Question

Radio waves from a star, of wavelength 282 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

Answer 1

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To start the waves from the same point draw a line from the reflection point perpendicular to the two rays from the star.
The distance the direct ray travels from this perpendicular line to the receiver is ,say, D.
The distance the reflected ray travels from the reflection point to the receiver is ,say, L.
The first interference minimum will occur when the extra distance " L-D" traveled by the reflected ray is a half wavelength (108 m)

L - D = 141

Now D , L and that perpendicular line form a right triangle with angle 67 deg opposite side "D"

So , D = LSin(68) = (.93)L

then;

141 = L - (.93)L = (.07)L

L = 2014 m

Then , as before, the cliff height is;

cliff height = LSin(22)

= 754.46m