Destructive interference: formula and equations, examples, exercise

The destructive interference , physical, occurs when two independent waves are combined in the same region of space are compensated. Then the crests of one wave meet the valleys of the other and the result is a wave with zero amplitude.

Several waves pass through the same point in space without a problem, and then each one goes its way unaffected, like the waves in water in the following figure:

Let us suppose two waves of equal amplitude A and frequency ω, which we will call y 1 and y 2 , which can be described mathematically by the equations:

1 = One sin (kx-ωt)

2 = One sin (kx-ωt + φ)

The second wave y 2 has a delay φ with respect to the first. When combined, as the waves can overlap perfectly, they result in a resultant wave called y R :

R = y 1 + y 2 = One sin (kx-ωt) + One sin (kx-ωt + φ)

Using trigonometric identity:

sin α + sin β = 2 sin (α + β) / 2. cos (α – β) / 2

The equation for and R becomes:

and R = [2A cos (φ / 2)] sin (kx – ωt + φ / 2)

Now this new wave has a resulting amplitude A R = 2A cos (φ / 2), which depends on the phase difference. When this phase difference acquires the values ​​+ π or –π, the resulting amplitude is:

R = 2A cos (± π / 2) = 0

Since cos (± π / 2) = 0. Precisely then is when destructive interference between waves occurs. In general, if the cosine argument is of the form ± kπ / 2 with k odd, the amplitude A  is 0.

Examples of destructive interference

As we have seen, when two or more waves pass through a point at the same time, they overlap, giving rise to a resulting wave whose amplitude depends on the phase difference between the participants.

The resulting wave has the same frequency and wave number as the original waves. In the animation below, two waves overlap in blue and green. The resulting wave is red in color.

Amplitude grows when interference is constructive, but cancels out when it is destructive.

Waves with the same amplitude and frequency are called coherent waves , as long as they maintain the same phase difference between the two. An example of a coherent wave is laser light.

Condition for destructive interference

When the blue and green waves are 180° out of phase at a given point (see figure 2), it means that as they move they have phase differences φ of π radians, 3π radians, 5π radians, and so on.

So dividing the resulting amplitude argument by 2 gives (π / 2) radians, (3π / 2) radians … And the cosine of such angles is always 0. So the interference is destructive and the amplitude becomes 0.

Destructive interference of waves in water

Suppose two coherent waves start in phase with each other. Such waves can be those that propagate through the water, thanks to two vibrating bars. If the two waves travel to the same point P over different distances, the phase difference is proportional to the path difference.

Since a wavelength λ is equal to a difference of 2π radians, it follows that:

1d 1 – d 2 │ / λ = phase difference / 2π radians

Phase difference = 2π x│d 1 – d 2 │ / λ

If the path difference is an odd number of half wavelengths, ie: λ/2, 3λ/2, 5λ/2 and so on, the interference is destructive.

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But if the path difference is an even number of wavelengths, the interference is constructive and the amplitudes are added at point P.

Destructive interference from light waves

Light waves can also interfere with one another, as Thomas Young revealed in 1801 through his celebrated double-slit experiment.

Young passed the light through a slit made in an opaque screen which, according to Huygens’ principle, in turn generates two secondary light sources. These sources continued their way through a second opaque screen with two slits and the resulting light was projected onto the wall.

The diagram is seen in the following image:

Young noticed a distinct pattern of alternating light and dark lines. When light sources interfere destructively, the lines are dark, but if they do so constructively, the lines are clear.

Another interesting example of interference is soap bubbles. They are very thin films, in which interference occurs because light is reflected and refracted on surfaces that border the soap film, above and below.

As film thickness is comparable to wavelength, light behaves in the same way as when it passes through the two Young slits. The result is a color pattern if the incident light is white.

This is because white light is not monochromatic, but contains all wavelengths (frequencies) of the visible spectrum. And each wavelength is seen as a different color.

Exercise solved

Two identical speakers driven by the same oscillator are 10 feet apart, and a listener is 6 feet from the midpoint of separation between the speakers at point O.

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It is then transferred to point P, at a perpendicular distance of 0.350 from point O, as shown in the figure. There he stops hearing the sound for the first time. What is the wavelength at which the oscillator emits?

Solution

The resulting wave amplitude is 0, so interference is destructive. If you have to:

Phase difference = 2π x│r 1 – r 2 │ / λ

By Pythagoras’ theorem applied to the shaded triangles in the figure:

1 = √1.15 2 + 8 2 m = 8.08 m; r 2 = √1.85 2 + 8 2 m = 8.21 m

1r 1 – r 2 │ = .08.08 – 8.21 │ m = 0.13 m

The minima occurs at λ / 2, 3λ / 2, 5λ / 2 … The first corresponds to λ / 2, so the formula for the phase difference is:

λ = 2π x│r 1 – r 2 │ / phase difference

But the phase difference between the waves must be π, so the amplitude A R = 2A cos (φ / 2) is zero, so:

λ = 2π x│r 1 – r 2 │ / π = 2 x 0.13 m = 0.26 m

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