00:01
In the given problem, there is a prism, a glass prism, like shown in the given figure.
00:12
This is a perpendicular of the prism.
00:15
This is the hypotenuse and this is the base of this prism.
00:22
The vertex angle is 30 degree.
00:26
This base angle is 60 degree.
00:30
Then there is a range.
00:35
Ray of light incident a laser beam is incident on this perpendicular surface.
00:43
So to show this refraction first of all we will draw a normal here at this straight surface perpendicular surface this is the normal drawn.
00:55
Then the laser beam is allowed to strike over it like this.
01:03
This is the laser beam striking over here this as it is entering from rarer to denser medium so it will deviate towards the normal now finally we want to have a total internal reflection here at this surface so that this beam does not come out of the hypotenuse so if we show one more refraction at this hypotenuse.
01:42
So we will have to draw another normal here.
01:46
This is the normal.
01:48
So this angle of incidence here should be at least it should be equal to critical angle means i see.
02:00
We can say it to be i c.
02:02
So when the air of light strikes at an angle equal to critical angle, total internal reflection takes place.
02:10
Means to say this ray of light will be reflected back at the same angle here like this.
02:20
So this will be reflected back at an angle of refraction r which will be exactly equal to this angle of incidence because this time this is not the refraction, this is reflection.
02:33
And hence, finally, it will strike at the base and will undergo one more refraction at the base like this, and then it will go away from the normal as the ray of light is passing from denser to rarer medium.
02:53
So finally, the ray of light will go like this.
02:59
In first part of the problem, we have to find this angle theta 1.
03:07
The maximum possible angle, the minimum possible angle, the smallest angle theta 1, for which the laser beam will undergo tir on the hypotenuse of the glass prism.
03:22
So to solve this first part, first of all, for the refraction at hypotenuse, using the expression for critical angle which says sine ic is equal to inubricing.
03:50
Verse of refractive index of the material through which the ray of light is trying to come out in the rarer medium.
03:58
This is the expression for critical angle.
04:01
Hence, as we know, the refractive index for the glass is 1 .5.
04:07
So it becomes 1 by 1 .5, which comes out to be 0 .67.
04:12
Hence, the value of this i see is equal to arc sine of 0 .67.
04:20
So finally, this i.
04:21
Here comes out to be 42 .07 degree.
04:28
So now this angle is 42 .07 degree...