jolly_math said:
I'm confused by how "the one below the water is out of phase by half a wavelength" is derived from the question.
The underlying reason is that, when a radio wave reflects off of the surface of water, the radio wave undergoes a (1/2)λ phase shift. This is taken into account by taking the "phantom" underwater point source to be emitting waves that are (1/2)λ out of phase with the above water point source.
Did you discuss phase shifts due to reflection in class? There is a rule for deciding whether or not a light wave has a (1/2)λ phase shift due to reflection. It involves the indices of refraction of the two media where the reflection takes place. The same rule applies to radio waves.
[EDIT: In introductory physics courses, it is often stated that there will be a (1/2)λ phase shift at reflection if the ratio of the indices of refraction ##n_2/n_1## is greater than 1. Actually, there are exceptions to this that depend on the angle of incidence and the polarization of the waves. But, this is not usually covered in introductory classes. So, I guess that in this question you were just expected to assume that there will be a (1/2)λ phase shift due to reflection since ##n_2/n_1## is greater than 1 for an air-to-water interface.]
jolly_math said:
Also, why is the path difference (1/2)λ?
In the first post you asked, "I don't understand why then d sin θ = λ - wouldn't it be d sin θ = (1/2)λ since it is out of phase?"
I wanted you to see that
if you take the path difference to be (1/2)λ, then the waves at the receiving station would not result in the receiving station losing contact with the ship. So, (1/2)λ path difference is not going to work.
jolly_math said:
When the question states that "radio contact is momentarily lost for the first time", is it looking for constructive or destructive interference?
The contact is lost because the waves from the two point sources "cancel out" at the receiving station. Would this be constructive or destructive interference?