00:01
Hi, we have a ray of light that is incident on an equilateral triangular prism, which has index of refraction 1 .23.
00:10
The ray is parallel to the base of the prism when approaches the present.
00:16
It is entering at the midpoint of one of its sides, so we need to find the direction of the ray when it emerges from the triangular prism.
00:29
So let's first look at the angles that we have here.
00:33
This is the equilateral triangle, so this angle is 60 degree.
00:38
If this is perpendicular to the side, this is 90 degree, which means this is 30 degree.
00:43
And from this geometry, we can see that if this is 60 degree, those are parallel, then this is also 60 degree.
00:50
And this is the normal to the side where the light ray enters.
00:55
That means we also have here 90 minus 60 which is 30 degrees so this angle with the normal of incident right is 30 degree let's try to understand what is the angle of refraction this theta r angle that is going inside of the prism so for that we are going to use smell slow n1 sign of theta 1 this is for the error is equal to n2 sign of theta 2 this is inside of the prism.
01:28
That means sine of theta 2 is equal to n1, sign of theta 1 over n2.
01:36
Now this is equal to n1 is for the air is 1 times sine of our angle 30 degree divided to the n2 is the refraction index for the prism which is 1 .23 given.
01:51
This is equal to 0 .4065...