6hmmm
interesting method
:D
In the arrangement shown wavelength of light used is \lambda . The distance b/w slits S1 and S2 is d ( <<D) . The distance b/w S3 and S4 is u = \lambda D / 3d . If the ratio of max to min intensity observed on screen is k. Find k.
http://img21.imageshack.us/i/45510347.png/
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7 Answers
1Ok i have memorized a general result which is very helpful in these problems
It states that if the distance between 2 slits is 'd' then at a distance of \lambdaD/3d from central maximum.. the intensity is 1/4th
So at S3 the intensity wud be 1/4th of that at S1 and S2..
If original intensity is 4I then at S3 it is I..and at S4 it is 4I..
Now as u know ImaxImin=(√I + √4I)2(√4I-√I)2...
Which is 9II..
therefore k=9..
1hmm thanks :D
U can check the assumption that i have used..
check out the answer of question no 30 b) of page 382 in H.C.Verma 1st part :)
49@aniket: It states that if the distance between 2 slits is 'd' then at a distance of D/3d from central maximum.. the intensity is 1/4th
Prove This! :D
1IImax=1/4
4a2Cos2(\phi2)4a2= 14
Where 'a' is the amplitude
so Cos2(\phi2)=14
Cos(\phi2)=12
(\phi2)=\Pi/3
Using the realation between phase diff and path diff..
we get path diff as \lambda /3..
therefore reqd distance is \lambda D /3d
:) :P