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HW 10 Question 13

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I have been reviewing and came across this question and wasn't sure exactly how it worked.

(a) If constructive interference occurs at a particular location of the sliding section, by what minimum amount should the slider be moved upward so that the destructive interference occurs instead?

If for maximum destructive interference Δr=(n+1/2)λ and λ= vsound/ f,

λ is found to equal 0.611m

How is Δr calculated? Which number for n is used? The answer for this part is equal to λ/4 and I'm not sure how I would reach that solution?

 

(b) What minimum distance from the original position of the sliding section will again result in constructive interference?
 

I am also having the same problem here. If Δr=nλ for maximum constructive interference, how is it that the answer is λ/2?

Thank you for your time.

asked Dec 10, 2014 in Intro College Physics by anonymous

1 Answer

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Best answer
For part (a), the actual n doesn't matter as we just need to find the difference between adjacent max and min. When the U-shaped slider moves up or down, it has two legs moving for same amount. So the total path increased or decreased will be twice as much. Therefore, ||\lambda/4|| makes ||\lambda/2||.

For part (b), you want the path difference be one ||\lambda||. This corresponds to the slider moving for half of that and thus ||\lambda/2||.
answered Dec 10, 2014 by PhysicsDude (21,750 points)
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