Let the x-z plane be the boundary between two transparent media. Medium 1 in `zge0` has a refractive index of `sqrt2` and medium 2 with `zlt0` has a refractive index of `sqrt3`. A ray of light in medium 1 given by the vector `vecA=6sqrt3hati+8sqrt3hatj-10hatk` is incident on the plane of separation. The angle of refraction in medium 2 is:

Explanation

Unit vector representing the normal to the plan `hat e_n=hat k`.
Component of the incident ray along the normal is `-1hat k`.
The unit vector that represents the plane of the incident ray and the normal
`hat e_(p) = ((6sqrt3 hati+8sqrt3hat j))/(sqrt((6sqrt3)^(2)+(8sqrt3)^(2))) =0.6hat i + 0.8hatj`
Angle between the incident ray and teh normal is given by
`cos theta= (6sqrt(3) hat(i) +8sqrt(3)hat(j)=10hat k)`
.`hat(k)//sqrt((6sqrt(3))^(2)+(8sqrt(3))^(2)+10^(2))`
or `cos theta=-0.5`
Therefore, teh angle `theta= 120^(@)`
The angle of incidence is `I=180^(@)-120^(@)= 60^(@)`
The angle of teh refracted beam is given by
`sqrt 2 sin(i) =sqrt3 sin(r)` or `r=45^@`
The equation of the emergent ray is `cos(r)hat(e)_n+sin(r)hat(e)_p`
`cos(45^(@))(-hat(k))+sin(45^@).(0.6hat(i)+0.8hat(j))`
`=1/(sqrt(2))(0.6hati+0.8hat(j)-hat(k))`
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