9no sid phase diff is always 0 were is it Î ?
for 2 coherent sources , I net =(√I1 + √I2)2 wen path diff = 0
now take these 2 and separate them like interference exp , you get max , min but avg intensity remains I1 + I2
now if u put these 2 sources together , path diff always 0 , hence always max given by eq abv . but here Inet ≠I1 + I2
why
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1first of all it is not possible to get them at a distance of d=0 as there always exists a minimukm distance in physics decribed by quantum mechanics...
9hmm rohan yes wat uve done abv is proof for first part of Q
ie wen d is small but ≠0
BUT im telling it doesnt match wen d=0 that ive assumed in second part
why is it so ??
see # 21
can u develop upon that idea of interference of energy at generation stage itself ??
ie nature gets energy from I1 ,I2 redistributes it uniformly from origin itself ????
how is it happening ??
is there an answer or is it a new mechanism of interaction ??
1hmm a good doubt celestine ...
you see the only difference between the two cases you mentioned is that the value of "d" is different
when you attempt to find the average intensity over a length L of the screen.. it will be given by
(0∫LIdx)/L where I is the intensity at a distance x
I=I1 + I2+2√(I1I2)cosφ where φ is the phase difference
we see that φ=k*xd/D
so let us integrate ..
we get it as
Iav=I1+I2+2√(I1I2)*sin(dkL/D)/(dkL/D)
obviously d cant be equal to zero other wise they will become the same source
so let us take d very small
here we can choose L as big as we like as the screen extends from 0 to ∞
if we take d=(1/∞)k then we can choose L as ∞k+1 (there is no harm in doing that as we are averaging over the whole thing to get the total actual average)
so then the term will become sinx/∞ = 0 ..
so Iav remains I1+I2
9see dis exp and verify if its ok
wen we r considerin putin 2 sources together its practically impossible cos wen u bring them together they act as 1 source and no more interference takes place.
this may be due to a new type of interference of energy ( not amplitude) at the generation of radiation stage itself .
9ok lemme be more clear
for 2 diff coherent sources
Inet = I1 + I2 + 2√(I1 X I2 ) cosθ
now wen u put these sources together , phase diff always 0
so cosθ always 1 so
Inet = I1 + I2 + 2√(I1 X I2 ) always
here input = I1 + I2 ≠output =Inet
why nishant sir
1What i meant 2 say was this.....
1) when we take interfence we get a series of pattern...in this there are dark n bright lines...now this is caused due to a phase diff ranging 4m 0 to Î ...so the avg intensity is I1+I2
2) when we put to sources together we are taking just on case of constant phase difference in which phase dif=0........dere can be another can in which phase dif is Î .....
so avg intensity 4 d first case is (√i1+√i2)2 ......and 4 d second case it is (√i1-√i2)2...
so d net avg intensity is i1-i2 same as above
11By " if u put these 2 sources together ", do you mean you are taking only one source?[7]
1also ur talkin about 2 different cases in the 2nd one when the 2 sources are together dere is max intensity while in interfernce we have 0 phase dif n pi phase difference at diff places....
a correct comparison would be interference with when d to souces are kept together at 0 phase dif n when at pi phase dif then the average comes out to be the same ..
think that serves the purpose...
1but i feel in interference there is no need 2 talk about average intensity.....we can get the intensity at any point n its just d redistribution of energy