Solving the Problem of Plane Height for Destructive Interference

  • Thread starter Thread starter atse1900
  • Start date Start date
  • Tags Tags
    Interference Slit
Click For Summary
SUMMARY

The discussion focuses on calculating the height of an airplane causing destructive interference for radio waves broadcast at 1500 kHz. The direct path from the radio station to the home receiver is 20 km, while the airplane is positioned approximately 100 m above the receiver. To achieve destructive interference, the path length difference must equal an odd multiple of half the wavelength. The wavelength for 1500 kHz is approximately 200 meters, leading to a specific calculation for the airplane's height based on the interference condition.

PREREQUISITES
  • Understanding of wave interference principles
  • Knowledge of radio frequency concepts
  • Familiarity with basic trigonometry
  • Ability to calculate wavelength from frequency
NEXT STEPS
  • Calculate the wavelength for 1500 kHz radio waves
  • Explore the concept of path length difference in wave interference
  • Study the effects of phase changes in wave reflections
  • Investigate real-world applications of destructive interference in telecommunications
USEFUL FOR

Students and professionals in physics, telecommunications engineers, and anyone interested in wave behavior and interference patterns.

atse1900
Messages
8
Reaction score
0
Waves broadcast by a 1500-kHz radio station arrive at a home reciever by two paths. One is a direct path, and the second is by reflection off an airplane directly above the home receiver. The airplane is approximately 100 m above the home receiver, and the direct distance from the station to the home is 20 km. What is the exact height of the airplane if destructive interference is occurring? (Assume that no phase change occurs on rfelective from the plane.)

I have no idea how would you solve this problem. Can someone help me out please? Thanks. =].
 
Physics news on Phys.org
To have destructive interfence at some point, the waves must be 180° out of phase, i.e. the maximum of one coincides with the minimum of the other. So the difference between path lengths must include one-half of the wave length, but not necessarily exactly one-half of a wavelength.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
13
Views
5K
  • · Replies 110 ·
4
Replies
110
Views
9K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
15K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 1 ·
Replies
1
Views
6K
  • · Replies 9 ·
Replies
9
Views
4K