Answer
$138\;nm$
Work Step by Step
Here, we consider only a part of the laser beam, which is within the central diffraction maximum.
The equation for the angular position of the edge of the central diffraction maximum is given by
$\sin\theta =\frac{1.22\lambda}{d}$,
where $\lambda$ is the wavelength and $d$ is the diameter of the aperture.
If the target is at a distance $D$ away from the source, the radius of the beam is
$r= D\tan\theta$
If $\theta$ is small, then $\tan\theta\approx\sin\theta$
Therefore,
$r= D\sin\theta$
or, $r= \frac{1.22D\lambda}{d}$
Therefore, the intensity of the laser beam at the target point is
$I=\frac{P}{\pi r^2}$
or, $I=\frac{Pd^2}{\pi(1.22D\lambda)^2}\;..............................(1)$
From the relation $(1)$, we obtain
$\lambda=\frac{d}{1.22D}\sqrt {\frac{P}{\pi I}}$
Putting known values, we obtain,
$\lambda=\frac{4}{1.22\times 3000\times10^3}\sqrt {\frac{5\times10^6}{\pi\times 1\times10^{8}}}\;m$
or, $\lambda=1.38\times10^{-7}\;m$
or, $\lambda=138\;nm$
Therefore, the maximum value of wavelength is $138\;nm$
