In a double slit experiment shown in figure, when light of wavelength $$400$$ nm is used, dark fringe is observed at $$P$$. If $$D = 0.2$$ m, the minimum distance between the slits $$S_1$$ and $$S_2$$ is $$\alpha$$ mm. Write the value of $$10\alpha$$ to the nearest integer.

Total path length of light from source to P through each slit:

$$x_{\text{path 1}} = \text{Source} \rightarrow S_1 \rightarrow P = \sqrt{D^2 + d^2} + \sqrt{D^2 + d^2} = 2\sqrt{D^2 + d^2}$$

$$x_{\text{path 2}} = \text{Source} \rightarrow S_2 \rightarrow P = D + D = 2D$$

Net path difference at point P:

$$\Delta x = x_{\text{path 1}} - x_{\text{path 2}} = 2\sqrt{D^2 + d^2} - 2D = 2D\left(1 + \frac{d^2}{D^2}\right)^{1/2} - 2D$$

Using binomial approximation ($$d \ll D$$):

$$\Delta x \approx 2D\left(1 + \frac{d^2}{2D^2}\right) - 2D = \frac{d^2}{D}$$

Condition for first minimum (dark fringe) for minimum distance $$d$$:

$$\Delta x = \frac{\lambda}{2} \implies \frac{d^2}{D} = \frac{\lambda}{2} \implies d = \sqrt{\frac{\lambda D}{2}}$$

$$d = \sqrt{\frac{4 \times 10^{-7} \times 0.2}{2}} = \sqrt{4 \times 10^{-8}} = 2 \times 10^{-4}\text{ m} = 0.2\text{ mm}$$

Given $$d = \alpha\text{ mm}$$: $$\alpha = 0.2 \implies 10\alpha = 2$$