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:

Options

A. `60^(@)`
B. `75^(@)`
C. `30^(@)`
D. `45^(@)`

Answer

D

Explanation

Incident angle from given plane
`theta_(1) = cos^(-1) ((vecA.veck)/(A))`
= `cos^(-1) ((-10)/(20)) = cos^(-1) ((-1)/(2)) = ((pi)/(3))`
Now by snell's law `n_(1) "sin"theta_(1) = n_(2) "sin"theta_(2)`
`"sin"theta_(2) = ((n_(1)/(n_(2))) "sin"theta_(1) = ((sqrt(2))/(sqrt(3)))((sqrt(3))/(2)) = (1)/(sqrt(2))`
`implies theta_(2) = 45^(@)`