding the frequencies that can be derived from the given setup.

Discussion Status

The discussion includes attempts to clarify the calculations and identify errors. Participants have provided feedback indicating that while the initial calculations are correct, there are additional frequencies to consider. There is an ongoing exploration of the implications of these findings.

Contextual Notes

Participants note the presence of errors in the original calculations and the need for further exploration of the frequency spectrum related to the interference problem. The original poster's confidence in their understanding is questioned, prompting a review of assumptions and calculations.

Flucky
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Hi all, I have a question here that I've done but it hasn't got an answer provided. I'm sure it's easy stuff but haven't looked at this material in a while so
just want to double check.

(The things in red are what I added to the original diagram)

intereferenceexampleQ.jpg


RE-DONE

d = √17

d + λ/2 = 2√5

λ/2 = 2√5 -d

λ = 4√5 -2(√17)

λ = 0.698m

Using v=fλ

∴f = 340/0.698 = 487 Hz

Is this correct?

Thanks in advance
 
Last edited:
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d=\sqrt{1^2+4^2}=\sqrt{17}
 
szynkasz said:
d=\sqrt{1^2+4^2}=\sqrt{17}

Woops, muppet at work. No idea how that came about, thanks.

Looking over it I've done a couple of very stupid errors such as line 4.

First post RE-DONE
 
Last edited:
Correct, but you get more frequencies:

(2n-1)\frac{\lambda}{2}=d\Rightarrow f=(2n-1)\frac{v}{2d},\,n=1,2,3,...
 
szynkasz said:
Correct, but you get more frequencies:

(2n-1)\frac{\lambda}{2}=d\Rightarrow f=(2n-1)\frac{v}{2d},\,n=1,2,3,...

Thanks a lot, big help
 

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