Maximum Width of Single Slit for No Diffraction Minima?

AI Thread Summary
The maximum width (D) of a single slit that results in no diffraction minima is determined by the relationship D * sin(θ) = mα, where α is the wavelength. For no minima to occur, the angle θ must reach 90 degrees, indicating that D must equal the wavelength α for the first minimum to appear. Thus, if D is greater than α, diffraction minima will not be observed. This geometric understanding of single slit diffraction is crucial for solving related proof problems. The conclusion is that the maximum slit width for no diffraction minima is equal to the wavelength.
leolaw
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Given a wavelength length \alpha, what is the maximum Width (D) of a single slit, which would have no diffraction minima?

It seems like a proof problem to me and I am trying to get a head start.
should I use D * sin (\theta) = m \alpha ?
 
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leolaw said:
Given a wavelength length \alpha, what is the maximum Width (D) of a single slit, which would have no diffraction minima?

It seems like a proof problem to me and I am trying to get a head start.
should I use D * sin (\theta) = m \alpha ?

Yes, that and what you know about the sine function.
 
that sin of zero degrees is 0
 
leolaw said:
that sin of zero degrees is 0

Yes, but at zero degrees you will never have a minimum. From the geometry of the single slit diffraction setup, to not find any minima after the slit, the angle \theta would have to be 90 degrees for the first minimum. So then what does

D * sin (\theta) = m \alpha

tell you about D?
 
I see, so D sin (90) = (1) \alpha, which is the first minimum, and D has to be equal to the wavelength \alpha.
 
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