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

Answer:

0.139 rad

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:

$sin\theta_2=\frac{n_1}{n_2}sin\theta_1$

And finally:

$\theta_2=arcsin(\frac{n_1}{n_2}sin\theta_1)$

We then insert our values:

$\theta_2=arcsin(\frac{n_1}{n_2}sin\theta_1)=Arcsin(\frac{1}{3.325}sin(0.478rad))=arcsin(0.13834714686 )=0.139 rad$