(Solved): \[ \left(90^{\circ}-\theta_{2}\right)+\Phi+\left(90^{\circ}-\theta_{3}\right)=180^{\circ}, \] so w ...

\[ \left(90^{\circ}-\theta_{2}\right)+\Phi+\left(90^{\circ}-\theta_{3}\right)=180^{\circ}, \] so we have \[ \theta_{3}=\Phi-\theta_{2} . \] A general form of Snell's law for light going through a prism where \( n \) is the index of refraction of the prism material is given by the following. \[ \sin \theta_{1}=n \sin \theta_{2} \] Solving for \( \theta_{2} \) for each wavelength of light, we have the following. (Express all calculated values to four significant figures.) \[ \begin{array}{l} \left(\theta_{2}\right)_{\text {violet }}=\sin ^{-1}\left(\frac{\sin 53.5}{1.66}\right)=4.0 \square 28.96 \\ \left(\theta_{2}\right)_{\text {red }}=\sin ^{-1}\left(\frac{1.62}{29.00} 20\right. \end{array} \] Step 4 For the ray exiting the glass, we have \[ \theta_{3}=60^{\circ}-\theta_{2} \] and, again using Snell's law, \[ \sin \theta_{4}=n \sin \theta_{3} . \] Solving for \( \theta_{4} \) for each wavelength of light, we have the following. (Express all calculated values to four significant figures.) \[ \left(\theta_{4}\right)_{\text {violet }}=\sin ^{-1}(1.66 \sin 4.0 \square \] \[ \left(\theta_{4}\right)_{\text {red }}=\sin ^{-1}(1.62 \sin 4.0 \square \]