# When is diffraction most pronounced?

My notes says the following things about diffraction:

The larger the ratio of wavelength to slit (or object) width, the more pronounced the diffraction is and the more spread out the wave energy is. Diffraction is most obvious when the wavelength and the dimensions of the obstacle/space are the same order of magnitude (approximately equal in size).

I find the second statement contradictory to the first statement. If diffraction is 'most pronounced' when $$\frac{\lambda}{d}$$ is the greatest, why is it 'most obvious' when $$\frac{\lambda}{d} \approx 1$$? (I interpreted the statement 'the wavelength and the dimensions of the obstacle/space are the same order of magnitude (approximately equal in size)' as $$\frac{\lambda}{d} \approx 1$$, not sure if this is correct.)

• In the limit $d \rightarrow 0$ you have a point source. The emitted light then goes in all directions. The precise pattern will depend on some details of the emission process but you can think of it as spherical waves going outwards from the source. This is the limiting case of diffraction. The range of angles into which the light is emitted is the entire range available. Jul 19, 2021 at 11:30

When you are talking about light and common day experiences the size of most apertures, $$d$$, are much, much larger than the wavelength of light. $$\lambda$$ which in turn means that $$\lambda$$ /$$d$$ is very, very small.
As this quotient gets bigger and approaches $$1$$ diffraction becomes more pronounced.

You could define the amount of diffraction by comparing the intensity of the diffracted light, ie the light that has changed direction as a result of passing through an aperture, as compared with the intensity of all the light which passes through the aperture.

Using this definition, when $$\lambda \approx d$$ almost all the incident light is diffracted.

Note that you can still observe diffraction if $$d\gg\lambda$$ as with the diffraction due to an edge but it can be more difficult to observe.

Here is an example of diffraction due to the edges of a razor blade. If the aperture size becomes much less that the wavelength of light very little light passes through the aperture and the light behaves as though the aperture was not there.

• Dear Farcher, two points in your answer are particularly interesting: 1. „the light that has changed direction as a result of passing through an aperture" and 2. according to the picture, the outer contour has no aperture but still deflects the light. Have you never thought about the interaction of the surfaces (e.g. induced phonons) with the photons as the cause of the diffraction effect? Jul 21, 2021 at 17:21

Sorry for my poor english. My native language is french.

Scalar theories of diffraction, those that are used most often, are not valid when the size of the aperture is of the order of the wavelength. The vectorial character of the electromagnetic field as well as the nature of the diffracting material should be taken into account.

So, most often, we speak of diffraction with openings significantly larger than the wavelength. But that doesn't mean the effect is negligible.

If you take an aperture of width $$a$$ illuminated in parallel light, the diffraction angle $$λ/a$$ may be small but the size of the diffraction figure on a screen at distance $$D$$ is $$λD/a$$ : it grows with $$D$$.

At the same time, the size of the projected shadow, predicted by geometric optics, is $$a$$.

When the width $$λD/a$$ of the zone disturbed by the diffraction is greater than the size $$a$$ of the image provided by geometric optics, the diffraction is no longer negligible : $$λD/a≫a$$ or : $$a≪√λD$$

This criterion is clearly less severe than the condition $$a≪λ$$ which is sometimes given!

In your notes 'pronounced' and 'obvious' could be referring to different things.

When $$d$$ is small the diffraction (amount of bending) is high and the spacing between maxima of the diffraction pattern becomes greater, so in that sense the maxima are easier to resolve if $$\frac{\lambda}{d}$$ is high.

If $$d$$ were very small however, the amount of light passing through the slit is low and it's spread over a wider region, so the pattern becomes dim, so in that sense is not so 'obvious' (easy to see).