In a Young's double slit experiment, the slit separation d is 0.3 mm and the screen distance D is 1 m. A parallel beam of light of wavelength 600 nm is incident on the slits at angle $\alpha$ as shown in figure. On the screen, the point O is equidistant from the slits and distance PO is 11 mm. Which of the following statement (s) is/are correct ?

Total geometric path difference for point P is

$∆x=d\sin \alpha +\frac{yd}{D}$

For option (1),

$\alpha =\frac{0.36}{\pi }$ degree

Since $\alpha$ is small, so

$$\begin{gathered} ∆x=\alpha d+\frac{yd}{D} \\ \Rightarrow ∆x=3900 nm \end{gathered}$$

Since $∆x=3900 nm$ is an odd multiple of $\frac{\lambda }{2}$ i.e., 300 nm.

$\Rightarrow \left(2n-1\right)\frac{\lambda }{2}=∆x$      (for destructive interference)

$\Rightarrow n=7$

For option (2),

Fringe width in all the above case reman unchanged.

For option (3),

$∆x_0=0$ and $\frac{yd}{D}=3300 nm$

As $\left(\alpha =0\right)$ (Destructive interference at P)

For option (4),

If $\alpha =\frac{0.36}{\pi }$ degree

Then $∆x_0=\alpha d=600 nm$

$\Rightarrow ∆x_0=n\lambda$ (for n=1)  (Constructive interference)