00:01
We know that when diffraction happens together with interference, the intensity has a profile that looks like i0 times kaw square 5 by 2 times sine beta by 2 by 2 by veta squared.
00:20
Where 5, notice that this is the diffraction term and this is the interference term.
00:33
That is, this 5 is due to the interference from the light, between the light coming from two different slits and this 5 is equal to 2 pi pi lambda a sine theta where a is the distance between the two slits and beta is 2 pi by lambda d sine theta where d is the width of the slit width of each slit which we suppose is the same now in our case we have the diffraction map pattern like this and within this we have an interference pattern which has exactly seven minima.
01:34
Let me draw this properly like this.
01:40
These two maxima's coincide.
01:44
Now as you can see that is one, two, three, four, five, six, seven maximuma.
01:49
Now we'll just consider this.
01:53
Now for the diffraction pattern, beta here, beta by two is equal to zero here and beta by two is equal to pi here because when beta by 2 is equal to pi sign pi becomes 0.
02:10
So beta goes from 0 to 2 pi within this length.
02:15
Whereas when we consider 5, we know that this is the core square curve and it goes 1, 2, 3, and 4 full cycles.
02:29
We know that cause square theta has a cycle of pi.
02:32
So this corresponds to pi.
02:35
And this whole thing corresponds to a change of 4 pi in phi 5 by 2 so that is 5 by 2 so fee goes from 0 to 8 pi within this interval when beta goes from 0 to 2 pi so beta by fee is 2 pi by 8 pi and we know that beta is 2 pi d d sine theta by lambda and fee is 2 pi as nc theta by lambda giving us d by a which gives us d is equal to a by 4 this is already given in the question though so just to make it clear let's call this angle theta 1 then beta 5 at theta 1 is equal to 2 pi by lambda a sine theta 1 which we know is 2 pi i'm sorry 8 pi and beta at theta 1 is 2 pi is 2 pi t sine theta 1 by lambda which is equal to 2 pi.
03:53
Now let's go to part a.
03:57
Similarly, in this case though, in this part a, within the diffraction minima, we have 5 instead of the 7, which means we have 1, 2, 3 on one side, 4 and 5, which gives us 3 full cycles of cost square theta, which means if this is theta...