Calculating Plane Height from Radio Wave Interference

AI Thread Summary
the problem involves calculating the minimum height of a plane causing radio wave interference at a frequency of 600kHz, received 10km from the transmitter. The interference results from the direct beam and a reflected beam from charged particles in the atmosphere. To approach the solution, the wavelength of the radio wave is calculated as 500m using the speed of light and frequency. The reflection occurs above the midpoint between the transmitter and receiver, leading to a distance equation that must exceed 10km by an odd number of half wavelengths. The discussion emphasizes using trigonometric relationships and interference formulas to derive the plane's height.
lektor
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Hi, first time posting, nice forums :)

Recently in my preparation for my scholarship exams later this year, i was approached by a question which has me rather confused.

Im not neccesarily asking for the final definate answer, but it would be nice for some help on how to approach the problem.

Radio waves of frequency 600kHz are received at a location 10Km from the transmitter. The radio reception temporarily fades due to the interference between the direct beam and that reflected without phase change from a horizontal layer of charged particles formed in the atmosphere by a passing plane. Calculate the minimum height of the plane



formulas suggested were λ = dx/L

and dsinθ = mλ


p.s how do you draw the symbols ?
 
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Welcome to the forum, Lektor.
Since the beam reaches the the receiver after the reflection, the reflection must have occurred somewhere above the spot halfway between the receiver & the transmitter (i.e., 5km away from either).(I reckon this is what you couldn't hit upon.)If the height of the plane is H, the distance traveled by the beam is ( because the reflection didn't change the phase) 2* sqrt( H^2 + 5^2). This distance must exceed 10km by an odd number of half wavelengths.
Regards,
Einstone.
 


Hi there, thank you for posting and welcome to the forums! This is a great question and a common one in physics and engineering exams. Let's break down the problem and see how we can approach it.

First, we need to understand the given information. We know that radio waves of frequency 600kHz are being received at a location 10km from the transmitter. We also know that the radio reception temporarily fades due to interference between the direct beam and a reflected beam from a horizontal layer of charged particles in the atmosphere. This interference is caused by a passing plane. We are asked to calculate the minimum height of the plane.

To solve this problem, we need to use the two formulas that were suggested: λ = dx/L and dsinθ = mλ. The first formula relates the wavelength (λ) of the radio wave to the distance between the transmitter and receiver (d) and the distance between the transmitter and the horizontal layer of charged particles (L). The second formula is the equation for constructive interference, where d is the distance between the two interfering waves, θ is the angle between them, and m is an integer representing the number of wavelengths in that distance.

To use these formulas, we need to find the wavelength of the radio wave. We can do this by using the formula c = fλ, where c is the speed of light (3x10^8 m/s) and f is the frequency of the wave (600kHz = 6x10^5 Hz). This gives us a wavelength of 500m. Now, we can use the first formula to find the distance between the transmitter and the horizontal layer of charged particles (L). Rearranging the formula, we get L = dx/λ = (10km x 500m)/500m = 10km.

Next, we need to find the angle (θ) between the direct beam and the reflected beam. This can be done by using basic trigonometry. We know that the distance between the transmitter and receiver is 10km, and we just found that the distance between the transmitter and the horizontal layer is also 10km. This forms a right angle triangle, where the angle opposite to the hypotenuse (θ) can be found using the formula sinθ = opposite/hypotenuse. Plugging in our values, we get sinθ = 10km/10km = 1. This means that θ = 90 degrees.

Now
 
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