Wave Optics (Interference due to reflection)

In summary, we have a radio-wave transmitter and receiver separated by a distance of 50.0m, with a height of 35.0m above the ground. The receiver can receive signals directly and indirectly from the transmitter. The longest wavelengths for constructive and destructive interference can be determined using the conditions δ = (m+1/2)λ and δ = mλ, where δ is the path difference and m is an integer. However, the wavelength must be positive, so only the longest possible wavelengths should be chosen as solutions.
  • #1
cryora
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3

Homework Statement



A radio-wave transmitter and a receiver, separated by a distance d = 50.0m, are both h = 35.0m above the ground. The receiver can receive signals directly from the transmitter and indirectly from signals that reflect from the ground. Assume the ground is level between the transmitter and receiver and a 180 degree phase shift occurs upon reflection. Determine the longest wavelengths that interfere a) constructively and b) destructively.

Homework Equations



Since there is a 180 degree phase shift of the reflected signal, the conditions for constructive interference is:
δ = (m+1/2)λ
and conditions for destructive interference is:
δ = mλ
where δ is the path difference of the two waves, and m is an integer.

Unlike Young's Double slit, however, δ ≠ dsinθ, where d is the distance between two coherent light sources, because the extrapolated reflected ray does not make a 90 degree angle with the incoming ray before reflection.

The Attempt at a Solution



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From the law of reflection, it can be deduced that the length of the path that the reflected ray takes from the source to the point of reflection is the same as the path it takes from the point of reflection to the receiver. Let's call this length l. Thus, we have an isosceles triangle where the sides are the paths of the direct ray and the ray that is reflected.

The point at which the ray is reflected is halfway (d/2) between the radio towers.

Knowing h, from the Pythagorean theorem, l = sqrt[h^2 + (d/2)^2]

Path difference δ = 2l - d
= 2sqrt[h^2 + (d/2)^2] - d

This is = to (m + 1/2)λ for constructive interference and mλ for destructive interference.

I have a feeling that I've already did something wrong at this point... but anyway the question asks for the greatest λ. Since h and d are constants, it seems the only thing I can vary is m. I could solve for λ and set m = 0 to maximize λ, but I get the wrong answer for constructive interference... and for destructive interference, it would mean λ → ∞.

So I'm quite confused about what I need to do. Any hints?

I just realized that m may not be variable but fixed, and unknown... This doesn't really improve things by much lol
 
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  • #2
Your work is correct, but you need to give physically possible solution. The wavelength of the radio wave must be positive. So find the m values which result in positive wavelengths and choose the longest ones.

ehild
 

Related to Wave Optics (Interference due to reflection)

1. What is wave optics?

Wave optics is a branch of physics that studies the behavior of light as a wave. It focuses on the interference, diffraction, and polarization of light, which are phenomena that cannot be explained by geometric optics.

2. What is interference due to reflection?

Interference due to reflection occurs when two or more waves reflected from different surfaces meet and interact with each other. This interaction can result in constructive or destructive interference, which affects the intensity of the reflected light.

3. How does interference due to reflection affect the appearance of objects?

Interference due to reflection can create patterns of bright and dark areas on the surface of an object, known as interference fringes. These patterns can alter the perceived color and brightness of the object, depending on the wavelength and angle of the incident light.

4. What factors affect interference due to reflection?

The main factors that affect interference due to reflection are the wavelength and angle of the incident light, the distance between the reflecting surfaces, and the reflectivity of the surfaces. These factors determine whether the interference will be constructive or destructive.

5. How is interference due to reflection used in practical applications?

Interference due to reflection is used in various applications, such as in anti-reflective coatings for lenses and solar panels, in the production of holograms, and in interferometers for precise measurements. It is also an important concept in understanding the colors and patterns produced by thin films, such as soap bubbles and oil slicks.

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