Answer
$\lambda=500nm$
Work Step by Step
We know that:
$\lambda=\frac{dsin\theta}{m}$
We plug in the known values to obtain:
$\lambda=\frac{5\times 10^{-6}sin(30^{\circ})}{m}=\frac{2.5\times 10^{-6}}{m}$
Therefore;
for $m=1$:
$\lambda=\frac{2.5\times 10^{-6}}{1}=2.5\times 10^{-6}m=2500nm$
for $m=2$;
$\lambda=\frac{2.5\times 10^{-6}}{2} =1250nm$
for $m=3$;
$\lambda=\frac{2.5\times 10^{-6}}{3} =833nm$
for $m=4$;
$\lambda=\frac{2.5\times 10^{-6}}{4} =625nm$
for $m=5$;
$\lambda=\frac{2.5\times 10^{-6}}{5} =500nm$
