Arun
Last Activity: 7 Years ago
Dear Amit
The incident angle is the angle the vector makes with the z-axis (that is the direction of the normal to the x-y plane). That angle θzi = arccos(k-component / |A|)
|A| = √[(6√3)² + (8√3)² + 10²)] = √[3*36 + 3*64 + 100] = 20
θzi = arccos(-10/20) = 60º
The refracted angle will be sin(θzr)/sin(θzi) = n1/n2 = √2/√3
sin(θzr) = √2/√3 * sin60º = √2/√3 * √3/2 = √2/2
θzr = 45º
cos45º = k-comp / |A|
for a unit vector, |A| = 1
k-comp = cos45º = √2/2
The other two components must be adjusted by a factor r to result in a unity magnitude:
√[r²*(3*36 + 3*64) + 0.5] = 1
300*r² = 0.5
r = 1/√600
The new vector is then
6/√200 i + 8/√200 j - √2/2 k
Regards
Arun (askIITians forum expert)
