hello, this is a question about optics, more specifically single slit
diffraction.  what is the percentage of total incident intensity
contained in the central peak of the diffraction pattern?  i would
prefer an answer like X% of incident intensity.  i already know the
equation on how to find the intensity for any angle, but this equation
is no good unless i know the intensity of the central peak.  so here's
what i know:  the incident power through the slit, and the equation
for intensity as a function of diffraction angle.  using these two
pieces of information, it seems straightforward that one should be
able to find the intensity of the central peak.  here is a useful link
http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinint.html#c3

thanks a lot.

Presumably you are asking about far-field (Fraunhofer) diffraction, in
which case the intensity of the diffraction pattern is given by:

     I = I0*[sin(b)/b]^2

where 

     I0 is the maximum intensity of the central peak
     b = (pi * D * x)/(L * R)
     D = diameter of the slit
     L = wavelength of the light
     R = distance from slit to projection plane
     x =  distance from axial position along projection plane.

(You don?t actually need to know all that to solve your problem
though; all you need is the equation for the intensity.)

The intensity pattern has minima at b = +/-pi, +/-2*pi, +/-3*pi?.

The fraction of the intensity in the central peak is simply the
integral of the intensity under that peak (i.e., from b = - pi to pi)
divided by the integral of the intensity from b = minus infinity to
positive infinity.  (Because of symmetry, we really only need to take
the ratio of the integrals from zero to the positive limits.)


F= [Integral from b=0 to pi of { [sin(b)/b]^2 db}]/[Integral from b =
0 to +infinity of { [sin(b)/b]^2 db}]

The integral from 0 to z of {[sin(z)/z]^2 dz} is given by:

  [cos(2*z) + 2*z*Si(2*z) ? 1]/(2*z)

where Si(t) is the so-called ?sine integral? = integral from 0 to t of
{sin(t)/t}  See, for instance
<http://mathworld.wolfram.com/SineIntegral.html>.  It?s a built-in
function in many numerical analysis packages, and values of the
integral are tabulated in basic mathematics handbooks.

For your problem, the numerical value of the integral of the intensity
from b = zero to pi is 1.41815, and the numerical value of the
intrgral from b = 0 to +infinity is 1.57087.  The ratio of these two
values is 0.90278, which is the result you are looking for (i.e.,
90.278% of the total transmitted intensity is in the central
Fraunhofer peak),

yes, that makes perfect sense.  i was so close, but gave up on account
of the crappy integral.  thank you so much, you are awesome.