How does the wavelength of a wave affect diffraction and why is it important?

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Homework Help Overview

The discussion revolves around the effects of wavelength on diffraction and the behavior of waves as they transition between different mediums, specifically from shallow to deep regions. The original poster presents a series of questions related to wave properties, diffraction, and the implications of wavelength in these contexts.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between wavelength and diffraction, questioning how changes in wavelength affect wave behavior when passing through openings of varying sizes. There are also inquiries about the implications of wavelength changes when transitioning between shallow and deep regions.

Discussion Status

Some participants express uncertainty about the questions posed, particularly regarding the relevance of certain concepts to their current syllabus. Others attempt to clarify their understanding of the physics involved, indicating a mix of exploration and confusion about the topic.

Contextual Notes

There is a mention of specific educational constraints, as some participants note that the content may not align with their current curriculum, which could impact their ability to engage fully with the questions raised.

primarygun
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I list the question one by one. I hope this is a good arrangment.
1.The blades of a fan is rotating in an anti-clockwise direction. Its motion is frozen with a strobe frquency of 1000 Hz without a change in appearance. Which of the following is not a possible rate at which the wheel is rotating?
There are six slits on the fan.
A.200 revolution per second.
B.500 rev. per second.
C.2000 revolution per second.
2.Why the wavelength of a wave increase when it passes from a shallow region to a deep region? Does the amplitude change?
3.Why the wavelength of a wave increase when it passes from a shallow region to a deep region? Does the amplitude change?
4. Why does the wavelength of a wave play a very important role in diffraction?
If the wavelength is bigger than the hole passing through, a very circular wave is produced while a very straight wave is produced if the hole is big enough?
 
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Have you read the 'sticky'? Show us what you have tried so we will know what kinds of hints you need.
 
OK. For the first one, actually, I suspect whether the choices are wrong.
For the other, the contents are not in my school's syllabus, so I don't have any idea.
Sorry!
 
Can anyone help me?
 
There is no paid staff waiting to help you. Please be patient.
 
primarygun said:
I list the question one by one. I hope this is a good arrangment.
1.The blades of a fan is rotating in an anti-clockwise direction. Its motion is frozen with a strobe frquency of 1000 Hz without a change in appearance. Which of the following is not a possible rate at which the wheel is rotating?
There are six slits on the fan.
A.200 revolution per second.
B.500 rev. per second.
C.2000 revolution per second.
This means that in 1/1000th of a second, the fan as moved an even multiple of 1/6th turns. So turns/sec = 1000* n/6.

Or, more formally:
[itex]\omega\Delta t = \omega/1000 = \theta = n*2\pi/6[/itex]
[itex]\omega = 2\pi f = n2\pi*1000/6[/itex].

So [itex]f = n*1000/6[/itex] where n is a whole number.

AM
 
Your answer is correct!
May I know what w and pi and theta are?
I don't know where I archieve to the wrong way, please tell me.
1/1000 s for one revolution, actually, our view is only blocked for 1/6000 s.
So there are 1/6000 s for the n complete wave to pass.
So I think (1/6000)/T =n T, the time required for one complete wave, Period.
Then I get f=6000n. (n: integer)
Why?
 
OK. I get it. However, would you explain it to me again?
According to the rule here, I shall tell you first in order to obey the law. :P
In (1/1000)s, the fan moves, ---> the slits moves a multiple of 1/6 n round(a complete revolution like a complete wave). That's mean there are (1/1000)s for it to move.
Period= (1/6 n)/(1/1000)s
 
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