In a Young's double slit experiment, the slit separation d is 0.3mm...

Path difference at $“P”$ , $∆ x = dsin \alpha + dsin \theta$
$\begin{cases}\theta, \alpha: smallangle, & sin \theta ≅ tan \theta = \dfrac{y}{D}\end{cases}$ and $sin \alpha \approx \alpha$
$⇒ ∆ x = d\alpha + \dfrac{dy}{D}$
\left(A\right) ∆ x = 3mm \times \dfrac{0.36}{\pi} \times \dfrac{\pi}{180} + 0 = 600nm$<br/>$600nm = n\lambda$<br/>$n = 1$<br/>Constructive interference$⇒ \left(A\right) is incorrect
\left(B\right) Fringe width does not depend on $\alpha . ⇒ \left(B\right)$ is incorrect
\left(C\right) ∆ x = 0.3mm \times \dfrac{0.36}{\pi} \times \dfrac{\pi}{180} + \dfrac{0.3mm \times 11mm}{1000} = 39 \times {10}^{- 4}mm$<br/>Now,$∆ x = 39 \times {10}^{- 4} = \left(2n - 1\right) \times \dfrac{600 \times {10}^{- 9} \times {10}^{3}}{2}$<br/>$⇒ n = 7$<br/>Destructive interference happened$⇒$correct <br/>\left(D\right)$\begin{cases}\alpha = 0 & ∴ ∆ x = dy / D = \dfrac{0.3 \times 11}{1000} = 33 \times {10}^{- 4}mm\end{cases}$$<br/>$∆ x$in terms of$\lambda = \dfrac{33 \times {10}^{- 4}}{600 \times {10}^{- 6}}\lambda = \dfrac{11\lambda}{2}$<br/>as$∆ x = \left(2n - 1\right)\dfrac{\lambda}{2}$<br/>There will be destructive interference.$⇒ \left(D\right)$ is incorrect.