Maximum Width of Single Slit for No Diffraction Minima?

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Homework Help Overview

The discussion revolves around determining the maximum width of a single slit that would result in no diffraction minima, specifically in the context of wave optics and diffraction phenomena.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the relationship between the slit width and the wavelength, questioning how the sine function applies to the diffraction formula. There is an examination of the conditions under which minima occur in single slit diffraction.

Discussion Status

The discussion is active, with participants offering insights into the geometric interpretation of the diffraction setup and the implications of the sine function at specific angles. Some guidance has been provided regarding the relationship between the slit width and the wavelength.

Contextual Notes

Participants are considering the implications of the angle at which diffraction minima occur and how this relates to the maximum width of the slit. There is an ongoing exploration of the conditions necessary to avoid minima.

leolaw
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Given a wavelength length [tex]\alpha[/tex], 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 [tex]D * sin (\theta) = m \alpha[/tex] ?
 
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leolaw said:
Given a wavelength length [tex]\alpha[/tex], 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 [tex]D * sin (\theta) = m \alpha[/tex] ?

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 [itex]\theta[/itex] would have to be 90 degrees for the first minimum. So then what does

[tex]D * sin (\theta) = m \alpha[/tex]

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

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