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Diffraction

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Diffraction

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.

The phenomenon of diffraction

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.

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.

Diffraction Two slit diffraction of light StudySmarterFigure 1. A wave is propagating towards an aperture. The arrows indicate the direction of the propagation, while the dotted lines are the wave fronts before and after the obstacle. Notice how the wave front briefly becomes circular but returns to its original linear shape as it leaves the obstacles behind. Source: Daniele Toma, StudySmarter.

Single slit aperture

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.

Diffraction Single slit aperture of light StudySmarterFigure 2. A wave passing through an aperture whose aperture length d is greater than the wavelength λ. Source: Daniele Toma, StudySmarter.

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:

n λ = d sin θ

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:

n (λ / 2) = d sin θ

Diffraction Single slit aperture of light StudySmarterFigure 3. Here, the energy is distributed on a wider wavelength as denoted by the distance between the blue lines. There is a slower transition between a maximum (blue) and a minimum (black) before the aperture. Source: Daniele Toma, StudySmarter.

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.

Diffraction caused by an obstacle

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.

Diffraction. StudySmarterFigure 4. A wave is generated below the obstacle, with the crests depicted in colour and the troughs in black. Source: Daniele Toma, StudySmarter.

The figure depicts a scenario in which the wave is always the same while the obstacles are increasingly wider.

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.

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.

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.

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.

Diffraction - key takeaways

  • Diffraction is the result of the border’s effect on the propagation of a wave when it encounters either an obstacle or an aperture.
  • The dimension of the obstacle has noticeable importance in diffraction. Its dimensions compared with the wavelength determine the pattern of crests and troughs once the wave has passed the obstacle.
  • The phase is altered by an obstacle that is big enough, thus causing the wave front to be bent.

Frequently Asked Questions about Diffraction

Diffraction is a physical phenomenon that occurs when a wave finds an aperture or an object in its path.

The cause of diffraction is a wave being affected by an object that is said to be diffracting.

The pattern of diffraction is affected by the width of the object compared to the wavelength of the wave.

Final Diffraction Quiz

Question

Which kind of obstacles do we consider when studying diffraction?

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Answer

Apertures and obstacles.

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Question

What is the effect of the borders of the obstacles on the direction of propagation?

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Answer

The borders of an object cause the wave to generate circular wave fronts.

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Question

What is another well-known phenomenon involved in diffraction?

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Answer

Interference.

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Question

How is interference related to diffraction?

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Answer

The obstacle creates phase shifts that generate constructive and destructive interference.

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Question

What is the formula to calculate the destructive interference points?

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Answer

n λ = d sin θ

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Question

If an obstacle is smaller than the wavelength, does it affect propagation?

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Answer

Yes, it does.

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Question

When there is an obstacle in the way of a wave, will diffraction occur?

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Answer

Yes, it will.

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Question

Can you explain the blurry effect we can observe after an obstacle?

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Answer

The wider the wavelength, the slower the transition between a maximum and a minimum. After the object, this translates into a blurry effect.

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Question

What is the effect of the interaction between a wave and an obstacle on the phase?

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Answer

Some points of the wave front are shifted.

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Question

A phase shift occurred. What does that mean?

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Answer

It means that there is a misalignment of the crests and troughs of the wave.

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Question

Which object quality affects diffraction?

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Answer

Its width.

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Question

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

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Answer

A maximum.

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Question

What is the formula for calculating the points where constructive interference occurs?

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Answer

n (λ / 2) = d sin θ

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Question

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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Answer

The width of the obstacle should be smaller than the wavelength.

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