refraction of light

**twistedmexican** · August 30th 2007, 08:00 PM

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?

**Jhevon** · August 30th 2007, 08:17 PM

Originally Posted by twistedmexican

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:
$n_1 \sin \theta_1 = n_2 \sin \theta_2$

well, more generally, it states that $\frac {\sin \theta_1}{\sin \theta_2} = \frac {v_1}{v_2} = \frac {n_2}{n_1}$ , but let's not worry about that

where $n_1$ is the refractive index of the initial material (in this case, that of the air)
$n_2$ is the refractive index of the final material (in this case, that of the plastic)
$\theta_1$ is the angle of incidence
$\theta_2$ is the angle of refraction

you are expected to know that the refractive index of air, $\theta_1 = 1.00029 \approx 1$ , just use 1, that's the regular convention

the only unknown here is $n_2$ (which is what we are asked to find obviously), so plug everything else in the formula and solve for it, we have:

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

i leave the rest to you

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