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