Short description of diffractionin physics

It is evident that the obstacle causes a misalignment of the wave front. Above the yellow line, there are two little crests that are unexpected and caused by the bending of the wave. This misalignment is observed in the sudden maximums after the obstacle has a phase shift.

In comparison to the wavelength, what should be the width of an obstacle in order for the wave to be disrupted but the wave front to remain unbroken?

We have a planar wave propagating towards an aperture. Right after the middle of that aperture, what do we expect to find?

We have a planar wave propagating towards an aperture. Right after the middle of that aperture, what do we expect to find?

Fraunhoferdiffraction

In comparison to the wavelength, what should be the width of an obstacle in order for the wave to be disrupted but the wave front to remain unbroken?

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We have a planar wave propagating towards an aperture. Right after the middle of that aperture, what do we expect to find?

Definitionof diffractionin Physics

The third case presents a complex pattern. Here, the wave front corresponding with the first crest (red line) is divided into three parts and features two minimums. The next wave front (blue line) has one minimum, and after that, we again see the difference between crests and troughs, even if they’re bent.

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When a wave propagates across an object, there is an interaction between the two. An example is a calm breeze moving the water around a rock that cuts through the surface of a lake. In these conditions, parallel waves are formed where there is nothing to block them, while right behind the rock, the shape of the waves becomes irregular. The bigger the rock, the bigger the irregularity.

Diffractiondefinition

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Diffractiondiagram

If we increase the wavelength of the wave, the difference between maximums and minimums is no longer evident. What happens is that the waves interfere with each other destructively according to the width d of the slit and the wavelength λ. We use the following formula to determine where the destructive interference occurs:

Diffraction is a phenomenon that affects waves when they encounter an object or an opening along their path of propagation. The way their propagation is affected by the object or the opening depends on the dimensions of the obstacle.

In comparison to the wavelength, what should be the width of an obstacle in order for the wave to be disrupted but the wave front to remain unbroken?

Diffraction ofwaves

The wave is disrupted by the smallest obstacle but not enough to break the wave front. This is because the width of the obstacle is small compared to the wavelength.

Typesof diffraction

The dimension of the aperture affects its interaction with the wave. In the centre of the aperture, when its length d is greater than the wavelength λ, part of the wave passes through unaltered, creating a maximum beyond it.

Keeping the same example but exchanging the rock for an open gate, we experience the same behaviour. The wave forms parallel lines before the obstacle but irregular ones while passing through and beyond the gate’s opening. The irregularities are caused by the gate’s edges.

Here, n = 0, 1, 2 is used to indicate the integer multiples of the wavelength. We can read it as n times the wavelength, and this quantity is equal to the length of the aperture multiplied by the sine of the angle of incidence θ, in this case, π/2. We, therefore, have constructive interference, which produces a maximum (the brighter parts in the image) at those points that are multiples of half the wavelength. We express this with the following equation:

Diffractionexamples

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A bigger obstacle, whose width is similar to the wavelength, causes a single minimum right after it (red circle, 2nd image from the left), which indicates that the wave front has been broken.

In comparison to the wavelength, what should be the width of an obstacle in order for the wave to be disrupted but the wave front to remain unbroken?

Our first example of diffraction was a rock in the water, i.e., an object in the way of the wave. This is the inverse of an aperture, but as there are borders that cause diffraction, let’s explore this, too. While in the case of an aperture, the wave can propagate, creating a maximum just after the aperture, an object ‘breaks’ the wave front, causing a minimum immediately after the obstacle.

Finally, n in the formula indicates not only that we are dealing with multiples of the wavelength but also the order of the minimum or maximum. When n = 1, the resulting angle of incidence is the angle of the first minimum or maximum, while n = 2 is the second one and so on until we obtain an impossible statement like sin θ must be greater than 1.