Single slit diffraction minima formula misunderstanding

In summary: With destructive interference, if one light is out of phase with another light then the first light will destructively interfere with the second light. However, with constructive interference, if one light is out of phase with another light then the two lights will both add together and the result will be less than the sum of the two original lights. This is why when dealing with m=2, the explanation is that the slit is split into four sections so there is a phase difference of pi for the paired up point sources and so destructive interference occurs. But if we used 2 sections for m=2 then the paired up point sources have a phase difference of 2pi and constructive interference occurs. Since the explanation for m=2 is that the paired
  • #1
HuaYongLi
16
0
Reading the derivation for the single slit diffraction minima formula from this site, I'm not convinced about its argument.
http://www.math.ubc.ca/~cass/courses/m309-03a/m309-projects/krzak/index.html"

What I'm having trouble understanding is the explanation of the formula for minima when m>1.
For example when dealing with m=2, the explanation is that the slit is split into four sections so there is a phase difference of pi for the paired up point sources and so destructive interference occurs. But if we used 2 sections for m=2 then the paired up point sources have a phase difference of 2pi and constructive interference occurs.
Also when m=3 (or an odd number), it seems we have to revert back to two sections to explain destructive interference which gives a phase difference of 3/2pi which is destructive interference.
Why does using 4 sections for m=2 give the 'right' answer of destructive interference and not 2 sections?
 
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  • #2
HuaYongLi said:
For example when dealing with m=2, the explanation is that the slit is split into four sections so there is a phase difference of pi for the paired up point sources and so destructive interference occurs. But if we used 2 sections for m=2 then the paired up point sources have a phase difference of 2pi and constructive interference occurs.
Just showing that paired points constructively interfere does not allow you to conclude anything interesting. Just because points A and A' and points B and B' constructively interfere does not mean that A and B do. A and B may still be out of phase.

On the other hand, if you can divide the slit in a way so you can show that each section is canceled by another, then that's that. That does allow you to conclude that all the light cancels and a dark fringe occurs.
 
  • #3
Thank You, I see it now.
I failed to take into account that constructive interference isn't like destructive interference when dealing with this pairing up business.
 

Related to Single slit diffraction minima formula misunderstanding

What is the single slit diffraction minima formula?

The single slit diffraction minima formula is a mathematical equation used to calculate the positions of the dark fringes (or minima) in a diffraction pattern created by a single slit. It is also known as the single slit diffraction pattern formula or the single slit interference formula.

How is the single slit diffraction minima formula derived?

The single slit diffraction minima formula is derived from the principles of wave interference and diffraction. It takes into account the wavelength of the incident light, the width of the slit, and the distance between the slit and the screen to calculate the positions of the minima in the diffraction pattern.

What is the most common misunderstanding about the single slit diffraction minima formula?

The most common misunderstanding about the single slit diffraction minima formula is that it only applies to single slit diffraction patterns. In reality, this formula can also be used to calculate the positions of minima in other types of diffraction patterns, such as double slit and multiple slit patterns.

What are the limitations of the single slit diffraction minima formula?

The single slit diffraction minima formula is only applicable to situations where the slit is much narrower than the wavelength of the incident light. It also assumes that the incident light is monochromatic (has a single wavelength) and that the slit is illuminated by a parallel beam of light.

How can the single slit diffraction minima formula be applied in real-world situations?

The single slit diffraction minima formula can be applied in various real-world situations, such as in the design and analysis of optical instruments like microscopes and telescopes, or in the study of diffraction patterns in everyday objects such as CDs and DVDs. It is also used in the field of spectroscopy to analyze the wavelengths of light emitted or absorbed by different substances.

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