Obliczenie: qx55JC - 1

d* sin(φ) =

( 2* k+1)* λ

/ 2

d* sin(φ) =

( 2* k+1)* λ

/ 2

$d\cdot sin(\phi)$ = $\frac{(2\cdot k+1)\cdot \lambda}{2}$

d* sin(φ) = k* λ+

λ

/ 2

$d\cdot sin(\phi)$ = $k\cdot \lambda+\frac{\lambda}{2}$

d =

( k* λ+

λ

/ 2

)

/ sin(φ)

$d$ = $\frac{k\cdot \lambda+\frac{\lambda}{2}}{sin(\phi)}$

d =

k* λ

/ sin(φ)

+

0.5* λ

/ sin(φ)

$d$ = $\frac{k\cdot \lambda}{sin(\phi)}+\frac{0.5\cdot \lambda}{sin(\phi)}$