00:03
So for the first question, let's say that o is the center of a sphere.
00:06
So when the oa is equal ob is equal oc, which is equal to r, when the theta ab is good theta a, b, and theta rb is equal theta a c, and theta a .b.
00:15
And theta a .b is equal theta rb.
00:20
So therefore, have theta a .c is equal theta ba.
00:23
We also know that theta a .a is equal to theta b .a.
00:28
And theta a .c is equal to theta b .c.
00:31
So therefore, we have theta b .c is equal to theta a .a.
00:36
So for the next question, well, so the angle deflection at point a should be equal to theta a minus theta b .a.
00:44
And the angle deflection at point b should be equal to pi minus 2 theta ab.
00:50
And we know theta b8 is equal to theta ab.
00:52
So therefore, have the angular deflection at point b is actually equal to pi minus 2 theta b .a.
01:01
And we also know that the angle deflection at point c is theta bc minus theta a .c, which is equal to theta a .a minus theta b .a.
01:11
So therefore, we have the total angle deflection, which is delta here, is equal to theta aa minus theta ba plus theta ba plus theta ba plus theta ba.
01:22
This will give us delta is equal to 2 theta a .a minus 4 theta b2 plus pi.
01:31
And for question c, we know n, which is the refractive index of the water, is equal to sine theta a .a over a sine theta b .a.
01:38
We know sine theta b a is equal to sine theta a over n.
01:42
So therefore we have theta b a is equal to arc sine sine sine theta a over n.
01:49
As you can tell, we can use the arc sine sine sine theta a over and substitute for the theta b.
01:54
So eventually we have delta is equal to 2 theta a a a minus four arc sine sine theta a over n plus pi.
02:02
So for the question d, while we know d d d d over d theta is equal to zero.
02:10
So therefore we also have d delta over d theta a is good to negative 4n cosine theta a a over n times square root minus sine square theta a plus n squared plus 2.
02:22
As you can tell n and n can be canceled out.
02:25
So this will give us negative 4 cosine theta a a over square root negative sine square theta a plus n squared plus 2 is good 0.
02:34
So say that theta a is good theta 1.
02:37
So we have negative 4 cosine theta 1 square root over square root, negative sine square theta 1 plus n squared is equal to negative.
02:46
So therefore we have cosine theta 1 over square root, negative sine square, theta 1 plus n square is equal to 1 half.
02:55
So we have cosine theta 1 is equal to 1 half times square root negative sine square theta 1 plus n square...