Answer
$560\;nm$
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 outer edge of the hole
$\tan\theta^{'}=\frac{50+10}{300}=\frac{1}{5}$
$\implies \theta^{'}=11.31^{\circ}$
Now,
$d\sin\theta^{'}=m\lambda^{'}$
for $m=1$
$d\sin\theta^{'}=\lambda^{'}$
or, $\lambda^{'}=d\sin\theta^{'}$
Substituting the known values
or, $\lambda^{'}=\frac{10^{-3}\times\sin11.31^{\circ}}{350}$
or, $\lambda^{'}=560\times10^{-9}\;m=560\;nm$
Therefore, the longest wavelength of the light that passes through the hole is $560\;nm$.
