According to the given diagram, the path difference for minima at P is given by

$∆x=2\sqrt{D^2+d^2}-2D ...\left(1\right)$

And, from the condition of minima, it follows that

$∆x=\frac{\lambda }{2} ...\left(2\right)$

Equations (1) and (2) imply that

$2\sqrt{D^2+d^2}-2D=\frac{\lambda }{2}$

$\Rightarrow \sqrt{D^2+d^2}-D=\frac{\lambda }{4}$

$\Rightarrow \sqrt{D^2+d^2}=\frac{\lambda }{4}+D$

$\Rightarrow D^2+d^2=D^2+\frac{{\lambda }^2}{16}+\frac{D\lambda }{2}$

$\Rightarrow d^2=\frac{D\lambda }{2}+\frac{{\lambda }^2}{16}$

$\Rightarrow d^2=\frac{0.2\times 400\times {10}^{-9}}{2}+\frac{16\times {10}^{-14}}{16}$

$\Rightarrow d^2\approx 400\times {10}^{-10}$

$\begin{array}{rcl}\therefore d& =& 20\times {10}^{-5} \mathrm{m}\\ & =& 0.20 \mathrm{mm}\end{array}$

Therefore, $10\alpha =2$.