00:01
So here we know that from snell's law of refraction, n sub 1, sign of theta sub 1 equals n sub 2 sign of theta sub 2.
00:13
We also know that n sub 2 of sine of theta sub 2 equals n sub 3 sign of theta sub 3.
00:22
So essentially by transitive property, n sub 1 sign of theta sub 1 equals n sub 3, sign of theta sub 1, equals n sub 3, of theta sub 3.
00:33
Now given this, this implies simply that theta sub 1 equals theta sub 3 when the refractive index of material 1 equals the refractive index sub 3.
00:47
So we are given that theta sub 1 equals 40 degrees and now we're going to analyze figure 3350a.
01:01
And essentially, once we find we're going to look for a point in where theta sub 3 equals 40 degrees analyzing figure 3350b.
01:26
And we find that here the refractive index n sub 1 is going to equal 1 .6 because this is where n sub 3 is equal in 1 .6.
01:40
So essentially, because of the transitive property making this relationship here, and we are given that theta sub 1 equals 40 degrees in figure 30, by analyzing figure 3350a, we go to 3350b, see the point where theta sub 3 is equaling 40 degrees, and this is where the refractive index sub 3 equals 1 .6, therefore n sub 1 also equals 1 .6.
02:08
So here, this would be our answer for part a...