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?:eek::eek::eek:

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- Aug 30th 2007, 08:00 PMtwistedmexicanrefraction of light
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?:eek::eek::eek:

- Aug 30th 2007, 08:17 PMJhevon
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