Radio waves from a star, of wavelength 224 m, reach a radio telescope by two sep

Question

Radio waves from a star, of wavelength 224 m, reach a radio telescope by two separate paths, as shown in the figure below (not drawn to scale). 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 theta = 28.0 degree above the horizon. Find the height of the cliff. (Assume no phase change on reflection. The image is not drawn to scale; assume that the height of the radio telescope is negligible compare to the height of the cliff.)

Explanation / Answer

For a minimum to occur, the path length difference is half a wavelength. The path length L1 is

Sin 28o = h/L1

The path length L2 is

Cos 56o = L2/L1

L2 = L1 Cos 56o

The optical path length L1 has an extra half wavelength.

L1 –(1/2) – L2 = (1/2)

L1– L2 =

L1– L1 Cos 56o =

L1(1–Cos 56o) = 224 m

L1= 224 m /(1–Cos 56o)

L1= 224 m /0.5592

L1= 400.58 m

This means the height is

h = L1 Sin 28o

h = 400.58 m *Sin 28o

h = 400.58 m *0.4695

h = 188.06 m

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