Understanding Minimum Path Difference for Waves Below the Centerline

In summary, the conversation discusses the concept of path difference in relation to interference patterns. It is established that for angles below the centerline, the minimum path difference should be 0.75 wavelengths instead of 0.25 wavelengths, and the path from A to the detector is greater than the path from B. The conversation also delves into how this affects the interference and phase of the waves, and concludes that the extra distance for A will make the phase of A "behind" the phase of B. Finally, it is determined that for points above and below the centerline, the distance from A will always be smaller than the distance from B, resulting in a 1/2 wavelength difference in phase between A and B.
  • #1
kent davidge
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Homework Statement



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Homework Equations

The Attempt at a Solution



(Sorry my poor English). I just don't understand why should the minimum path difference be 0.75 wavelengths for angles below the centerline instead 0.25 wavelenghts as it is for angles above the centerline.
 
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  • #2
For angles below the centerline, is the path length from A to the detector greater or less than the path length from B?
 
  • #3
It's greater, but how would it change the expression for interference?
 
  • #4
Yes, the the path from A is greater when the angle is below the center line.

The wave from A has a "head start" compared to the wave from B due to the vibration of the speaker at A being 1/4 period ahead of the vibration at B. When the waves arrive at the detector, does the extra distance for A tend to make the phase of the wave from A even farther ahead of B or not as much ahead?

You might consider the case where the extra distance for A is 1/4 wavelength. A sketch will help.
 
  • #5
(Sorry my poor English). Okay, but instead adding some value, it seems we have to subtracting one so that the right side of the equation equals to 0.75 wavelength

Xa - (1/2)λ - Xb = λ/4
Xa - Xb = 3λ/4
 
  • #6
kent davidge said:
(Sorry my poor English). Okay, but instead adding some value, it seems we have to subtracting one so that the right side of the equation equals to 0.75 wavelength

Xa - (1/2)λ - Xb = λ/4
Xa - Xb = 3λ/4
I'm sorry, but I'm not quite following what you did here.

The extra distance that the wave from A travels will shift the phase of A in a direction that makes the phase of A "behind" the phase of B. But the wave from A started out 1/4 λ ahead of the wave from B. So, if the extra path distance that A has to travel is 1/4 λ, the waves from A and B will end up in phase at the detector.

You need to find the path difference in order for A and B to end up out of phase at the detector.
 
  • #7
oh thank you TSny I understand it now. For any point above the centerline the distance from A will be always smaler than the distance from B. If waves from B start 1/4λ behind of waves from A, and if the distance from B is 1/4λ greater than distance from A, we add 1/4λ to find the position of the wave from B, which will be 1/4λ + 1/4λ = 1/2λ out of phase with waves from A. Analogous situation occurs from points below the centerline. Am I right?
 
  • #8
kent davidge said:
oh thank you TSny I understand it now. For any point above the centerline the distance from A will be always smaler than the distance from B. If waves from B start 1/4λ behind of waves from A, and if the distance from B is 1/4λ greater than distance from A, we add 1/4λ to find the position of the wave from B, which will be 1/4λ + 1/4λ = 1/2λ out of phase with waves from A. Analogous situation occurs from points below the centerline. Am I right?
Yes. That's right.
 
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What is wavelength difference?

Wavelength difference is the variance in the length of a wave between two different points. It is measured as the distance between two corresponding points on the wave, such as the peak or trough.

How is wavelength difference calculated?

To calculate wavelength difference, you need to know the distance between two corresponding points on the wave (wavelength) at one point and the distance between the same corresponding points at a different point. The difference between these two distances is the wavelength difference.

What affects wavelength difference?

The wavelength difference can be affected by the medium through which the wave travels, the frequency of the wave, and the speed of the wave. Different mediums can cause the wave to travel at different speeds, which can alter the wavelength difference. Additionally, the frequency of the wave can also impact the wavelength difference.

What is the relationship between wavelength difference and frequency?

There is an inverse relationship between wavelength difference and frequency. As the frequency of a wave increases, the wavelength difference decreases. This means that shorter wavelengths have a larger wavelength difference compared to longer wavelengths.

How is wavelength difference used in science?

Wavelength difference is used in various scientific fields, such as physics, astronomy, and chemistry. It is a crucial factor in understanding the behavior of waves and can help scientists make predictions and calculations in their research. For example, in astronomy, the wavelength difference of light can provide information about the composition and movement of celestial objects.

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