The XY plane is the boundary between two tranparednt media. Medium 1 with `z ge 0` has a refraxtive index of `sqrt2` and medium 2 with `z le 0` has a refractive index of `sqrt3`. A ray of light in medium 1 given by the vector `6sqrt(3)hat(i)+8sqrt(3)hat(j)-10hatk` is incident on teh plane of separation. Find the unit vector in the direction of teh refracted ray in medium 2.

Explanation

Unit vector representing the normal to the plane `hate_a=hatk`.
Component of the incident ray along the normal is `-10 hatk`
The unit vector that represents the plane of the incident ray and the normal.
` hate_p = ((6 sqrt(3) hati + 8 sqrt(3) hatj))/( sqrt((6 sqrt(3))^2 +(8 sqrt(3))^2))=0.6 hati + 0.8 hatj`
Angle between the incident ray and the normal is given by
` cos theta = ( 6 sqrt(3) hati + 8 sqrt(3) hatj - 10 hatk ) . hatk // sqrt((6 sqrt(3))^2 +(8 sqrt(3))^2 + 10^2)`
(or) ` cos theta =- 0.5 `
therefore the angle ` theta = 120^@`
The angle of incidence is ` theta = 180^@ - 120^@=60^@`
the angle of the refracted beam is given by
` sqrt(2) sin ( theta) = sqrt(3) sin (r) or r= 45^@ `
The equation of the emergent ray is
` cos (45) (- hatk ) + sin (45) [0.6 hati + 0.8 hatj]`
`= 1// sqrt(2) ( 0.6 hati + 0.8 hatj - hatk)`