Obliczenie: UQX3sL - 1
d* sin(φ) =
| ( 2* k+1)* λ |
| / 2 |
|
d* sin(φ) =
$$d\cdot sin(\phi)$$ = $$\frac{(2\cdot k+1)\cdot \lambda}{2}$$
| ( 2* k+1)* λ |
| / 2 |
d* sin(φ) = k* λ+
$$d\cdot sin(\phi)$$ = $$k\cdot \lambda+\frac{\lambda}{2}$$
| λ |
| / 2 |
d =
$$d$$ = $$\frac{k\cdot \lambda+\frac{\lambda}{2}}{sin(\phi)}$$
( k* λ+
| |||
| / sin(φ) |
d =
+
$$d$$ = $$\frac{k\cdot \lambda}{sin(\phi)}+\frac{0.5\cdot \lambda}{sin(\phi)}$$
| k* λ |
| / sin(φ) |
| 0.5* λ |
| / sin(φ) |