高校物理 波動 ~ 光

$$c = 2.99792458 \times 10^8 \ \mbox{m/s} \fallingdotseq 3.00 \times 10^8 \ \mbox{m/s}$$

c = 2.99792458 \times 10^8 \ \mbox{m/s} \fallingdotseq 3.00 \times 10^8 \ \mbox{m/s}

$$n_{12} = \frac{\sin i}{\sin r} = \frac{v_1}{v_2} = \frac{\lambda _1}{\lambda _2}$$

n_{12} = \frac{\sin i}{\sin r} = \frac{v_1}{v_2} = \frac{\lambda _1}{\lambda _2}

$$\frac{\sin i_0}{\sin 90 ^\circ} = \sin i_0 = n_{12} = \frac{n_2}{n_1}$$

\frac{\sin i_0}{\sin 90 ^\circ} = \sin i_0 = n_{12} = \frac{n_2}{n_1}

$$\begin{eqnarray*} \frac{1}{a} + \frac{1}{b} = \frac{1}{f} \\ \\ m = \left| \frac{b}{a} \right| \end{eqnarray*}$$

\frac{1}{a} + \frac{1}{b} = \frac{1}{f}

m = \left| \frac{b}{a} \right|

$$\begin{eqnarray*} \text{明線:}\ |L_1 – L_2| &=& m \lambda = \frac{\lambda}{2} \cdot 2m \ (m=0, 1, 2, \cdots) \\ \text{暗線:}\ |L_1 – L_2| &=& \left( m + \frac{1}{2} \right) \lambda = \frac{\lambda}{2} \cdot (2m +1) \ (m=0,1,2, \cdots ) \end{eqnarray*}$$

\text{明線:}\ |L_1 – L_2| &=& m \lambda = \frac{\lambda}{2} \cdot 2m \ (m=0, 1, 2, \cdots)

\text{暗線:}\ |L_1 – L_2| &=& \left( m + \frac{1}{2} \right) \lambda = \frac{\lambda}{2} \cdot (2m +1) \ (m=0,1,2, \cdots )

$$d \sin \theta = m \lambda = \frac{\lambda}{2} \cdot 2m \ (m=0,1,2, \cdots )$$

d \sin \theta = m \lambda = \frac{\lambda}{2} \cdot 2m \ (m=0,1,2, \cdots )