Let the x-y plane be the boundary between two transparent media . Medium 1 in `z ge0` has a refractive index of `sqrt(2)` and medium 2 with `z lt 0 ` has a refractive index of `sqrt(3)`. A ray of light in medium 1 given by the vector `vecA = 6 sqrt(3hati) + 8sqrt(3hatj) - 10 hatk` is incident on the plane of separation . The angle of refraction in medium 2 is :-

Answer

`(1)/(5sqrt2) (3hati + 4 hatj - 5hatk)`

Explanation

Incident ray
`vecA = 6sqrt3hati + 8sqrt3hatj - 10 hatk = (6sqrt3 hati + 8sqrt3hatj) + (-10)hatk`

`QvecO + PvecQ` (As shown in figure )
Note that `QvecO` is lying on x-y plane .
Note , OQ' and Z-axis are mutually perpendicular .
Hence , we can show them in two- dimensional figure as below .

Vector `vecA` makes an angle i with z-axis , given by
`i = cos^(-1).{(10)/(sqrt((10)^(2) + (6sqrt3)^(2) + (8sqrt3)^(2)))} = cos^(-1){(1)/(2)}`
`i = 60^(@)`
Unit vector in the direction of `QOQ'` will be
`hatq = (6sqrt3hati + 8sqrt3hatj)/(sqrt((6sqrt3)^(2) + (8sqrt3)^(2))) = (1)/(5)(3hati + 4 hatj)`
Snell's law gives `(sqrt3)/(sqrt2) = ("sini")/("sinr") = ("sin"60^(@))/("sinr")`
`therefore "sinr" = (sqrt3//2)/(sqrt3//sqrt2) = (1)/(sqrt2) therefore r = 45^(@)`
Now , we have to find out a unit vector in refracted ray's direction OR . say it is `hatr` whose magnitude is 1 .
Thus `hatr = (1"sinr")hatq - (1 - "cosr") hatk = (1)/(sqrt2)(hatq - hatk)`
`= (1)/(sqrt2) [(1)/(5) (3hati + 4 hatj) - hatk]`
`hatr = (1)/(5sqrt2). (3hati + 4hatj - 5hatk)`