How will the change in wavelength affect diffraction in television signals?

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In summary, diffraction affects the transmission of signals through openings such as those between buildings. When signals have a shorter wavelength (such as with future digital television signals), diffraction can cause the central diffraction maximum to occupy a larger angle.
  • #1
poison_ivy
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I know I'm just a newbie, but i need your help guyz :blushing:
This is a matter of life and death Thanks!



--> In conventional television, signals are broadcast from towers to
home receivers. Even when a receiver is not in direct view of a tower because of a hill or building, it can still intercept a signal if the signal diffracts enough around the obstacle, into the obstacle's “shadow region”. Current television signals have a wavelength of about 50cm, but future digital television signals that are to be transmitted from towers will have a wavelength of about 10mm. (a) Will this change in wavelength increase or decrease the diffraction of signals into the shadow regions of obstacles? Justify your answer with physics. Now, assume that a signal passes through an opening of 5.0m width between too adjacent buildings. What is the angular spread of the central diffraction maximum (ie how large of an angle does the central maximum occupy) for wavelengths of (b) 50cm, (c) 10mm?
 
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  • #2
poison_ivy said:
I know I'm just a newbie, but i need your help guyz :blushing:
This is a matter of life and death Thanks!



--> In conventional television, signals are broadcast from towers to
home receivers. Even when a receiver is not in direct view of a tower because of a hill or building, it can still intercept a signal if the signal diffracts enough around the obstacle, into the obstacle's “shadow region”. Current television signals have a wavelength of about 50cm, but future digital television signals that are to be transmitted from towers will have a wavelength of about 10mm. (a) Will this change in wavelength increase or decrease the diffraction of signals into the shadow regions of obstacles? Justify your answer with physics. Now, assume that a signal passes through an opening of 5.0m width between too adjacent buildings. What is the angular spread of the central diffraction maximum (ie how large of an angle does the central maximum occupy) for wavelengths of (b) 50cm, (c) 10mm?
To begin, tell us what you know about diffraction.

AM
 
  • #3
Diffraction is the spreading of waves around obstacles or squeezing through a small hole. It occurs with sound waves, light, atoms, and subatomic particles. Its like if light waves from 2 slits can interfere
with each other, the light from different parts of the same slit also interfere.

I'm really not good in Physics, i already answered 2/4 of my homework but i can't do this and the other one. I really need help coz this is 15% of my total grade, and this is due tomorrow. Thanks...
 
  • #4
For question (a), you need only ask yourself "when is the effect of diffraction most noticeable?" (Think in terms of wavelength). For questions (b) and (c) you will need a formulae. What formulae do you know which can describe the diffraction pattern of a single slit?
 
  • #5
Oh ok, i already passed my HW this morning. So, thanks anyway ;)
 

1. What is diffraction and why is it important?

Diffraction is a phenomenon that occurs when waves encounter an obstacle or pass through an opening. It causes the waves to spread out and bend around the edges of the obstacle or opening. Diffraction is important because it allows us to study the properties of waves and understand how they interact with their surroundings.

2. What are some real-world applications of diffraction?

Diffraction is used in many areas of science and technology, including optics, acoustics, and radio waves. It is essential in the design and functioning of devices such as telescopes, microscopes, and antennas. Diffraction is also used in the study of crystal structures and in the production of holograms.

3. How does the size of the obstacle or opening affect diffraction patterns?

The size of the obstacle or opening has a direct impact on the diffraction pattern produced. A smaller obstacle or opening will result in a wider diffraction pattern, while a larger obstacle or opening will produce a narrower pattern. This is due to the wavelength of the wave, as well as the distance between the obstacle or opening and the screen where the diffraction pattern is observed.

4. Is there a mathematical formula for calculating diffraction patterns?

Yes, there are several mathematical formulas that can be used to calculate diffraction patterns. The specific formula used depends on the type of wave (e.g. light, sound, radio waves) and the shape of the obstacle or opening. However, these calculations can be complex and are usually done using computer simulations.

5. How does diffraction differ from other wave phenomena, such as refraction or reflection?

Diffraction is different from refraction and reflection in that it does not involve a change in the direction of the wave. Instead, it causes the wave to spread out or bend around the edges of an obstacle or opening. Refraction, on the other hand, involves a change in the direction of the wave as it passes through different mediums, while reflection is the bouncing of a wave off a surface.

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