00:01
In this problem, we have three sheets of plastic of unknown indices of reflection.
00:06
We have sheet one placed on top of sheet two, and there's a beam of a laser that strikes it so that it strikes the interface, it strikes the interface at 26 .5 degrees.
00:18
Then the experiment is repeated with sheet three on top of sheet two, with the same angle of incidence, with the refracted beam making an angle of 36 .7 degrees, and for the third one, it's sheet one on top of sheet two, with the same angle of incidence, with the refracted beam making an angle of 36.
00:30
Of sheet 3 and what's the expected angle of refraction.
00:34
So here i've drawn all the three different placements.
00:37
So i have n1 on n2, n3 on n2, and n1 on n3.
00:42
So for the first case, n2 is equal to n1 sine theta 1 over sine theta 2, which equals n1 sign of 26 .5 degrees divided by sign of 31 .7 degrees.
00:55
For the second placement, we have n3 is equal to n2 sine theta 2, divided by sine theta 1.
01:03
So we have an expression for n2.
01:05
So we can put that in for our n2...