# Wave interferance and determining an unknown height

• chris097
In summary, the conversation discusses the use of a radio telescope to measure the height of a cliff by analyzing radio waves from a star. The first minimum of destructive interference occurs at an angle of 23.7° above the horizon, leading to the calculation of the cliff's height as 550.7 meters. The conversation also mentions using a figure to better understand the problem and suggests double-checking the calculations.
chris097

## Homework Statement

Radio waves from a star, of wavelength 231 m, reach a radio telescope by two separate paths. One is a direct path to the receiver, which is situated on the edge of a cliff by the ocean. The second is by reflection off the water. The first minimum of destructive interference occurs when the star is 23.7° above the horizon. Calculate the height of the cliff. (Assume no phase change on reflection.)

The attempt at a solution

Since its destructive interference, the difference between the two lengths is half a wavelength (=115.5 m).
So let's say L-D=115.5 m.
D = Lsin(90-23.7) = Lsin(66.3)
L - (0.916)L = 115.5 m
0.0843L = 115.5 m
L = 1370 m

Height = Lsin(23.7)
= 550.7 m

Any idea what I'm doing wrong? Thanks :)

chris097 said:
The attempt at a solution

Since its destructive interference, the difference between the two lengths is half a wavelength (=115.5 m).
So let's say L-D=115.5 m.
D = Lsin(90-23.7) = Lsin(66.3)
I don't think 23.7 is the angle to use here. Did you draw a figure to show what is going on?

The rest of it looks okay:
L - (0.916)L = 115.5 m
0.0843L = 115.5 m
L = 1370 m

Height = Lsin(23.7)
= 550.7 m

Any idea what I'm doing wrong? Thanks :)

I would first like to commend you for your attempt at solving this problem using wave interference principles. It shows that you have a good understanding of the concept. However, there are a few things that could be improved upon in your solution.

Firstly, the equation D = Lsin(90-23.7) is incorrect. The correct equation should be D = Lsin(180-23.7), since the reflected wave travels an additional distance of L before reaching the receiver. This would give us D = Lsin(156.3).

Secondly, the value of L that you have calculated (1370 m) seems to be incorrect. Using the correct equation above, we can calculate L as follows:

L = D/sin(156.3)
= 115.5 m / sin(156.3)
= 1372.7 m

This value of L should be used in the final calculation for the height of the cliff.

Therefore, the correct height of the cliff would be:

Height = Lsin(23.7)
= (1372.7 m) * sin(23.7)
= 552.2 m

In summary, it seems that the incorrect equation for D and a minor calculation error for L led to the incorrect height calculated in your solution. But overall, your approach using wave interference principles is correct. Keep up the good work!

## 1. What is wave interference?

Wave interference is the phenomenon that occurs when two or more waves meet and interact with each other. This interaction can result in either constructive interference (where the waves reinforce each other) or destructive interference (where the waves cancel each other out).

## 2. How is wave interference used to determine an unknown height?

Wave interference can be used to determine an unknown height by measuring the wavelength of the waves and the distance between the two wave sources. By using the known relationship between wavelength, distance, and height, the unknown height can be calculated.

## 3. What factors can affect wave interference?

The factors that can affect wave interference include the wavelength of the waves, the amplitude of the waves, the distance between the wave sources, and the medium through which the waves are traveling.

## 4. How can we minimize unwanted wave interference?

To minimize unwanted wave interference, we can use techniques such as shielding (blocking out certain wavelengths), damping (reducing the amplitude of the waves), or adjusting the distance between the wave sources.

## 5. What are some practical applications of wave interference?

Wave interference has many practical applications, such as in radio and television broadcasting, noise-cancelling headphones, and medical imaging techniques like ultrasound. It is also used in engineering and physics experiments to study the behavior of waves and their interactions.

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