Answer
$470\;nm$
Work Step by Step
Light will pass through the hole, if any diffraction maxima occurs at the position of hole.
Here,
$d=\frac{1}{350}\;mm=\frac{10^{-3}}{350}\;m$
$D=30\;cm=300\;mm$
For the inner edge of the hole
$\tan\theta=\frac{50}{300}=\frac{1}{6}$
$\implies \theta=9.46^{\circ}$
Now,
$d\sin\theta=m\lambda$
or, $m=\frac{d\sin\theta}{\lambda}$
Substituting the known values
or, $m=\frac{10^{-3}\times\sin9.46^{\circ}}{350\times\lambda}$
or, $m=\frac{470\times10^{-9}}{\lambda}$
The wavelength of white light is $\gt400\;nm=400\times10^{-9}\;m$
Therefore, m = 1 is allowed, when $\lambda=470\;nm$ which is the shortest wavelength of the light that passes through the hole.
