00:01
Okay, so this is problem 24 .15.
00:03
And this problem, we have an interference problem.
00:07
And in this interference setup, we have actually like two experiments.
00:13
In the first experiment, we have a monochromatic light with wavelength, 418 nanometers.
00:19
Okay.
00:20
And this wavelength, this light is going through the tool slits that are, we don't know the distance between the toolstlets, but we know that the peak, the third maximum peak, is located 600 millimeter away from the central peak, okay, and the distance between the toollets and the screen where we detect the interference pattern is 1 .6 meters.
00:51
So the only thing that we don't know here is just d.
00:56
So we can solve for d first, right, to get the separation distance between the two slits.
01:02
So we know the condition for interference better is just d sine theta is equal to m lambda.
01:10
And because we are solving for d, we can get d to be m lambda over sine theta.
01:20
And again, sine theta would be just y over r because the angle is like this.
01:28
So the sign of this angle will be just this distance, which is y, over this, the length of the hypotenus, but because the angle here is very small, we can approximate the length of the hypotenus to be the length of this side, which is just r here, right? so, d is equal to m lambda over y over r, and when we substitute with the values that we have here three times 480 times 10 minus 9 meters over y which is 16 times 10 power minus 3 over r which is 1 .6 this will give us 1 .4 4 times m power minus 4 meters so the separation distance between the two slits is 1 .44 times 10 par minus 4 meters...