Physics

580 nm light shines on a double slit with d=0.000125 m.
What is the angle of the third dark interference minimum(m=3)?

$\begin{array}{rcll} Let\ d &=& 0.000125\ m \\ d &=& 125 \cdot 10^{-6}\ m \\\\ Let\ \lambda &=& 580\ nm \\ \lambda &=& 580 \cdot 10^{-9}\ m \\ \end{array}$

$\begin{array}{|rcll|} \hline d\sin(\theta) &=& m \lambda \\ \sin(\theta) &=& m\cdot \frac{\lambda}{d} \quad & | \quad m=3 \\ \sin(\theta) &=& 3\cdot \frac{580 \cdot 10^{-9}\ m }{125 \cdot 10^{-6}\ m} \\ \sin(\theta) &=& 3\cdot \frac{580}{125} \cdot 10^{-9+6} \\ \sin(\theta) &=& \frac{1740}{125} \cdot 10^{-3} \\ \sin(\theta) &=& 13.92 \cdot 10^{-3} \\ \sin(\theta) &=& 0.01392 \\ \theta &=& \arcsin(0.01392) \\ \theta &=& 0.79758300970^{\circ} \\ \hline \end{array}$