Radio waves from a star, of wavelength 212 m, reach a radio telescope by two sep

Question

Radio waves from a star, of wavelength 212 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 = 26.0° 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

Use ideas from Fraunhofer diffraction:
Destructive interference occurs when
path difference = (odd integer) * (wavelength / 2)
For the first minima, this odd integer is 1.

To find the path difference:
Drop a perpendicular from the point of reflection R onto the direct path. Say it intersects the direct path at P.

Then the paths of direct ray till P and of reflected ray till R are equal.

Path difference = RT - PT where T is the location of the telescope.
PT = RT cos (2)
and
RT = h / sin , where h is the height of the cliff.

Substitute to get
h (1 - cos(2)) / sin = (wavelength/2)

Substitute and wavelength, and solve for 'h'.

Therefore, h (1 - cos(52)) / sin26 = (290/2)

h (1 – 0.6156) / 0.4383 = 145

h (0.8770) = 145

h = 145 /(0.8770)

h = 165.33 m

Therefore,

The height of the cliff is 165.33 m

Related Questions

Navigate

• Browse (All)
• Browse R
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.