In the first application of
interferometric methods in radio
astronomy, Australian 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.
Assume that radio waves have a frequency of 6.0 × 107 Hz, and that the radio receiver is
25 m above the surface of the sea. What is the smallest angle θ above the horizon that
will give destructive interference of the waves at the receiver?
A converging lens has a focal length f, and a diverging
lens has a focal length –f, which has the same magnitude
as the converging lens. They are separated by a distance
D which is greater than f, as shown. Parallel light enters
from the left. Will the light be brought to a focus, and if
so where? Imagine that the parallel light rays in Problem 3 enter from the right. Explain why the
light will be brought to focus without using any equations.
4 answers
c = 3*10^8 m/s
distance = rate * time
so T = 1/f = lambda /3*10^8
lambda = 3*10^8/6*10^7 = .5*10^1 = 5 meters
So we want a path difference of at least 2.5 meters.
I despair of describing the geometry of my path length difference
The reflected ray hits the water at angle Theta and bounces to your eye at theta
The distance from your eye to the reflection point on the water is 25/sin theta
Now draw perpendicular from that point on water to the incoming ray to the tower.
The angle down from horizontal at tower to point on water is theta
The angle up from horizontal to ray from sun is also theta
so that angle between the two rays at the tower is 2 theta (as anyone who uses a bubble sextant for navigation knows)
so we have a right triangle with hypotenuse 25/sin theta
and angles 2 theta and 90 and 90-2theta
we want the difference betwen the straight leg to the tower and the hypotenuse to be at least 2.5 meters
I am going to leave the rest of the rig to you.
2 answers