00:01
In this problem, we have been given that there is a light ray which is falling and it strikes a piece of glass at an angle of 84 degree.
00:12
So basically here, the angle of incidence, which the incident ray makes with the normal, that's 84 degree.
00:19
And we need to determine the angle between the reflected as well as the refracted rays, provided that we have also been given the index of refraction, that's the refractive.
00:31
Index of glass, which is 1 .5.
00:35
So here we observe that according to the law of reflection, the angle of reflection will be equal to the angle of incidence.
00:44
And according to that, the reflected angle or the reflected ray will make 84 degree with respect to the normal.
00:52
And let's see the refracted ray, which will definitely pass through the medium.
00:58
And here let's consider r as the angle of refraction.
01:02
So applying snell's law here, mu times sine i is equal to mu two times sine r.
01:12
So basically we multiply the refractive index with the sign of angle that's made with the normal in each medium.
01:19
So here as it is air because the light ray is falling from air whose a refractive index is generally one.
01:26
So substituting the value here and considering this as mu one, which is the refractive index of the first medium, that's one times, sine 84 degree and that's equal to mu 2, which is the refractive index of the glass, that's 1 .5 times sine r.
01:44
So from here we observe that sine r, it will be equal to sine 84 degree upon 1 .5...