Single slit diffraction question

Question

I'm curious about single slit diffraction and noticed on this site they say the following:

note that the width of the central diffraction maximum is inversely proportional to the width of the slit. If we increase the width size, a, the angle T at which the intensity first becomes zero decreases, resulting in a narrower central band. And if we make the slit width smaller, the angle T increases, giving a wider central band.

That would seem to imply that more aperture/source points of the wave is associated with a more "particle-like" distribution (higher central density, less pronounced fringes) and that less source points equals a more wave-like distribution (fringes are all closer in density).

However, if waves have an infinite amount of source points wouldn't this logic be counter-intuitive? As the amount of source points approaches infinity, shouldn't the the distribution approach a particle-like pattern? Wouldn't a particle pattern be exhibited regardless of the slit width, as all widths would have an infinite amount of source points as waves are non-discreet?

Answer 1

This can be explained mathematically when looking at the overall phase difference in the single slit. Phase difference between two waves of equal frequency is given by:

$\delta = 2\pi /{\lambda} *d\sin \theta$

where d is the width of the slit and $\theta$ is the angle that subtends the distance from the center point on the screen. For the first minimum, $\delta$ must be $\pi$ (this becomes apparent when you look at a phasor diagram for the single slit which I suggest you do). This means that the position of the first minimum on the screen is given by:

$\lambda / {2d} = \sin\theta$

Now by using the small angle approximation $\sin\theta \approx y/L$ where y is the distance on the screen and L is the distance of the slit from the screen we can write the width of the central maximum as

$2y = \lambda L /2d$

As you can see the width is indeed inversely proportional to the width of the slit. I suggest you take a look at the hyperphysics site on Single slit diffraction, it helps to develop an intuitive understanding of phasor diagrams and the nature of single slit diffraction. This is the link for it: http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinint.html

Answer 2

You have to make a theoretical distinction between decreasing slit width and adding more slits, for example, to make a diffraction grating. the fringe spacing is roughly the same, regardless of how many slits are added

Having an infinite number of source points should not reduce the need to model light as a wave, in fact it should enhance the wave-like characteristics. The fringe spacing is not even approximately equal to the slit spacing. We need diffraction to explain the interference pattern in a classical setting.