Physical Optics

[BlackSails]

For a multi-slit diffraction:

a=slit width
d=distance between slits

The minima are given by a*Sin(theta)=m*wavelength?
The "missing ranks" are given by d/a=m/n


d*sin(theta)=n*wavelength
what is n? Are they the principal maxima?

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[smeagol]

I'm not sure what you mean by principal maxima but values of n are the integer values (e.g. 1, 2, 3..) which will change theta such that it marks the location of other maxima.

For instance, in this picture:
http://en.wikipedia.org/wiki/Image:Ebohr1.png

The white/bright bands on the right are related to the n values. The reason n is an integer is because you need the differences in path lengths of your light sources to be equal to a multiple of a wavelength in order to observe constructive interference.

 

[BlackSails]

I'm not sure what you mean by principal maxima but values of n are the integer values (e.g. 1, 2, 3..) which will change theta such that it marks the location of other maxima.

For instance, in this picture:
http://en.wikipedia.org/wiki/Image:Ebohr1.png

The white/bright bands on the right are related to the n values. The reason n is an integer is because you need the differences in path lengths of your light sources to be equal to a multiple of a wavelength in order to observe constructive interference.

The principal maxima are the large peaks.

Ill rephrase my question: What is the effect of changing the slit width, and changing the slit distance on the intensity?

Do the peaks depend on d, and the missing orders on a, or vice versa?

 

[smeagol]

The principal maxima are the large peaks.

Ill rephrase my question: What is the effect of changing the slit width, and changing the slit distance on the intensity?

Do the peaks depend on d, and the missing orders on a, or vice versa?

Hrm ok. I'm not too familiar with multi-slit diffraction but perhaps some principles of single-slit diffraction may still apply?

I know that decreasing slit width, a, in single slit diffraction will increase the intensity/size of the large peak: since, asin(.&#1012 = m.λ, .by decreasing a you must increase sin(.&#1012 in order to achieve the same value of ..m..λ. Increasing .. .ϴ will increase the separation of the dark bands from destructive interference, thereby increasing the large peak.

I currently can't conceive of why having multiple slits would change that, so I think the same concept would apply.

From the relationship d/a = m/n, perhaps we can manipulate it to give a = dn/m, thus. asin(.&#1012 = ..dn/m.sin(.&#1012. It seems then to me that d would also affect the intensity/size of the large peak.

Again, I'm not too sure about this and I hope someone can confirm that this is alright.
.

 

[MooseKing]

--i don't know how much of this will help, but here is it anyways

d*sin(theta)= the path length difference <---- always true

n*wavelength=d*sin(theta) when the waves are in phase and if n is an integer then its a point of constructive interference (assuming same frequency for both waves)

n*wavelength will not always be equal to d*sin(theta) at constructive points. it goes back to the equation y(x,t)=asin(ect. ect. ect.) where you want change in TOTAL phase to be an even number times pi for constructive odd times pi for destructive and every thing else partial interference.