Ripple Tank, Double Slit: Waves

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Homework Statement


A double-slit interference experiment is done in a ripple tank (a water tank using a vibrating rod to produce a plane wave on the surface of the water). The slits are 6.00 cm apart, and a viewing screen is 2.15 m from the slits. The wave speed of the ripples in water is 0.012 m/s, and the frequency of the rod producing the ripples is 5.20 Hz. How far from the centerline of the screen will a second order minimum be found? The second order minimum is the second time that destructive interference happens.


Homework Equations


Y = (m + 1/2)(λ*R/d)
λ= v/f



The Attempt at a Solution



Y = (2 + .5) * ((.012*2.15)/(5.2*.06)) = 2.06E-1 m

If I'm understanding this equation correctly, Y = distance from central fringe. M = xth order minimum/maximum. R= distance from slits to screen. d= distance between slits.
I'm doing something wrong though, somehow!

Thanks for the help.
 

Answers and Replies

  • #2
Ibix
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Do you know the expression for the locations of the maxima? There must be a minimum between each maximum and the next.
 
  • #3
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Yeah. To find a max it should be:

Y = mRλ/d

I'm trying to visualize how that's helpful.

So, if ~~~ is destructive and whitespace is constructive, this is what I'm imagining:

| |~~~~~| |~~~~~| |

If finding the location of the 2nd order fringe (aka max) would be measuring from the first | to the 3rd | (so measuring from the beginning of the first fringe to the beginning of the second fringe).

How can I combine the distances between fringes to find location of the min? Having a hard time wrapping my mind around this one for some reason.

EDIT: Well, I'm slightly confused. The equation specified in the OP may or may not only be used to find the location of an nth order FRINGE, while creating destructive interference, and the equation in this post might serve the same function, but for when the waves are creating constructive interference. I'm not sure, frankly :/ trying to think it out.
 
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  • #4
Ibix
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If the first maximum is on the axis at Y=0, and the second at Y=Rλ/d, and the third at Y=2Rλ/d, where must the first and second minima lie?
 
  • #5
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3Rλ/2d

Since the minimum is halfway between both maximums. Meaning, Rλ/d + Rλ/2d = 3Rλ/3d

Thanks, I appreciate it. Brain is about quitting on me, sorry I didn't get it more quickly!
 
  • #6
Ibix
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No problem. The basic problem is that "second order minimum" is slightly ambiguous given that the indexing parameter starts at zero. Your original m=2 is not unreasonable if you count zeroth-first-second, but neither is m=1 if you argue m=0 is first, m=1 is second.
 

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