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.

Answer

A, B, C, D

Explanation

Figure 1 shows vector OM^`=`=6sqrt3i+8sqrt3hatj`
Figure 2 shows vector `vecA=6sqrt3hati+8sqrt3hatj-10hatk`
The perpendicular to line `MOM^`` is Z-axis which has a unit vector of `hatk`.
Angle between vextor ``vec(MP)` and `vec(OP)` can be found by dot product. `vec(MP),vec(OP)=(MP)(OP) cosi`
`((6sqrt3hati+8sqrt3hatj-10hatk).(-hatk))/(sqrt((6sqrt3)^(2)+(8sqrt3)^(2)+(-10)^(2)+sqrt((-1)^(2))))=cosi` `implies``i=60circ`
Unit vector in the firection of `MOM^(`)` from fig. (1) is
`hatn=(6sqrt3hati+8sqrthatj)/([(6sqrt3)^(2)+(8sqrt3)^(2)]^((1)/(2))`),hatn=(3)/(5)hati+(4)/(5)hatj`
To find the angle of refraction, we use Snell's law
`(sqrt3)/(sqrt2)=(sini)/(sinr)=(sin60^circ)/(sinr)implies45^circ`
Now, `hatr=(sinr)hatn-(cosr)hatk`
`=(sin45^circ)[(3)/(5)hati+(4)/(5)hatj]-(cos45^circ)hatk`
`=(1)/(5sqrt2)[3hati+4hatj-5hatk]`