Aprēķins: qybrmS - 1

d+sin(φ) =

( 2* k+1)* λ

/ 2

d+sin(φ) =

( 2* k+1)* λ

/ 2

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

d =

( 2* k+1)* λ

/ 2

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

d = k* λ+

λ

/ 2

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

d = k* λ+ 0.5* λ-sin(φ) $d$ = $k\cdot \lambda+0.5\cdot \lambda-sin(\phi)$