00:01
So for this problem, we are told that a thin film of soap solution is illuminated by the white light at an angle of incidence that is given.
00:09
And that angle of incidence is equal to the sign of minus 1 of 4 divided by 5.
00:23
And in the reflected light, two dark consecutive overlapping fringes are observed corresponding to the wavelength.
00:30
The wavelength that we're going to call the wavelength 1, that is equal to 6 .1 times 10 to the minus 7 meters, and a second wavelength for 6 times 10 to the minus 7 meters.
00:58
The refractive index of soap solution is also given, and that index is equal to 4 divided by 3.
01:07
So the question is to calculate the thickness of the film.
01:14
So the first thing that we are going to do is to write the sign of the angle of incidence.
01:24
So from the expression given, we will obtain that that is just simply 4 divided by 5.
01:32
Now, with that said, we can obtain the sign of the reflective reflective, because that is going to be the sign of the incidence divided by the index of refraption of the medium.
01:51
So that will be 4 divided by 5, which is this value in here, divided by the index of refraption that we are given, 4 divided by 3.
01:58
So in here we can simplify this and then this will be 0 .6.
02:04
Now we are going to obtain the cosine of the reflected angle and that's because we are going to use this value later in the.
02:13
Expression for the thickness.
02:16
So that will be the cosine of the reflective in this is the square root of one minus the sign square of the reflected angle.
02:25
So that will be the square root of 1 minus 0 .6 to the square so that will give us a value of 0 .8.
02:35
In here we have used the this property, the trigonometry property that is the cosine square of theta.
02:44
Plus the sine square of the angle theta is equal to 1, and we just solve for the cosine of the angle...