Answer
$\lambda=600nm$
Work Step by Step
We know that:
$dsin{\theta}=m\lambda$
We plug in the known values to obtain:
$dsin(90)=4\times (450\times 10^{-9})=1.8\times 10^{-6}m$
Now,
$dsin{\theta}=m\lambda$
We plug in the known values to obtain:
$(1.8\times 10^{-6})sin(90)=3\times (\lambda)$
$\lambda=600\times 10^{-9}m=600nm$
