00:01
In this problem, we have three monochromatic light waves.
00:04
The first one, the wave equation of this one, is e cosine omega t.
00:08
The second one is e cosine omega t plus phi, and the third one is e cosine omega t minus phi.
00:16
So here, for part a, we get the total wave equation of the interference pattern.
00:25
So here, y adjust to equal to y1 plus y2 plus y3 and this is equal to ecosine omega t plus e cosine plus e cosine omega t minus phi so here y and just equal to e cosine omega t plus e e cosine omega t cosine phi minus e sine omega t this this is for the second term here for the third term we have plus e cosine omega t cosine phi plus e sine omega t cosine phi plus e sine omega t sine phi.
02:19
So now this term cancels with this term and here we would have y to be equal to e times 1 plus 2 cosine of phi multiplied by cosine of omega -t.
02:55
So here the amplitude of this wave adjust e times 1 plus 2 cosine of phi.
03:11
This is for part a now for part b we have the intensity just equal to the magnitude or the amplitude squared.
03:26
So this is equal to e squared times 1 plus 2 cosine of phi squared.
03:43
But now for the maximum interference pattern, we must have here cosine of phi to be equal to 1, such that this bracket is just equal to 3 squared which is equal to 9 for the maximum interference pattern so here i0 which is corresponding to the intensity of the maximum interference pattern would be equal to 3 squared which is 9 times e squared so this is 9 e squared so now e squared is just equal to the maximum intensity divided by 9.
04:31
So now for the intensity at some point p, ip is just equal to i0 divided by 9 multiplied by 1 plus 2 cosine of 5 squared.
04:56
This is for part p now for part c for the lesser maximum we must have phi here or this term to be equal to 0 or to be equal to minus 1 in case if it's 0 we would have here i p is equal to i 0 over 9 and in case cosine 5 is equal to negative 1 so here we would have minus 1 squared which is 1 times i 0 over 9 this is related to the lesser maxima so here phi must be equal to 0 which implies that sorry here cosine 5 must be equal to 0...