00:01
Hi, here in this given problem, critical angle for the liquid air interface that is given as i, c, the symbol for critical angle and this is equal to 42 .5 degree.
00:36
So, index of refraction of the liquid will be given by n is equal to inverse of sign of critical angle means this is 1 by sine 42 .5 degree or we can say this is 1 by 0 .676 so finally index of refraction for the liquid comes out to be 1 .48 now in the first part of the problem for the refraction from liquid to air using snell's law, the ratio of the index of refraction of the second medium, and here in this case, second medium is air.
01:57
So the ratio of index of refraction of air to the index of refraction of the first medium means liquid, that is equal to sine i by sine r.
02:10
And in this first part of the problem, sine i means angle of incidence in the liquid.
02:17
Is given as 35 .0 degree.
02:20
So we can say this is 1 by 1 .48 we have found the index of refraction for the liquid to be 1 .48 then this is sign of 35 degree divided by sine r angle of refraction in the air which we have to find.
02:41
So we get an expression for sign r that is equal to sine 3 35 degrees.
02:49
Degree multiplied by 1 .48...