- Examples of destructive interference
- Condition for destructive interference
- Destructive interference of waves in the water
- Destructive interference of light waves
- Exercise resolved
- Solution
- References
The destructive interference, in physics, is when two independent waves are combined in the same region of space are offset. Then the crests of one of the waves meet the valleys of the other and the result is a wave with zero amplitude.
Several waves pass without problem through the same point in space and then each one continues on its way without being affected, like the waves in water in the following figure:
Figure 1. Raindrops produce ripples on the surface of the water. When the resulting waves have zero amplitude, the interference is said to be destructive. Source: Pixabay.
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 means of the equations:
y 1 = A sin (kx-ωt)
y 2 = A sin (kx-ωt + φ)
The second wave y 2 has an offset φ with respect to the first. When combined, since the waves can easily overlap, they give rise to a resulting wave called y R:
y R = y 1 + y 2 = A sin (kx-ωt) + A sin (kx-ωt + φ)
Using the trigonometric identity:
sin α + sin β = 2 sin (α + β) / 2. cos (α - β) / 2
The equation for y R becomes:
and R = sin (kx - ωt + φ / 2)
Now this new wave has a resultant 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:
A R = 2A cos (± π / 2) = 0
Since cos (± π / 2) = 0. It is precisely then that destructive interference occurs between the waves. In general, if the cosine argument is of the form ± kπ / 2 with odd k, the amplitude A R 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 following animation two waves in blue and green colors are superimposed. The resulting wave is in red.
The amplitude grows when the interference is constructive, but cancels out when it is destructive.
Figure 2. The blue and green colored waves are superimposed to give rise to the red colored wave. Source: Wikimedia Commons.
Waves that have the same amplitude and frequency are called coherent waves, as long as they keep the same phase difference φ fixed between them. An example of a coherent wave is laser light.
Condition for destructive interference
When the blue and green waves are 180º 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.
In this way, dividing the argument of the resulting amplitude by 2, results in (π / 2) radians, (3π / 2) radians… And the cosine of such angles is always 0. Therefore the interference is destructive and the amplitude becomes 0.
Destructive interference of waves in the water
Suppose that 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, traveling different distances, the phase difference is proportional to the path difference.
Figure 3. The waves produced by the two sources travel in the water to point P. Source: Giambattista, A. Physics.
Since a wavelength λ is equal to a difference of 2π radians, then it is true that:
│d 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, that is: λ / 2, 3λ / 2, 5λ / 2 and so on, then the interference is destructive.
But if the path difference is an even number of wavelengths, the interference is constructive and the amplitudes add up at point P.
Destructive interference of light waves
Light waves can also interfere with each other, as Thomas Young showed in 1801 through his celebrated double slit experiment.
Young made light pass through a slit made on an opaque screen, which, according to Huygens' principle, 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 a wall.
The diagram is seen in the following image:
Figure 4. The pattern of light and dark lines on the right wall is due to constructive and destructive interference, respectively. Source: Wikimedia Commons.
Young observed a distinctive 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 light.
Another interesting example of interference is soap bubbles. These are very thin films, in which the interference occurs because light is reflected and refracted on the surfaces that limit the soap film, both above and below.
Figure 5. An interference pattern forms on a thin film of soap. Source: Pxfuel.
As the thickness of the film is comparable to the wavelength, the light behaves the same as it does when it passes through the two Young's 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 looks like a different color.
Exercise resolved
Two identical speakers driven by the same oscillator are 3 meters apart and a listener is 6 meters away from the midpoint of separation between the speakers, at point O.
It is then translated to point P, at a perpendicular distance of 0.350 from point O, as shown in the figure. There you stop hearing the sound for the first time. What is the wavelength at which the oscillator emits?
Figure 6. Diagram for the resolved exercise. Source: Serway, R. Physics for Science and Engineering.
Solution
The amplitude of the resulting wave is 0, therefore the interference is destructive. It has to:
Phase difference = 2π x│r 1 - r 2 │ / λ
By the Pythagorean theorem applied to the shaded triangles in the figure:
r 1 = √1.15 2 + 8 2 m = 8.08 m; r 2 = √1.85 2 + 8 2 m = 8.21 m
│r 1 - r 2 │ = │8.08 - 8.21 │ m = 0.13 m
The minima occur in λ / 2, 3λ / 2, 5λ / 2… The first corresponds to λ / 2, then, from the formula for the phase difference we have:
λ = 2π x│r 1 - r 2 │ / Phase difference
But the phase difference between the waves must be π, so that the amplitude A R = 2A cos (φ / 2) is zero, then:
λ = 2π x│r 1 - r 2 │ / π = 2 x 0.13 m = 0.26 m
References
- Figueroa, D. (2005). Series: Physics for Science and Engineering. Volume 7. Waves and Quantum Physics. Edited by Douglas Figueroa (USB).
- Fisicalab. Wave interference. Recovered from: fisicalab.com.
- Giambattista, A. 2010. Physics. 2nd. Ed. McGraw Hill.
- Serway, R. Physics for Science and Engineering. Volume 1. 7th. Ed. Cengage Learning.
- Wikipedia. Thin film interference. Source: es.wikipedia.org.