**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 E = 0.38 radians

Part F = 1900 m

**Solutions Below:**

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

- 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

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

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

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

Express your answer in radians, to two significant figures.

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

0.38 radians

Part F

_{m}= 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 y

_{m}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.

Express your answer in meters to two significant figures.

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):

y_{m} = R * (m + 1/2)λ / d

y_{m} = 5000 * (0 + 1/2) * 300 / 400

y_{m} = 1875

1900 m

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

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(θ

_{1}) + sin(θ_{2}) = 2sin([θ_{1}+ θ_{2}] / 2) * cos([θ_{1}- θ_{2}] / 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!