Destructive interference of direct and reflected rays

[adelianne]

QUESTION:
Astronomers observed the interference between a radio wave arriving at their antenna directly from the Sun, and on a path involving reflection from the surface of the sea. (http://i62.photobucket.com/albums/h101/kaikibana/problem1.jpg)
Assume that the radio waves have a frequency of 6.0 x 10^7Hz, and that the radio receiver is 25m above the sea's surface. What is the smallest angle θ above the horizon that will give destructive interference of the waves at the receiver?

EQUATIONS USED:
y'_m = (m + 1/2)(λL/d) where m = 0, 1, 2...
λ = v/f

ATTEMPTED SOLUTION:
The problem looked similar to the double slit diagrams, and the question's asking for destructive interference. So I figured using the equation for the position of dark fringes from the double slit would solve the problem..
I let m=0 to find the 'smallest angle'. I let d=50m (I drew a vertical line from the point of reflection off the sea up to a point on the top ray. I assumed both the rays from the sun were nearly parallel.. hence 50m is equivalent to the 'distance between two slits.') for lambda, I found it to be 5 m.

25 = (0 + 1/2)(5L/50)
L = 500 m

Then using L (distance from point of reflection to the receiver), I found theta:

arctan(25/500) = 2.86 degrees

OKAY. so i think this answer's wrong.. I realized that after being reflected, the bottom ray was 1/2λ out of phase. But the equation I used is for two rays which aren't originally out of phase (I think?). I'm not even sure if letting m=0 would be finding the smallest angle..
So... help? please? any hints would be appreciated >_<

 

[Carid]

Rays from the Sun are parallel.
You have the right idea. The 50m vertical is the place from which the top ray and the bottom ray have equal path lengths going to the receiver.
Since the reflection gives us a half wavelength path difference (which guarantees destructive interference) then the two paths from the Sun to the 50m vertical will need to differ by a whole number of wavelengths.
For the first destructive fringe this number will be one, not zero. My reasoning, which may be false, is that the half wavelength is due to a reflection and not a real difference in path length, so geometrically it doesn't make sense to consider the lower reflected ray as traversing exactly the same physical distance as the upper one.

 

[thomate1]

RESPONSE:
Your approach to using the double slit equation is a good start, but there are a few things that need to be adjusted.

Firstly, the equation you used is for the position of dark fringes in a double slit experiment, where the two sources of light are coherent and in phase with each other. In this case, the two sources (direct and reflected rays) are not coherent and will not produce a pattern of dark and light fringes. Instead, we need to use the equation for destructive interference, which is:

d sinθ = mλ

Where d is the distance between the two sources, θ is the angle between the two sources and the receiver, m is an integer representing the order of the interference, and λ is the wavelength of the waves.

In this case, we can rearrange the equation to solve for the angle θ:

θ = sin^-1(mλ/d)

Since we are looking for the smallest angle, we can let m=0, as you did, and solve for θ:

θ = sin^-1(0.5λ/d)

Now we need to find the value of λ. You correctly calculated the wavelength of the waves using the equation λ = v/f. However, you used the speed of light, which is not the appropriate speed for radio waves. Instead, we need to use the speed of radio waves, which is approximately 3 x 10^8 m/s.

λ = (3 x 10^8 m/s) / (6.0 x 10^7 Hz) = 5 m

Now we can plug in the values for λ and d (which is 50m, as you correctly identified) into our equation for θ:

θ = sin^-1(0.5(5m) / (50m)) = 5.7 degrees

Therefore, the smallest angle that will produce destructive interference of the waves at the receiver is approximately 5.7 degrees above the horizon.

One thing to note is that this calculation assumes that the waves are perfectly parallel and that there is no diffraction or other effects that may change the angle of interference. In reality, the angle may be slightly different due to these factors, but this calculation gives a good estimate.