From Snell's law:

$\overline{){{\mathbf{\eta}}}_{\mathbf{a}\mathbf{i}\mathbf{r}}{\mathbf{s}}{\mathbf{i}}{\mathbf{n}}{{\mathbf{\theta}}}_{\mathbf{a}\mathbf{i}\mathbf{r}}{\mathbf{=}}{{\mathbf{\eta}}}_{\mathbf{r}\mathbf{e}\mathbf{d}}{\mathbf{s}}{\mathbf{i}}{\mathbf{n}}{{\mathbf{\theta}}}_{\mathbf{r}\mathbf{e}\mathbf{d}}}$

The angle of refraction of red light:

${\mathit{\theta}}_{\mathbf{r}\mathbf{e}\mathbf{d}}\mathbf{=}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(\frac{{\eta}_{air}sin{\theta}_{air}}{{\eta}_{red}}\right)$

Similarly, for the blue light, the angle of refraction will be:

${\mathit{\theta}}_{\mathbf{b}\mathbf{l}\mathbf{u}\mathbf{e}}\mathbf{=}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\frac{{\mathbf{\eta}}_{\mathbf{air}}{\mathbf{sin\theta}}_{\mathbf{air}}}{{\mathbf{\eta}}_{\mathbf{b}\mathbf{l}\mathbf{u}\mathbf{e}}}\mathbf{\right)}$

A beam of white light is incident on the surface of a diamond at an angle *?*

a.(Figure 1) Since the index of refraction depends on the light's wavelength, the different colors that comprise white light will spread out as they pass through the diamond. The indices of refraction in diamond are *n*_{red}=2.410 for red light and *n*_{blue}=2.450 for blue light. The surrounding air has *n*air=1.000. Note that the angles in the figure are not to scale.

Derive a formula for *?*, the angle between the red and blue refracted rays in the diamond.

Express the angle in terms of *n*red, *n*blue, and *?*a. Use *n*_{air}=1. Remember that the proper way to enter the inverse sine of *x* in this case is asin(x).

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