Radio waves from a star, of wavelength 288 m, reach a radio telescope by two sep
ID: 1305064 • Letter: R
Question
Radio waves from a star, of wavelength 288 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 ? = 22.0 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
The path length difference = lambda/2 = 144 m.
The solution involves analysis of two right triangles.
Draw the diagram showing two parallel rays of a plane wave at theta = 22 d from horizontal, one going directly to the telescope at point T and the other reflected off the water at point R, and then forming a 44 d angle at T. Draw a perpendicular to the incoming rays that passes through R. Label the line's upper-ray intersection point A. This is a line of equal phase of the plane wave, and it completes a right triangle connecting A, R and T. The path length difference of the direct and reflected rays is the reflected ray length RT, which is the hypotenuse of the right triangle, minus the adjacent side AT. So the path length difference 144 m = RT(1-cos(44 d)). Solve this for RT.
A second right triangle below this one is made up of the cliff height H, RT and the distance from R to the base of the cliff. The angle at R is 25.3 d. Cliff height H = RTsin(22 d).
I got RT = 513.07 m and cliff height H = 192.199 m.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.