00:01
In the question we are given this figure, this is angle 39 .5 degree.
00:08
This is the angle of incidence.
00:10
So it will be 90 minus 39 .5 that is 50 .5 degree.
00:16
And the refractive index of glass is given to be 1 .56.
00:21
Now when the ray of light will go towards glass, it will undergo refraction and it will bend towards the normal.
00:28
Let us say this angle is r1.
00:30
And when it will go towards water it will enter water the ray of light will go away from the normal let us say this angle is r2 so by snail's law we can write when the ray of light is going from air to glass we have n a sine i is equals to n g sine r one this implies sine r1 is n a sine i upon n g now n a the refractive index of air is one i is 50 .5 degree and n g is 1 .56 so on solving we get 0 .716 upon 1 .56 that is equals to 0 .49 degree hence the angle r1 is sorry this is not degree this is 0 .49 so r1 will be sine inverse of 0 .49 that is 29 .34 degree.
01:45
Hence the angle in the glass is 29 .34 degree.
01:53
Now as the ray of light goes from glass to water, the angle of incidence is now r1 that is 29 .34 degree.
01:56
Now as the ray of light goes from glass to water, the angle of incidence is now r1 that is 29 .3.
02:04
So, by snell's law we can write ng sine r1 is equals to nw, sign r.
02:15
This implies sine r2 is ng sine r1 upon nw.
02:24
So on substituting the values we get 1 .56 into sine of 29 .34 upon nw, the refracture active index of water is 1 .333.
02:40
So on simplification we get 0 .7644 upon 1 .333 that is equals to 0 .573.
02:50
This implies r2 is sine inverse of 0 .573 that is 34 .96 degree...