- #1

adelianne

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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 >_<