Fraunhofer diffraction for a small obstracle

[Mahasweta]

1. Two towers built on hills are 50 km apart and the line joining those passes above a hill midway between them. what is the longest wavelength of radio waves which can be sent between the towers without serious diffraction effects caused by the central hill?


2. Is the width of the obstrucle in necessary?



3. I think the problem can be related to Fraunhofer diffraction for a small obstracle. The condition for this type of diffraction is R≥ a^2/λ. But this requires the maximum width of obstracle.

 

[Simon Bridge]

If the transmitter used extremely short wavelengths - what would happen?
Notice that the line joining the towers passes above the hill.

 

[Mahasweta]

But in case I want some approximate numerical value of the wavelength that can pass without any diffraction effect? What should I do?

 

[Simon Bridge]

But in case I want some approximate numerical value of the wavelength that can pass without any diffraction effect? What should I do?

What you do, to start with, is answer the question in post #2. You should also follow suggestions.
I'll only show you one step at a time because you are the one who has to take the steps, and everyone's path is different.

 

[paranormal]

I would like to provide some clarification and additional information regarding Fraunhofer diffraction and its application to this scenario. Firstly, Fraunhofer diffraction is a type of diffraction that occurs when the source of light is at an infinite distance from the diffracting object, resulting in a parallel beam of light incident on the object. In this case, the two towers can be considered as the source of light and the central hill as the diffracting object.

To answer the first question, the longest wavelength of radio waves that can be sent between the towers without serious diffraction effects caused by the central hill can be calculated using the Fraunhofer diffraction equation R≥ a^2/λ. Here, R represents the distance between the towers (50 km), a represents the width of the central hill, and λ represents the wavelength of the radio waves. By rearranging the equation, we can calculate the maximum wavelength of radio waves that can be used without significant diffraction effects: λ = a^2/R. Therefore, if we assume the width of the central hill to be 1 km, the longest wavelength of radio waves that can be used is 1 km^2/50 km = 0.02 km or 20 meters.

Regarding the second question, the width of the obstacle is indeed necessary in this scenario as it affects the diffraction of the radio waves. The wider the obstacle, the stronger the diffraction effects will be, causing a decrease in the maximum wavelength that can be used without significant diffraction. Therefore, it is important to consider the width of the obstacle when calculating the maximum wavelength of radio waves that can be used.

Lastly, as you have correctly pointed out, this problem can be related to Fraunhofer diffraction for a small obstacle. However, it is important to note that the Fraunhofer diffraction equation R≥ a^2/λ is only valid for a small obstacle, where the width of the obstacle is much smaller than the distance between the source and the obstacle. In this scenario, we can assume that the width of the central hill is much smaller than the distance between the towers, making the Fraunhofer diffraction equation applicable. However, if the width of the obstacle is comparable to the distance between the towers, other diffraction equations and principles may need to be considered.

In conclusion, the longest wavelength of radio waves that can be sent between the two