## Related questions with answers

A beam of light traveling in an unknown material encounters a boundary with flint glass (n = 1.66). If the critical angle for total internal reflection is $32^{\circ}$, what is the index of refraction of the unknown material?

Solution

VerifiedSnell Law states:

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

Where $n_1$ and $n_2$ are the indexes of refraction respectively of the medium 1 and 2, $\theta_1$ is the angle of the ray to the normal in medium 1 and $\theta_2$ is the angle of the ray to the normal in medium 2.

Total internal reflection can happen when light is trying to enter a medium with a lower index of refraction, we call the first angle at which we observe this phenomenon the critical angle $\theta_c$, at this angle, the light travels parallel to the surface separating the two mediums yielding a $90^o$ degree angle with the normal.

We can then assert the following:

$n_1\sin\theta_c=n_2\sin 90^o$

$n_1=\frac{n_2}{\sin\theta_c}=\frac{1.66}{\sin 32^o}=\boxed{3.13}$

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