00:01
The refractive index of the rod is given to be 1 .37 and the refractive index of air is 1 .00.
00:12
Incident ray gets reflected, sorry, refracted when it enters the rod and this refracted ray is totally internally reflected along the walls of the rock.
00:27
For total internal reflection to occur, the answer, the antire is totally internally reflected along the walls of the rock.
00:31
Angle of refraction must at least be 90 degrees as the ray travels from the rod to air.
00:48
Ray diagram can be drawn as applying snail's law at point b.
01:02
We get n -s -sign theta -2 equals n -a -sign sine, 90 degrees since sine 90 is 1 n sine theta 2 is equal to n a now rearranging for theta 2 we get sine inverse of n a over n substituting we get sine inverse of 1 .00 over 1 .37.
01:46
On calculating we get 46 .88 degrees.
01:53
Thus, theta 2 must be greater than or equal to 46 .88 degrees.
02:04
From triangle a .o .b in the figure, we can say that theta 1 plus 90 degrees plus theta 2 equals 180 degrees.
02:21
Simplifying for theta 1 we get 90 degrees minus theta 2...