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Phys lo7

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Shawn Zhang

This document discusses the concept of interference that occurs when waves from two sources overlap. It defines constructive and destructive interference, which depends on whether the peaks and troughs of the waves overlap or not. An example problem is shown where the path difference between two waves is calculated using trigonometry to determine if there will be constructive or destructive interference at a specific point.

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Shawn Zhang LC2
When waves propagate from 2 sources, and come in contact with each other, interference occurs It is a 2D representation of the interference that arise from waves overlapping
2 sound source placed next to each other Other signals distorting your WiFi
Constructive When waves overlap their peaks with peaks, and troughs with troughs Destructive When waves overlap peaks with troughs, and vice versa
Phys lo7
As you can see from the previous picture, the waves overlapped The graph above the waves show areas of constructive and destructive interference To overlap peaks with peaks and troughs with troughs, the phase difference must be 2 Destructive interference is caused by phase difference of
Also, we can look at only one portion of the wave to determine the interference at a specific point in space
To determine the path difference and interference pattern at point P on the previous slide, we can solve it by using trigonometry If the distance and/or angle is given, we will solve for the unknown and after obtaining the distances, we will divide by the wavelength to see the phase difference caused by path difference is
In this example, the phase difference can be seen from the graph at the overlapping point If only numerical info were given, we will need to solve it via trig Practice time!
Using the same pic as a reference, the angle that wave 1 is 60 degrees from the slit Wave 2s slit is 10m below wave 1 Point P is 50m away from the wall where the slits are The wave length of the waves are 5m What kind of interference occurs at point P?
The information that we have are that S1 and S2 are 10m apart, while P is 50m away from the wall, and also that wave 1 propels at a 30 degree angle From the info we are given we can make a new pic
Using trig, we can determine that the distance travelled by Wave 1 is 100m, and the vertical distance from S1 to point P is 86.60m The vertical distance from S2 to point P will be 86.60 10 = 76.60m Once the vertical distance from S2 to point P is obtained, we can calculate the distance travelled by wave 2 sqrt(76.60^2 + 50^2) = 91.4744m 91.4744/5 = 18.2949 wavelengths 100/5 = 20.00 wavelengths 20-18 = 2 wavelengths apart = Constructive
From the calculations on the previous page, it can be seen that at point P, there is constructive interference, because the path difference is an integer number of wavelengths a part That concludes my LO for this week

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Phys lo7

  • 2. When waves propagate from 2 sources, and come in contact with each other, interference occurs It is a 2D representation of the interference that arise from waves overlapping
  • 3. 2 sound source placed next to each other Other signals distorting your WiFi
  • 4. Constructive When waves overlap their peaks with peaks, and troughs with troughs Destructive When waves overlap peaks with troughs, and vice versa
  • 6. As you can see from the previous picture, the waves overlapped The graph above the waves show areas of constructive and destructive interference To overlap peaks with peaks and troughs with troughs, the phase difference must be 2 Destructive interference is caused by phase difference of
  • 7. Also, we can look at only one portion of the wave to determine the interference at a specific point in space
  • 8. To determine the path difference and interference pattern at point P on the previous slide, we can solve it by using trigonometry If the distance and/or angle is given, we will solve for the unknown and after obtaining the distances, we will divide by the wavelength to see the phase difference caused by path difference is
  • 9. In this example, the phase difference can be seen from the graph at the overlapping point If only numerical info were given, we will need to solve it via trig Practice time!
  • 10. Using the same pic as a reference, the angle that wave 1 is 60 degrees from the slit Wave 2s slit is 10m below wave 1 Point P is 50m away from the wall where the slits are The wave length of the waves are 5m What kind of interference occurs at point P?
  • 11. The information that we have are that S1 and S2 are 10m apart, while P is 50m away from the wall, and also that wave 1 propels at a 30 degree angle From the info we are given we can make a new pic
  • 12. Using trig, we can determine that the distance travelled by Wave 1 is 100m, and the vertical distance from S1 to point P is 86.60m The vertical distance from S2 to point P will be 86.60 10 = 76.60m Once the vertical distance from S2 to point P is obtained, we can calculate the distance travelled by wave 2 sqrt(76.60^2 + 50^2) = 91.4744m 91.4744/5 = 18.2949 wavelengths 100/5 = 20.00 wavelengths 20-18 = 2 wavelengths apart = Constructive
  • 13. From the calculations on the previous page, it can be seen that at point P, there is constructive interference, because the path difference is an integer number of wavelengths a part That concludes my LO for this week