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

Answer

A

Explanation

The angle between `+Z-` axis and `vecA`
`cos theta=((vecA).(hatk))/(|vecA||hatk|)`
`=((6sqrt3hati+8sqrt3hatj-10hatk).hatk)/((sqrt((6sqrt3)^(2)+(8sqrt3)^(2)+(-10)^(2))))`
`=(-10)/(sqrt(108+192+100))`
`=-(10)/(20)=-(1)/(2)`
`rArr" "theta=cos^(-1)(-(1)/(2))=120^(@)`
`i+theta=180^(@)`
`rArr" "i=180^(@)-120^(@)`
`=60^(@)`
Applying Snell.s law
`mu_(1)sin i=mu_(2)sinr`
`sqrt2sin 60^(@)=sqrt3 sin r`
`sqrt2xx(sqrt3)/(2)=sqrt3sinr`
`rArr" "sinr=(1)/(sqrt2)`
`therefore" "r=45^(@)`