RF_FAN Messages 2 Reaction score 0 Thread starter Dec 4, 2014 #1 how does this aproximation reduse the integral
mfb Mentor Insights Author Messages 37,488 Reaction score 14,378 Dec 4, 2014 #2 Where are lambda and k_z in the integral? The approximation simply assumes the integrand to be constant, evaluates it at the midpoint (z'=0) and multiplies it with the range z' takes.
Where are lambda and k_z in the integral? The approximation simply assumes the integrand to be constant, evaluates it at the midpoint (z'=0) and multiplies it with the range z' takes.
RF_FAN Messages 2 Reaction score 0 Dec 5, 2014 #3 so its because r'=0 ?? then why we need l<<lambda kz<<pi ?
mfb Mentor Insights Author Messages 37,488 Reaction score 14,378 Dec 5, 2014 #4 The exponential function should not change its argument too much, so kzl/r should be small compared to 1.
The exponential function should not change its argument too much, so kzl/r should be small compared to 1.