a beam of light traveling in air is incident on a transparent plastic material at an angle of incidence of 50 degrees. The angle of refraction is 35 degrees. What is the index of refraction of the plastic?
we use Snell's law here:
it states that:
$\displaystyle n_1 \sin \theta_1 = n_2 \sin \theta_2$
well, more generally, it states that $\displaystyle \frac {\sin \theta_1}{\sin \theta_2} = \frac {v_1}{v_2} = \frac {n_2}{n_1}$, but let's not worry about that
where $\displaystyle n_1$ is the refractive index of the initial material (in this case, that of the air)
$\displaystyle n_2$ is the refractive index of the final material (in this case, that of the plastic)
$\displaystyle \theta_1$ is the angle of incidence
$\displaystyle \theta_2$ is the angle of refraction
you are expected to know that the refractive index of air, $\displaystyle \theta_1 = 1.00029 \approx 1$, just use 1, that's the regular convention
the only unknown here is $\displaystyle n_2$ (which is what we are asked to find obviously), so plug everything else in the formula and solve for it, we have:
$\displaystyle n_2 = \frac {n_1 \sin \theta_1}{\sin \theta_2}$
i leave the rest to you