00:01
All right, so let's say we have a tall building like this, and there's a shoreline here, and a submarine is a certain distance below this, and it sends out a laser beam, kind of like this, and this laser beam is refracted a little bit as it enters the water and strikes this tall building.
00:26
So we'll call this angle theta 1.
00:28
We'll call this angle, i suppose, phi.
00:31
This angle is theta 2 and let's see we're told that this distance is let's see 210 meters and the submarine is 300 meters from the shore like this so if the index of a fraction of water is 1 .33 we want to know what is the angle of incidence of the beam striking the air water interface so what is theta 1 basically? so first let's find, sorry, this is 100 meters too.
01:21
We're told this is 100.
01:23
So that means theta 1, this is the same as theta 1 here.
01:26
So what we can see is the tangent of theta 1 is just going to be 300 divided by 100 or 3.
01:34
So that gives us theta 1 is the inverse tangent of this.
01:38
This is 71 .57 degrees.
01:41
And then that's part b.
01:45
For part c, what we should have is the index of a fraction of water times the sign of this angle should equal the sine theta 2 because theta 2 represents the interface of air.
01:59
So if we solve for that, theta 2, of course, is the inverse sign of the sine of 71 .57 over...
02:09
Sorry, it's 1 .33 times the sine.
02:14
Of 71 .57...