We will now investigate the total illumination distributed over the area of the circle of radius r. We have
| I² = | π²R4 | · | 4J1²(z) | (19), |
| 벃² | z² |
where
z = 2πRr/λf (20).
Thus
| 2π ∫ I²rdr = | λ²f² | ∫ I²zdz = πR²· 2 ∫ z−1J1²(z)dz. |
| 2πR² |
Now by (17), (18)
z-1J1(z) = J0(z) − J1′(z);
so that
| z-1J1²(z) = − ½ | d | J0² − ½ | d | J1²(z), |
| dz | dz |