Answer
$2.09^{\circ}$
Work Step by Step
Here, $d=\frac{1}{180}\;mm=\frac{10^{-3}}{180}\;m$
When $\lambda_1=400\;nm$, for second order maxima,
$d\sinθ_1=2λ_1$
or, $\sinθ_1=\frac{2λ_1}{d}$
or, $θ_1=\sin^{-1}(\frac{2λ_1}{d})$
Substituting the known values
$θ_1=\sin^{-1}(\frac{2\times400\times10^{-9}\times180}{10^{-3}})$
or, $θ_1=8.28^{\circ}$
When $\lambda_2=500\;nm$, for second order maxima,
$d\sinθ_2=2λ_2$
or, $\sinθ_2=\frac{2λ_2}{d}$
or, $θ_2=\sin^{-1}(\frac{2λ_2}{d})$
Substituting the known values
$θ_2=\sin^{-1}(\frac{2\times500\times10^{-9}\times180}{10^{-3}})$
or, $θ_2=10.37^{\circ}$
Therefore, the angular separation between the second order maxima is $(10.37^{\circ}-8.28^{\circ})=2.09^{\circ}$
