Problem 16.31

จาก \(\frac{\sin \theta}{\sin \theta _{2}}=\frac{n_{2}}{n_{1}}\)

\(\frac{\sin \theta}{\sin (180° - (\theta  + \beta ))}=\frac{n_{2}}{n_{1}}\)

\(\frac{\sin \theta}{\sin (\theta  + \beta )}=\frac{n_{2}}{n_{1}}\)

\(n_{1}\sin \theta = n_{2}\sin (\theta + \beta )\)

\(n_{1}\sin \theta = n_{2}(\sin \theta \cos \beta + \sin \beta \cos \theta )\)

\(n_{1}\sin \theta = n_{2}\cos \theta (\frac{\sin \theta}{\cos \theta}\cos \beta + \sin \beta )\)

\(n_{1}\frac{\sin \theta}{\cos \theta}= n_{2}(\frac{\sin \theta}{\cos \theta}\cos \beta + \sin \beta )\)

\(n_{1}\tan \theta = n_{2}\tan \theta \cos \beta + n_{2}\sin \beta\)

\(n_{1}\tan \theta - n_{2}\tan \theta \cos \beta = n_{2}\sin \beta\)

\(\tan \theta (n_{1} - n_{2}\cos \beta ) = n_{2}\sin \beta\)

\(\tan \theta =\frac{n_{2}\sin \beta}{n_{1} - n_{2}\cos \beta}\;\blacksquare\)

แทน \(β = 90°\) จะได้ \(tanθ = \frac{n_{2}\sin (90°)}{n_{1} - n_{2}\cos (90°)}\)

\(\tan \theta =\frac{n_{2}}{n_{1}}\)

สอดคล้องกับกฎของบริวสเตอร์ \(tanθ_{p}=\frac{n_{2}}{n_{1}}\;\blacksquare\)