Problem 17.24 Challenge

a) จาก \(α + θ_{4} + 90° = 180°\)

\(\alpha + \theta _{4} + 90° = \alpha + \theta _{3} + \gamma + \beta\)

\(\beta = 90° + \theta _{4} - \theta _{3} - \gamma\)

\(\beta = 90° + \theta _{4} - \theta _{3} - (\theta _{1} - \theta _{2}) ; \gamma = \theta _{1} - \theta _{2}\)

\(\beta = 90° + \theta _{4} - \theta _{2} - (\theta _{1} - \theta _{2}); \theta _{3} = \theta _{2}\)

\(\beta = 90° + \theta _{4} - \theta _{1}\cdots \cdots (1)\)

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

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

จาก \(\frac{\sin \theta _{3}}{\sin \theta _{4}}=\frac{n_{1}}{n_{2}}\)

\(\sin \theta _{2} =\frac{n_{1}}{n_{2}}\sin \theta _{4}\cdots \cdots (3) ; \theta _{3} = \theta _{2}\)

\((2) = (3)\) จะได้ \(sinθ_{1} = sinθ_{4}\)

\(\theta _{1} = \theta _{4}\cdots \cdots (4)\)

\((4)\) แทนใน \((1)\) จะได้ \(β = 90° + θ_{4} - θ_{4}\)

\(\beta = 90°\)

∴ เนื่องจาก \(β\) เป็นมุมแย้งภายในของมุม \(90°\) ดังนั้นลำแสงที่หักเหออกมาจาก

ตัวกลางดัชนีหักเห n₂ ขนานกับลำแสงที่หักเหเข้ามายังตัวกลางดัชนีหักเห \(n_{2} \;\blacksquare\)

b) จาก \(sinγ = \frac{d}{\ell}\)

\(d = \ell \sin (\theta _{1} - \theta _{2})\)

\(d =\frac{t}{\cos \theta _{2}}\sin (\theta _{1} - \theta _{2}); \cos \theta _{2} =\frac{t}{\ell}\)

\(d =\frac{t}{\cos \theta _{2}}(\sin \theta _{1}\cos \theta _{2} - \sin \theta _{2}\cos \theta _{1})\)

\(d = t(\sin \theta _{1} - \cos \theta _{1}\tan \theta _{2})\)

\(d = tcos\theta _{1}(\tan \theta _{1} - \tan \theta _{2})\)

จาก \((2)\) จะได้ \(d = tcosθ_{1}[tanθ_{1} - tan(sin^{-1}(\frac{n_{1}}{n_{2}}\sin \theta _{1}))] \;\blacksquare\)