# Relationship of slit, wavelength, and intensity

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1. Nov 15, 2016

### nso09

1. The problem statement, all variables and given/known data
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 $\lambda=50 nm.$
$d=.01mm$
2. Relevant equations
$dsin\theta=m\lambda$
$dsin\theta=(m+.5)\lambda$

3. 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?

2. Nov 15, 2016

### Staff: Mentor

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?

3. Nov 15, 2016

### nso09

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.

4. Nov 15, 2016

### Staff: Mentor

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.

5. Nov 15, 2016

### nso09

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?

6. Nov 15, 2016

### Staff: Mentor

Yes. That's it. The diffraction pattern disappears when d < λ.