Radio waves of wavelength 163 m from a galaxy reach a radio telescope by two separate paths as shown in the figure below. One is a direct path to the receiver, which is situated on the edge of a tall cliff by the ocean, and the second is by reflection off the water. As the galaxy rises in the east over the water, the first minimum of destructive interference occurs when the galaxy is θ = 29.5° above the horizon. Find the height of the radio telescope dish above the water.

To find the height of the radio telescope dish above the water, we can use the concept of path difference and interference.

First, let's define the variables:
λ = wavelength of the radio waves (163 m)
θ = angle of the galaxy above the horizon (29.5°)

Now, let's analyze the interference that occurs when the radio waves reach the telescope. The direct path and the reflected path should have a path difference of half a wavelength (λ/2) for destructive interference to occur.

From the figure, we can see that the path difference (d) can be represented as:

d = 2h tan(θ)

Where h is the height of the radio telescope dish above the water.

To achieve the first minimum of destructive interference, the path difference should be equal to λ/2:

λ/2 = 2h tan(θ)

Rearranging the equation, we have:

h = (λ/4) / tan(θ)

Substituting the given values, we get:

h = ((163 m)/4) / tan(29.5°)

Calculating the value, we find:

h ≈ 70.7 m

Therefore, the height of the radio telescope dish above the water is approximately 70.7 meters.

To find the height of the radio telescope dish above the water, we need to understand the concept of interference and how it relates to the given information.

Interference occurs when two or more waves meet at the same point in space. Depending on their relative phase, they can either add up constructively (creating a larger wave) or cancel each other out destructively (creating a smaller or no wave).

In the given scenario, the radio waves from the galaxy reach the radio telescope by two separate paths - one direct path and the other by reflection off the water. The interference between these two paths creates a pattern of constructive and destructive interference.

The first minimum of destructive interference occurs when the path difference between the two waves is equal to half of the wavelength. In this case, the wavelength of the radio waves is given as 163 m.

According to the given information, the galaxy is at an angle of θ = 29.5° above the horizon when the first minimum occurs. Let's denote the height of the radio telescope dish above the water as h.

To solve for h, we can use the trigonometric relationship between the angle of incidence, angle of reflection, and the path difference.

The angle of incidence is equal to the angle of reflection, which is θ in this case. The path difference can be represented as 2h, as the waves travel twice the height h.

Now, we can set up the trigonometric relationship as follows:
2h = (n + 1/2)λ

where n is an integer representing the order of the minimum (in this case, the first minimum, so n = 0) and λ is the wavelength.

Plugging in the values, we get:
2h = (0 + 1/2) * 163
2h = 1/2 * 163
2h = 81.5
h = 81.5 / 2
h = 40.75 m

Therefore, the height of the radio telescope dish above the water is 40.75 meters.