Snell's Law and Index of Refraction

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Physics › Snell's Law and Index of Refraction

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A light ray is traveling through air hits a transparent material at an angle of from the normal. It is then refracted at . What is the speed of light in the material?

$1.7 \times 10^{8} \frac{m}{s}$

$2.2 \times 10^8 \frac{m}{s}$

$3.1 \times 10^8 \frac{m}{s}$

$1.3 \times 10^8 \frac{m}{s}$

Explanation

This problem requires Snell's Law and the corresponding equation:

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

We know that the index of refraction of air is:

$n_1 = n_{air} = 1.0003$

We also know that:

$\theta_1 = 30$ and $\theta_2 = 16.25$

Now we can plug in these values into the Snell's Law equation to find the index of refraction for the transparent material.

$n_2 = \frac{n_1\sin\theta_1}{\sin\theta_2} = \frac{(1.003)\sin30}{\sin16.25} = 1.792$

Finally, we need to calculate the speed of light moving through this transparent material now that we know the index of refraction for it. To do that, we need to use this equation:

$V = \frac{c}{n}$

Where is the speed of light and is the index of refraction. We plug in our known values and get:

$V = \frac{c}{n} = \frac{3.0\times10^8}{1.792}= 1.7\times10^8\frac{m}{s}$