Relationship of slit, wavelength, and intensity

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SUMMARY

The discussion focuses on the relationship between slit width, wavelength, and intensity in diffraction patterns, specifically addressing a slit width of 0.01 mm. It concludes that when the slit width (d) is less than the wavelength (λ), there are no points in the diffraction pattern where the intensity is zero. This is due to the breakdown of the equations governing diffraction, which results in no real solutions for the angle, leading to a continuous intensity rather than distinct minima. The wavelengths discussed include blue light (500 nm), infrared light (10.6 μm), and ultraviolet light (50 nm).

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  • Understanding of single-slit diffraction principles
  • Familiarity with the equation for diffraction minima: dsinθ = mλ
  • Knowledge of coherent electromagnetic radiation
  • Basic concepts of wave interference
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nso09
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Homework Statement


Coherent electromagnetic radiation is sent through a slit of width 0.01 mm. For which of
the following wavelengths will there be no points in the diffraction pattern where the intensity is zero?
A. Blue light of wavelength ##\lambda=500 nm##

B. Infrared light of wavelength ##\lambda=10.6 μm##

C. Ultraviolet light of wavelength ##\lambda=50 nm.##
##d=.01mm##

Homework Equations


##dsin\theta=m\lambda##
##dsin\theta=(m+.5)\lambda##

The Attempt at a Solution


I'm not sure exactly what the question is asking. What does it mean diffraction pattern? Is it just talking about the central maximum or the rest of the pattern where there are maxima and minima?
I assume this is a single slit so the minima would be where intensity is 0 so an equation that makes sense is ##dsin\theta=m\lambda##
I checked the solutions though and it said that
"If the slit width d is less than the wavelength, there are no points at the diffraction pattern at which the
intensity is zero." What does it mean that there are no points at the pattern where the intensity is 0? And why does d have to be less than the wavelength?
 
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Take your relevant equations and move the 'd' to the right hand side of them (so that the relationship between d and λ is clear). What does d being less than λ imply?
 
If d is less than ##\lambda##, then ##sin\theta\geq1##. So the domain doesn't fit ##sin\theta.## But how can I make use of the intensity since I don't have an angle to plug in ##I##=##I_0####((sin(\beta/2)/(\beta/2)##)^2 and set it equal to 0. Or is that the whole point? I don't have an angle so therefore, there are no points where intensity is 0? If so, what pattern are we talking about? The central maxima? I just need help visualizing what this whole pattern looks like.
 
Sorry for the delay in responding.

You've got the right idea. When the equations break down because there is no real solution for the angle then there will be no distinct minima. There may be a graduated continuum, but no locations where the intensity falls to zero.
 
gneill said:
Sorry for the delay in responding.

You've got the right idea. When the equations break down because there is no real solution for the angle then there will be no distinct minima. There may be a graduated continuum, but no locations where the intensity falls to zero.
Oh I see. So basically there is no point in the diffraction pattern where intensity is 0. The smallest intensity may be really tiny but never 0, therefore no waves will completely destructively interfere. Is that it?
 
Yes. That's it. The diffraction pattern disappears when d < λ.
 
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