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Physics: Post your doubts here!

usama321

usama321

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Princess Raven said:
How

do you know I is 9?? I don't get what u just said
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Read the part c) They say that the intensity of wave A at point P is I. Now intensity = A^2 and amplitude is the point of highest displacement on a wave. Seeing wave A, we can see that the amplitude is 3/-3. Thus, squaring it, we get 9, which the question states is I. We have to solve the rest of the parts in terms of this I
 
K

kitkat <3 :P

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actually i think i got it :p look
we have the path difference that is 28
and the wavelength is 8.25
and its asking fo minima which is cuz of the destructive interference
so we will use the frmula pathdiff = (n+0.5)* lamda
= 2.89
n has to be in whole number cuz u know u cant have a half minima so n=2
m not sure thouh :oops:
DeViL gURl B) said:
I'm sorr


Sorry gave the link :)
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usama321

usama321

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Posted this a while ago DeViL gURl B)
kitkat <3 :p said:
m sorry :( i tried alot but i cant solve it :(
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robinhoodmustafa said:
I also wanted to ask this question. #same problem
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usama321 said:
The wave from S1 has to travel 100cm. Using Pythagoras theorem, S2 will have to travel 128.1 cm.There is a path difference of 28.1 cm

Using v =f lambda, we know that
lambda= 330/1000 to 330/4000 Thus
8.25cm <= lambda <= 33 cm



Now, we know that a minima is formed when S2 is at its trough, while S1 at its crest. As both waves have the same frequencies, they will have travelled the same amount of distance in their respective first 100cm. However, in the next 28.1 cm, S2 must have travelled at least half, 1.5, 2.5 and so on cycles to cause destructive interference and thus a minima. We will check this with different possible wavelengths.

28.1 = 0.5 lamda
lamda = 56 cm. This is outside our range

28.1 = 1.5 lamda
lamda = 18.73. This is in our range

28.1 = 2.5 lamda
lamda = 11.24 cm This is in our range
28.1/3.5 = 8.02 Outside our range
answer = 2
Click to expand...
 
DeViL gURl B)

DeViL gURl B)

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kitkat <3 :p said:
actually i think i got it :p look
we have the path difference that is 28
and the wavelength is 8.25
and its asking fo minima which is cuz of the destructive interference
so we will use the frmula pathdiff = (n+0.5)* lamda
= 2.89
n has to be in whole number cuz u know u cant have a half minima so n=2
m not sure thouh :oops:
Click to expand...
Can u please elaborate a bit more .. Please :)
 
usama321

usama321

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Asad Moosvi said:
Need help. Quick! Q5 part c.
Someone please explain it ASAP! http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_21.pdf
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Calculate the path difference between the two waves. We get .15m Divide it by lambda and you get 6. This means that the wave from source S2 has travelled 6 whole waves more than S2 before reaching the point P. Furthermore, as 6 is a whole number, we also know that there is going to be constructive interference at P, leading to a maxima. Now, we also know that there would be maximas formed when wave S2 has travelled 5,4,3,2,1 extra waves. Thus we get 6 maximas in total. Include the maxima at O and you get 7 maximas with 6 minimas in between
 
Asad Moosvi

Asad Moosvi

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usama321 said:
Calculate the path difference between the two waves. We get .15m Divide it by lambda and you get 6. This means that the wave from source S2 has travelled 6 whole waves more than S2 before reaching the point P. Furthermore, as 6 is a whole number, we also know that there is going to be constructive interference at P, leading to a maxima. Now, we also know that there would be maximas formed when wave S2 has travelled 5,4,3,2,1 extra waves. Thus we get 6 maximas in total. Include the maxima at O and you get 7 maximas with 6 minimas in between
Click to expand...

Thank you. Could you please explain the equation we use to get 6? Why did you you divide the path difference by lambda?
 
usama321

usama321

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Path difference is the extra path S2 is covering. S2-S1 would be that extra path. One wavelength covers .025m. How many will there be in .15m. Simply divide them
 
robinhoodmustafa

robinhoodmustafa

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usama321 said:
Calculate the path difference between the two waves. We get .15m Divide it by lambda and you get 6. This means that the wave from source S2 has travelled 6 whole waves more than S2 before reaching the point P. Furthermore, as 6 is a whole number, we also know that there is going to be constructive interference at P, leading to a maxima. Now, we also know that there would be maximas formed when wave S2 has travelled 5,4,3,2,1 extra waves. Thus we get 6 maximas in total. Include the maxima at O and you get 7 maximas with 6 minimas in between
Click to expand...
Tough Part tha. Not everyone could do it
 
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