Theo đề bài: i = 300; \(sinr_1 = \dfrac{1}{2n}\)
i2 = 900 (HÌnh 28.4G); r2 = igh --> \(sinr_2 = \dfrac{1}{n}\)
Nhưng r1 = A – r2 – 600 - igh
\(\begin{array}{l}
\Rightarrow \dfrac{1}{{2n}} = \frac{{\sqrt 3 }}{2}.\dfrac{{\sqrt {{n^2} - 1} }}{n} - \dfrac{1}{n}.\dfrac{1}{2}\\
\Rightarrow n = \sqrt {1 + \dfrac{4}{3}} = \sqrt {\frac{7}{3}} \approx 1,53
\end{array}\)
