Homework Help Forum: Physics

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

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.

• Physics - What equations - Sam, Sunday, January 25, 2009 at 8:30pm

  What equations would you use for those problems?

  
  

• Physics - Damon, Sunday, January 25, 2009 at 8:55pm

  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.

  
  

• Physics - Damon, Sunday, January 25, 2009 at 8:57pm

  I need the picture for the second one, sorry.

  
  

• Physics - k, Thursday, January 29, 2009 at 11:32pm

  to give destructive interference, you must have path difference, x- y = (2n + 1) * lamda/2 where n = (0, 1, 2, ....)

  
  

Answer this Question

First Name:
School Subject:
Answer: