More slits, farther maxima of interference patterns, why?

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Interference patterns exhibit more intense and widely spaced maxima with an increasing number of slits due to constructive interference. The primary maxima occur at specific angles determined by the equation mλ = d sin(θ), where m is an integer, and this relationship holds regardless of the number of slits. In multi-slit setups, the cancellation of non-constructive interference leads to sharper and more intense maxima, as intensity is proportional to the square of the electric field, increasing with the number of slits (N²). The diffraction grating's ability to direct specific wavelengths to precise locations enhances its utility in spectral measurements. Understanding these principles is crucial for analyzing both multi-slit and single-slit diffraction patterns.
Cardinalmont
When observing interference patterns, one will notice that the maxima from a double slit are more intense and farther apart than the maxima of a single slit. Likewise, the maxima from a diffraction grating are more intense and farther apart than maxima of a double slit. Why is this?

I understand that the minima are created when light with a phase difference of half a wavelength interact with each other, and more slits means more opportunity for half a wavelength phase changes, but doesn't it also present more opportunity for light of different phase shifts to hit those spots? Wouldn't it also present more opportunities for everything, not just minima?

I ask because I put a red laser pointer through a diffraction grating and the maxima were extremely far apart and extremely distinct! Unbelievable!
 
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For the two slit case as well as the multi-slit or multi-line case, the important thing is the primary maxima, where constructive interference occurs at ## m \lambda =d \sin{\theta} ## with ## m ##=integer. Whether it is just two slits or many equally spaced slits, the primary maxima for a given ## d ## occur in the same place. When you shine a laser onto a grating, it is likely you used ## N=100 ## or more of the slits at the same time. When they are equally spaced, if slit 1 constructively interferes with slit 2, and slit 2 constructively interferes with slit 3, then you can say all of them constructively interfere. The same logic applies, even if you have 100 or more slits. In the case of many slits, the many slits generally have a cancellation effect for the most part, anywhere that they do not constructively interfere, (unlike the two slit case where the maxima are rather broad), so that the primary maxima are very intense, and very little intensity outside of them. In addition, because the intensity is proportional to the square of the electric field, the intensity at the maxima for the case of a grating will be proportional to ## N^2 ##. The fact that the diffraction grating can put a specific wavelength at a very specific location makes it very useful in making spectral measurements. ## \\ ## I could also add a little about the single slit diffraction pattern: The most important feature there is the width of the central maximum, where the first zeros from the center peak occur for ## m=+1 ## and ## m=-1 ## at location ## m \lambda =b \sin{\theta} ##. The same diffraction/interference principles are responsible for the single slit diffraction pattern that cause the multi-slit interference. For multi-slit interference, oftentimes narrow slits are used , making a wide pattern, so the diffraction from individual slits can often be ignored in the multi-slit interference. ## \\ ## The derivation of the single slit pattern uses a continuum of sources across the slit where the phase ## \phi ## is a function of position inside the slit. (## \phi=\phi(x)=\frac{2 \pi x \sin{\theta}}{\lambda} ## ). The sources are added together (i.e. the electric fields from each in the far-field are superimposed) in integral form. The source at location ## x ## is ## E_o \cos(\omega t +\phi(x)) \, dx ##. ## \\ ## Meanwhile, the multi-slit interference pattern uses point sources that have a phase ## \phi ## that also depends on their location: ## \phi_n= \frac{2 \pi (n-1)d \sin{\theta}}{\lambda} ## where ## n ## is the slit number for a given slit. The sources for this case are summed with an arithmetic sum. The source for a given slit is ## E_o \cos(\omega t+\phi_n ) ##. ## \\ ## In both cases, the intensity ## I(\theta) ## (in watts/m^2) is proportional to the square of the amplitude of the resultant ## E ## field.
 
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