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Physics: Post your doubts here!

Choi WW

Choi WW

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eliyeap said:
Im not sure about the point sources but the way i look at is that the middle of XY will have the highest intensity(highest peak) because its a zero order. Then as you go further from the middle, lets say to the first order the intensity gradually decreases. The 2 speakers are coherent sources, meaning they have the same amplitude and phase difference. Its the path difference between them that forms then interference pattern on the screen XY
Click to expand...

But the answer is D, can you please explain me the reason why they don't have complete destructive interference (not zero intensity)?
 
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eliyeap

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Choi WW said:
But the answer is D, can you please explain me the reason why they don't have complete destructive interference (not zero intensity)?
Click to expand...
https://www.google.com/search?q=you...nx.org%2Fcontent%2Fm42508%2Flatest%2F;875;521
The interference pattern on the screen should like this in terms of intensity. The peaks are where constructive interference occurs while the troughs between each peak is where destructive interference occurs. For path difference, when constructive interference occurs, the path diff must be n(lambda) where n can be 1,2,3 and so on. n is basically the no of order. When n=0, which would be in the middle you would have the maximum intensity of a bright fringe. As the n increases, the intensity of the bright fringes decreases. Dark fringes have no intensity as it is already dark, you can't make it darker.
 
Choi WW

Choi WW

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eliyeap said:
https://www.google.com/search?q=young's double slit experiment&rlz=1C1KMZB_enMY614MY614&tbm=isch&tbo=u&source=univ&sa=X&ei=xKVQVd-KF6SwmAWLoYDoDg&ved=0CDAQsAQ&biw=1241&bih=584#imgrc=mdwZhBYE58VwmM%3A;hZqAo6hjXQTxXM;http%3A%2F%2Fcnx.org%2Fresources%2Fdc45f72de39c65b1e7f2cf1a0ec14598%2FFigure_28_03_06a.jpg;http%3A%2F%2Fcnx.org%2Fcontent%2Fm42508%2Flatest%2F;875;521
The interference pattern on the screen should like this in terms of intensity. The peaks are where constructive interference occurs while the troughs between each peak is where destructive interference occurs. For path difference, when constructive interference occurs, the path diff must be n(lambda) where n can be 1,2,3 and so on. n is basically the no of order. When n=0, which would be in the middle you would have the maximum intensity of a bright fringe. As the n increases, the intensity of the bright fringes decreases. Dark fringes have no intensity as it is already dark, you can't make it darker.
Click to expand...

However, for the question, D, which is answer has no position having zero intensity right?
 
qwertypoiu

qwertypoiu

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Choi WW said:
View attachment 53476 The answer is D so I thought it's because there is the difference in the path travelled by the waves, so there is difference in intensities, amplitudes of the waves, so there would be no complete destructive interference. But then, for the young's double slit experiment, there should be also no complete destructive interference according to what I thought, but they consider it as complete destructive interference, right?

So, I was searching for it, then, I saw the explanation on the webstite, http://physics-ref.blogspot.com/2015/03/physics-9702-doubts-help-page-93.html.

They said it's because the speaker isn't point source.... what does it mean by that and how the factor 'point source' affect the result.....

Also, is my thought in the beginning wrong?
Past Exam Paper – November 2014 Paper 11 Q26
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I'm not sure, but I think that a point source will cause two completely in phase waves with same speed to be produced (because of diffraction), but perhaps the two loudspeakers will produce waves that are slightly out of phase with each other. My imagination is not good enough to imagine what effects this may have, but perhaps this may help you arrive at your explanation. Maybe somehow two waves with a phase difference are not able to cause complete destruction.
 
buckle_crackk

buckle_crackk

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The Young modulus of the metal of a wire is 0.17 TPa. The cross-sectional area of the wire is 0.18 mm2

what does the letter "T" in TPa here stand for?
 
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Studydayandnight

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Guys please help with this question!!!! PLEASE.

A ball is thrown with an initial velocity of 20m/s at an angle of 30 to the horizontal. Calculate the horizontal distance travelled by the ball.

I've found the horizontal and vertical components.

Vv= 10 m/s
Vh= 17.3 m/s

I can't seem to figure out how to find the time! :(
 
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Sahan27

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Studydayandnight said:
Guys please help with this question!!!! PLEASE.

A ball is thrown with an initial velocity of 20m/s at an angle of 30 to the horizontal. Calculate the horizontal distance travelled by the ball.

I've found the horizontal and vertical components.

Vv= 10 m/s
Vh= 17.3 m/s

I can't seem to figure out how to find the time! :(
Click to expand...

Right so...
In order to find Horizontal distance travelled
You need Vh and time taken..
Vh= 20cos 30
Use equations of motion to find time taken.
Use S=ut +0.5at^2
where:
S=0
U=10
a=-9.8
t= ?
S is 0 because overall vertical displacement is 0.
so you have 10t-4.9t^2=0
sove for t
you get 10/4.9
substitute back in orginal equation
Horizontal distance = Vh * 10/4.9= 20 cos30* 10/4.9= 35.3
 
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