## Mastering Physics Solutions: Understanding Two-Source Interference

Understanding Two-Source Interference

Part A = An observer stands on a road that runs five kilometers away from the two synchronized transmitting towers for a radio station.
Part B = 2 only
Part C = 1 and 2 only
Part D = 1 and 3 only
Part F = 1900 m

Solutions Below:

For solving two-source interference problems, there exists a standard set of equations that give the conditions for constructive and destructive interference. These equations are usually derived in the context of Young’s double slit experiment, though they may actually be applied to a large number of other situations. The underlying assumptions upon which these equations are based are that two sources of coherent, nearly monochromatic light are available, and that their interference pattern is observed at a distance very large in comparison to the separation of the sources. Monochromatic means that the wavelengths of the waves, which determine color for visible light, are nearly identical. Coherent means that the waves are in phase when they leave the two sources.

In Young’s experiment, these two sources corresponded to the two slits (hence such phenomena are often called two-slit interference). Under these assumptions, the conditions for constructive and destructive interference are as follows:

for constructive interference:

dsin(θ) = mλ (m = 0, ±1, ±2, …),

and for destructive interference

dsin(θ) = (m + 1/2)λ (m = 0, ±1, ±2, …),

where d is the separation between the two sources, λ is the wavelength of the light, m is an arbitrary integer, and θ is the angle between a line perpendicular to the line segment connecting the sources and the line from the midpoint of that segment to the point where the interference is being observed. These equations are often spoken of in terms of visible light, but they are, in fact, valid for any sort of waves, as long as the two sources fit the other criteria given.

Part A

Which of the following scenerios fits all of the criteria for the two-source interference equations to be valid?

• An observer is standing far away from two red LED signal lights.
• Light from an incandescent bulb shines onto a screen with a single slit; then the light shines onto a screen with two slits in it and the light from the two slits finally shines onto a far-away screen.
• An observer stands on a road far away from two neighboring radio towers for different radio stations.
• Light from an incandescent bulb shines onto a screen with a single slit; then the light shines onto a screen with two slits in it and the light from the two slits finally shines onto a nearby screen.
• An observer stands on a road that runs five kilometers away from the two synchronized transmitting towers for a radio station.

An observer stands on a road that runs five kilometers away from the two synchronized transmitting towers for a radio station.

Part B

Which of the following statements explain why the two-source interference equations are not valid for an observer far away from two red LED signal lights?

1. not monochromatic sources
2. incoherent sources
3. observed from a distance similar to or smaller than the separation between the sources

• 1 only
• 2 only
• 3 only
• 1 and 2 only
• 1 and 3 only
• 2 and 3 only
• all three

2 only

Part C

Why are the two-source interference equations not valid for an observer on a road far away from two neighboring radio towers for different radio stations?

1. sources emit at different frequencies (i.e., not monochromatic sources)
2. incoherent sources
3. observed from a distance similar to or smaller than the separation between the sources

• 1 only
• 2 only
• 3 only
• 1 and 2 only
• 1 and 3 only
• 2 and 3 only
• all three

1 and 2 only

Part D

Why are the two-source interference equations not valid for light from an incandescent bulb that shines onto a screen with a single slit, and then the light shines onto a screen with two slits in it and the light from the two slits finally shines onto a nearby screen?

1. not monochromatic sources
2. incoherent sources
3. observed from a distance similar to or smaller than the separation between the sources

• 1 only
• 2 only
• 3 only
• 1 and 2 only
• 1 and 3 only
• 2 and 3 only
• all three

1 and 3 only

Part E

Consider a road that runs parallel to the line connecting a pair of radio towers that transmit the same station (assume that their transmissions are synchronized), which has an AM frequency of 1000 kilohertz. If the road is 5 kilometers from the towers and the towers are separated by 400 meters, find the angle θ to the first point of minimum signal (m = 0). Hint: A frequency of kilohertz corresponds to a wavelength of 300 meters for radio waves.

Since the problem asks for the minima, use the equation for destructive interference:

dsin(θ) = (m + 1/2)λ
400 * sin(θ) = (0 + 1/2) * 300
400 * sin(θ) = 150
sin(θ) = 0.375
θ = 0.384

Part F

If the angle θ in the two-source interference equations is small, then using small-angle approximations yields the equation ym = R * (m + 1/2)λ / d, where R is the distance from the sources to the points where they are being detected (in Young’s experiment the screen, in this example the road), and ym is the distance from the central maximum to the minimum of order m. Use this equation to find the distance from the central maximum to the minimum in the previous part.

This problem is kind of a pain because it doesn’t tell you what to use for “R”. Mastering Physics wants you to use the distance to the road (5000 m):

ym = R * (m + 1/2)λ / d
ym = 5000 * (0 + 1/2) * 300 / 400
ym = 1875

1900 m

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### 2 Responses to Mastering Physics Solutions: Understanding Two-Source Interference

1. avid subedi says:

hello, i am from Nepal and i am a bit confused in interference of light. I want to know the equations of constructive and destructive interference of light using cos

• Mastering Physics Solutions says:

Well we will try and help you, but we’re not exactly sure what you wish to know, so we will give you some general information about wave interference and hope it’s what you need! You know that light is an electromagnetic wave, and there are peaks (maxima) and troughs (minima) in the wave. If the wave is purely sinusoidal, you ordinarily would not need to use cos. But you can incorporate cos (for waves that are not entirely sinusoidal) by using trig identities.

When there are two waves interacting, you can use the expressions for two waves to find an overall wave:

y = D*sin(kx – ωt) + D*sin(kx – ωt + φ)

Where φ is the phase shift between the two waves.

And using trig identities, we can get:

sin(θ1) + sin(θ2) = 2sin([θ1 + θ2] / 2) * cos([θ1 - θ2] / 2)

Which gives:

y = 2D * cos(φ / 2) * sin(kx – ωt + φ / 2)

For constructive interference, amplitude is a maximum, so |cos(φ / 2)| = 1 and therefore φ = 0, 2π, etc.

For destructive interference, amplitude is a minimum (zero), so |cos(φ / 2)| = 0 and therefore φ = π, 3π, 5π, etc.

Note that these equations give you the overall form of the wave.

Hope this helps!