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March 31, 2015

March 31, 2015

Posted by **Sam** on Sunday, January 25, 2009 at 8:23pm.

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.

- Physics - What equations -
**Sam**, Sunday, January 25, 2009 at 8:30pmWhat equations would you use for those problems?

- Physics -
**Damon**, Sunday, January 25, 2009 at 8:55pmDifference 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.

- Physics -
**Damon**, Sunday, January 25, 2009 at 8:57pmI need the picture for the second one, sorry.

- Physics -
**k**, Thursday, January 29, 2009 at 11:32pmto give destructive interference, you must have path difference, x- y = (2n + 1) * lamda/2 where n = (0, 1, 2, ....)

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