Answer
$21.1^{\circ}$
Work Step by Step
Here, $d=\frac{1}{180}\;mm=\frac{10^{-3}}{180}\;m$
When $\lambda_1=400\;nm$, for $m_1$th order maxima,
$d\sinθ_1=m_1λ_1$
or, $\sinθ_1=\frac{m_1λ_1}{d}\;.............(1)$
When $\lambda_2=500\;nm$, for $m_2$th order maxima,
$d\sinθ_2=m_2λ_2$
or, $\sinθ_2=\frac{m_2λ_2}{d}\;.............(2)$
Equating eq. (1) and (2), we obtain
$\frac{m_1λ_1}{d}=\frac{m_2λ_2}{d}$
or, $m_1=\frac{m_2λ_2}{λ_1}$
Putting the known values,
$m_1=\frac{m_2\times500}{400}$
or, $m_1=1.25m_2$
The values of $m_1$ and $ m_2$ are integer.
Thus, the lowest possible case, when the resulting maxima are superimposed is
$m_2=4$ and so, $m_1=5$
Thus, the angle is
$θ_1=\sin^{-1}(\frac{m_1λ_1}{d})$
or, $θ_1=\sin^{-1}(\frac{5\times400\times10^{-9}\times180}{10^{-3}})$
or, $θ_1=21.1^{\circ}$
Therefore, the smallest angle at which two of the resulting maxima are superimposed
is $21.1^{\circ}$
