Does the thickness of a slit play a role in the fringe pattern of either the double or single slit experiment?

Question

So it is widely known that the width of the slit affects the interference pattern. And the double slit experiment is a demonstration of the wave nature of light. But i could casually see how the thickness of the slits would effect the particle nature of light, if we supposed them as the classical billiard balls.

But I looked around and could not find any information relating the effect of the thickness of the slits when they are particularly small. Thickness is defined in the attached photo.

My first thought would be to explore something that is the thickness on the order of the wavelength. But i know it would no longer be appropriate to think about it classically.

Image for clarity

Answer 1

It will not affect the pattern but it may affect the quality. If it gets too thick you won’t see anything at all. EDIT 5-3-21 It does not affect a double or multiple slit pattern but it does affect a single slit pattern. (Wavelength) X (distance to screen) / (thickness of slit)= first dark fringe and ( 2 x Wavelength) X (distance to screen) / (thickness of slit)= second dark fringe Etc.

Answer 2

But I looked around and could not find any information relating the effect of the thickness of the slits when they are particularly small.

When the slits have small width (size in direction perpendicular to propagation), then the depth of the slit (size in direction of propagation) can have a pronounced effect on the intensity of output light. Namely, suppose the diaphragm with the slits is made of metal. Let incident light wave be polarized so that $\vec E$ vector is parallel to the slit, and let the wavelength be larger than twice the slit width. Then the wave wouldn't be able to be a standing wave between the slit walls, it would have to be "compressed" to have smaller wavelength to pass through still satisfying boundary conditions. This means that wave vector component in the direction perpendicular to propagation will be higher. The wave still has to maintain its frequency, so total wave vector magnitude must remain the same (because $|\vec k|=\omega/c$). Since (in two dimensions) we have

$$\vec k^2=k_x^2+k_y^2,$$

then to have $|k_x|>|\vec k|$, we must have $k_y^2<0$. This leads to an effect known as evanescent wave, where the wave, instead of oscillating like a sinusiod, falls off exponentially. So, as the depth of the thin slits increases, intensity of outgoing light will decrease exponentially.

If the wave is polarized in the direction parallel to slit width, there'll be no such effect, and the wave will pass virtually without change in amplitude, since boundary conditions on the metallic slit walls are "easier" in this case.

The shape of outgoing wave, unlike amplitude, won't change much: since the source is smaller than half wavelength, it's effectively a point source.

Answer 3

There are effects, but they depend on the light source. If the light source is very highly collimated (which implies that it also has high spatial coherence, light can pass through the slits with almost no scattering off the walls of the slits. Any light that does scatter off the walls will affect the interference fringe pattern downstream mostly by changing the fringe contrast.

Answer 4

Yes it does. But the pattern itself will not change in shape, or sharpness, only the intensity and visibility will change.

Let's start with thickness from very thick. What you will see is that the intensity of the pattern is decreased because of attenuation, and total internal reflection (less visible).

Now if the slits are not so thick, smaller then the wavelength, you can see the pattern well visible, with high intensity.

Now when the thickness reaches almost 0, you will have something called the plasmon.