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:

Answer

`vec(r)=(3)/(5sqrt(2))hat(i)+(2sqrt(2))/(5)hat(j)-(1)/(sqrt(2))hatk ("angle of incidence"=60^(@);r=45^(@)`

Explanation

For angle of incidence
`(oversetrarrA. hatN) = A"cosi"`
`cosi=((6sqrt(3)hati+8sqrt(3)hat(j)-10hat(k))(-hat(k)))/(sqrt((6sqrt(3))^(2)+(8sqrt(3)^(2))+(-10)^(2)))`
`cos i=(10)/(20) rArr cos i(1)/(2)`
`i=60^(@)` From the law of refraction
`mu_(1)sin i=mu_(2)sin r`
`sqrt(2) sin 60^(@)=sqrt(3) sin r`
`sqrt(2)xx(sqrt(3))/(2)=sqrt(3) sin r`
`r=45^(@)`
Due to refraction only normal conponents changes
`vec(B)=6sqrt(3) hat(i)+8sqrt(3)hat(j)-zhat(k)`
`vec(B).hat(N)=Bcosr`
`6sqrt(3) hat(i)+8sqrt(3)hat(j)-zhat(k).(-hat(k))=sqrt(6(sqrt(3))^(2)+(8sqrt(3))^(2)+(-z)^(2))xx(1)/(sqrt(2))`
`sqrt(2)z=sqrt(300+z^(2))`
`2z^(2)=300+z^(2) rArr z=10sqrt(3)`
`vec(B)=6sqrt(3)hat(i)+8sqrt(3)hat(j)-10sqrt(3)hat(k)`
`(6sqrt(3)hat(i)+8sqrt(3)hat(j)-10sqrt(3)hat(k))/(10sqrt(6))`
`hat(B)=(3)/(5sqrt(2))hati+(2sqrt(2))/(5)hat(j)-(1)/(sqrt(2))hat(k)`