Question

You have an incident ray from a medium with a n1=1, through a medium with a n2 =3.325. If the incident angle is equal to 0.478 rad, what is the refraction angle? (Answer is in radians)

Answer 1

0.139 rad

Step-by-step explanation:

We use Snell's law
`n_1sin\theta_1=n_2sin\theta_2`, where if
`n_1` is the refractive index of the medium containing the incident ray,
`\theta_1` would be the incident angle, and if
`n_2` is the refractive index of the medium containing the refracted ray,
`\theta_2` would be the refraction angle, which we want, so we do:

`sin\theta_2=(n_1)/(n_2)sin\theta_1`

And finally:

`\theta_2=arcsin((n_1)/(n_2)sin\theta_1)`

We then insert our values:

`\theta_2=arcsin((n_1)/(n_2)sin\theta_1)=Arcsin((1)/(3.325)sin(0.478rad))=arcsin(0.13834714686 )=0.139 rad`