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.
What equations would you use for those problems?
Difference in path length = 1/2 wavelength
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.
I need the picture for the second one, sorry.
to give destructive interference, you must have path difference, x- y = (2n + 1) * lamda/2 where n = (0, 1, 2, ....)
To find the minimum angle θ above the horizon that will give destructive interference of the waves at the receiver, we can use the concept of path difference.
The interference between the direct wave from the Sun and the reflected wave from the sea occurs when the path difference between the two waves is an odd multiple of half the wavelength. This condition leads to destructive interference.
The path difference can be calculated by considering the difference in distance traveled by the two waves. The direct wave travels a distance equal to the straight line distance from the Sun to the antenna, while the reflected wave travels a distance equal to the sum of the straight line distance from the Sun to the sea surface and the distance from the sea surface to the antenna.
Let's denote the distance from the sea surface to the antenna as h. The path difference Δd can be expressed as:
Δd = 2h * sin(θ)
where θ is the angle above the horizon. The condition for destructive interference is Δd = (2n + 1) * λ/2, where n is an integer.
Plugging in the values:
Δd = 2 * h * sin(θ) = (2n + 1) * λ/2
Since we are looking for the smallest angle θ, we can substitute n = 0 (for the first destructive interference) and λ = c/f, where c is the speed of light and f is the frequency of the radio waves.
2 * h * sin(θ) = λ/2 = c/2f
Simplifying,
sin(θ) = c / (4hf)
Using the given values, the frequency f = 6.0 × 10^7 Hz, and the radio receiver is 25 m above the sea:
sin(θ) = (3.0 × 10^8 m/s) / (4 * (6.0 × 10^7 Hz) * 25 m)
Now, to find the smallest angle θ, we take the inverse sine of the result:
θ = sin^(-1) [ (3.0 × 10^8 m/s) / (4 * (6.0 × 10^7 Hz) * 25 m) ]
Now, calculating this value will give you the smallest angle above the horizon that will give destructive interference of the waves at the receiver.