In a double slit experiment shown in figure, when light of...

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

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

Equations (1) and (2) imply that

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

$⇒ \sqrt{{D}^{2} + {d}^{2}} - D = \dfrac{\lambda}{4}$

$⇒ \sqrt{{D}^{2} + {d}^{2}} = \dfrac{\lambda}{4} + D$

$⇒ {D}^{2} + {d}^{2} = {D}^{2} + \dfrac{{\lambda}^{2}}{16} + \dfrac{D\lambda}{2}$

$⇒ {d}^{2} = \dfrac{D\lambda}{2} + \dfrac{{\lambda}^{2}}{16}$

$⇒ {d}^{2} = \dfrac{0 . 2 \times 400 \times {10}^{- 9}}{2} + \dfrac{16 \times {10}^{- 14}}{16}$

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

$\begin{cases}∴ d & = & 20 \times {10}^{- 5}m \\ & = & 0 . 20mm\end{cases}$

Therefore, $10\alpha = 2$.