Problem 17.27 Challenge

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

\(\frac{\sin \theta}{\sin φ}=\frac{n}{1}\)

\(\sin φ =\frac{\sin \theta}{n}=\frac{r}{nR}; \sin \theta =\frac{r}{R}\)

จาก \(tanφ = \frac{\sin φ}{\cos φ}\)

\(\tan φ =\frac{\sin φ}{1 - \sin ^{2}φ}\)

\(\tan φ =\frac{r}{n^{2}R^{2} - r^{2}}\approx \frac{r}{nR}; \sin φ =\frac{r}{nR}\)

และ \(tanθ = \frac{r}{R^{2} - r^{2}}\approx \frac{r}{R}\)

จาก \(tan(θ - φ) = \frac{\tan \theta  - \tan φ}{1 + \tan \theta \tan φ}\)

\(\frac{r - x}{R^{2} - r^{2}}=\frac{(\frac{r}{R})-(}{1 + (\frac{r}{R})(\frac{nRr)}{nRr})}\)

\(\frac{r - x}{R}=\frac{rnR^{2}(n - 1)}{nR(nR^{2} + r^{2})}\)

\(r - x =\frac{rR^{2}(n - 1)}{nR^{2} + r^{2}}\)

\(x = r -\frac{rR^{2}(n - 1)}{nR^{2}}; nR^{2} + r^{2} \approx nR^{2}\)

\(x = r -\frac{r(n - 1)}{n}\)

\(x=\frac{r}{n}\;\blacksquare\)