00:01
To solve this problem fully, we're going to apply snell's law, both between material x and water, and between water in the air.
00:10
For part a, we just need to apply snell's law between material x and the water.
00:16
And so we're going to let n .a.
00:19
Be the index of a fraction for material x here.
00:24
And nb will be the index of a fraction of water in w.
00:30
And we know this is 1 .333, since we know the index of a fraction of water.
00:37
Now we know theta a, the angle normal in material x is equal to 25 degrees, and theta b, we're told, is 48 degrees.
00:53
And so now let's apply snell's law to figure out what n .a is.
01:00
And so we can get n -a by taking nb and sine of theta b, and just dividing by sine of theta a, like that.
01:16
And we know what nb is.
01:18
Nb is just nw.
01:22
Sign of theta b is sine of 48 degrees over sine of 25 degrees...