Recently physicists have managed to build "attosecond lasers", lasers which emit pulses 10^-18 seconds long, interrupted by much longer periods of darkness (at least 10^-14 seconds). Before them, lasers emitting femtosecond (10^-15 seconds) pulses have been around. Assuming they produce visible light, what colour is it?

The solution can be worked out by first determining the periodicity of the waveform.  At first glance, the period seems to be related to 10^-18s.  However, this is not true because the waveform is not a square pulse.  If it were then it should repeat every 2*10^-18s.

Now, the period of repeating is actually 10^-14s approximately.  If one assumes 10^-14s is the period of darkness then the actual period is 10^-14s + 10^-18s ~ 10^-14s.

The fourier series for the pulse function f(t) which repeats every 10^-14s is f(t)= A0/2 +sum{n=1 to n=infinity}[An(cos(2*pi*f0*t)) + Bn(sin(2*pi*f0*t))] where An = 2f0*integral{t=0 to t=(1/f0)}(f(t)*cos(2*pi*f0*t)dt) for even n and Bn = 2f0*integral{t=0 to t=(1/f0)}(f(t)*sin(2*pi*f0*t)dt) for odd n

Given I0= light intensity of waveform over the whole spectrum

Given tau= 10^-18s

Given f0= 10^14 cycles/second, f(t)= I0 from t=0 to t=tau and f(t)=0 from t=tau+ to t=(1/f0),

Bn = 2f0*(I0/(2*pi*n*f0))*(1 - cos(2*pi*n*f0*tau))

=2f0*(I0/(2*pi*n*f0))*((2*pi*n*f0*tau)^2/2) by taylor expansion of cosine since tau<<(1/f0)

=2*I0*n*pi*(f0*tau)^2

An = 2f0*(I0/(2*pi*n*f0))*(sin(2*pi*n*f0*tau))

=2f0*(I0/(2*pi*n*f0))*(2*pi*n*f0*tau) by taylor expansion of sine since tau<<(1/f0)

= 2*I0*(f0*tau)

Since the odd terms depend on the (f0*tau)^2 term, they can be neglected compared to the even terms which depend on (f0*tau) (which is a much larger quantity)

Note: (f0*tau)=10^-4

Now, the visible spectrum is between 4.0 * 10^14 Hz and 7.89 * 10^14 Hz.

Consequently, the only even terms in this range is at 4*10^14Hz and 6*10^14Hz.  Once again, the term at 5*10^14Hz and 7*10^14Hz are much smaller in magnitude so they are neglected.

Assuming the speed of light is approximately the same in air as a vaccuum,

(4*10^14Hz)*wavelength0=c=3*10^8m/s

(6*10^14Hz)*wavelength1=c=3*10^8m/s

wavelength0= 7.5*10^-7m = 750nm (red)

wavelength1=5*10^-7m= 500nm (green)

Now, the human eye is much more sensitive to green than extreme red (almost infrared).  Thus, I would say it appears green since the "color" is a human perception.  However, since the intensity is so small, I would say you might not even be able to see the light at all unless the total laser emission intensity is very high.

Now, since the (f0*t) and (f0*t)^2 terms are so small, the spectrum is very wide.  This explains the near independence of the even terms on n.  As the space between the pulses becomes wider then the spectrum gets even wider.  This also explains why each spectral component has such a small intensity.  This is because the total intensity I0 is spread among many frequencies pretty evenly.  So, each spectral component has a small intensity itself.