00:01
In this problem we have these two identical light sources and we are going to derive an expression for the irradiance of the interference pattern.
00:14
That may be observed, let's see, on a screen like this, as a function of this angle theta.
00:29
But we have something unusual here compared to the usual, you know, double -silt experiment.
00:35
We are given that these two light sources are out of face by some angle given by a difference of phi 02 and phi 01.
00:50
So let's try to understand the situation in terms of the electric field.
00:57
We have to deal with the electric field explicitly because after all the electric field is the dominant component in a light ray.
01:06
Irradiance or the intensity is given simply by the electric to total electric field squared at any point in space so now we have this first light ray from the first source to this point p and there will be this corresponding electric field so we have this amplitude e0 and we know that these fields vary in time and they oscillate in time so we have this sign of omega t and we include some phase 5 -0 here then we have the second light ray coming from the other source to this point p and there will be this corresponding electric field given by e0 times sine of omega -t plus this other phase 502.
02:13
Then let us have a look at the total or the let us have look at these electric fields reaching at this point p.
02:22
So the first field will be e0 times sine of omega t plus 501.
02:36
And during this motion or during this travel of this light rate, there will be, will be some face shift developed naturally due to this pet.
02:49
So let's call it sample 5 -0.
02:52
It will be a constant and it will not matter at the end and we are going to see that.
02:57
But i'm including this for the sake of completeness of this physical situation.
03:06
Now if you have this other component, other field, and it will become e0 times sine of omega -t plus 502 plus again there is some phase shift that will develop naturally as this wave as this light rate travels and there will be this additional shift due to the pet difference between these two arrays now let's consider this pet difference it is very similar to the pet difference in the usual double -stit experiment so we have this length delta and is given by a times sine of theta and the corresponding angle this phi is related to this quantity delta l through this relation fai equal to 2 pi times delta l over the wavelength.
04:23
We have everything to attack this problem.
04:27
So let us start with the total electric field at this point p...