\sin α = \sqrt{n^{2} - {[\frac{(m_{max} - 1) λ - \frac{λ}{2}}{2 h}]}^{2}} = \sqrt{1, 5^{2} - {[\frac{(465 - 0, 5) \times 6, 438 \cdot 10^{- 7}}{2 \times 10^{- 4}}]}^{2}} = 0, 098

${\sin \alpha= \sqrt{n^2 - \left[\frac{(m_\text{max}-1) \lambda - \frac{\lambda}{2}}{2h}\right]^2} = \sqrt{1,5^2 - \left[\frac{(465-0,5) \times 6,438 \cdot 10^{-7}}{2 \times 10^{-4}} \right]^2 } = 0,098}$

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