Two ideal slits S_1 and S_2 are at a distance | vedclass

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(C) The total path difference $\Delta$ between the two waves reaching point $O$ is the sum of the path difference before reaching the slits and the pa

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$\sqrt {\frac{{\lambda D}}{2}} $

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(C) The total path difference $\Delta$ between the two waves reaching point $O$ is the sum of the path difference before reaching the slits and the path difference after passing through the slits.Let $\Delta_1$ be the path difference between the waves from $S$ to $S_1$ and $S$ to $S_2$. Since $S$ is aligned with $S_2$,the path $S-S_2$ is $D$,and the path $S-S_1$ is $\sqrt{D^2 + d^2}$. Thus,$\Delta_1 = \sqrt{D^2 + d^2} - D$.Let $\Delta_2$ be the path difference between the waves from $S_1$ to $O$ and $S_2$ to $O$. The path $S_2-O$ is $D$,and the path $S_1-O$ is $\sqrt{D^2 + d^2}$. Thus,$\Delta_2 = \sqrt{D^2 + d^2} - D$.The total path difference is $\Delta = \Delta_1 + \Delta_2 = 2(\sqrt{D^2 + d^2} - D)$.Using binomial expansion for $d \ll D$,$\sqrt{D^2 + d^2} = D(1 + \frac{d^2}{D^2})^{1/2} \approx D(1 + \frac{d^2}{2D^2}) = D + \frac{d^2}{2D}$.Therefore,$\Delta \approx 2(D + \frac{d^2}{2D} - D) = \frac{d^2}{D}$.For destructive interference (darkness) at $O$,the path difference must be an odd multiple of $\frac{\lambda}{2}$,i.e.,$\Delta = (2n - 1)\frac{\lambda}{2}$ where $n = 1, 2, 3, ...$.Setting $\frac{d^2}{D} = (2n - 1)\frac{\lambda}{2}$,for the minimum distance $d$,we take $n = 1$:$\frac{d^2}{D} = \frac{\lambda}{2} \Rightarrow d^2 = \frac{\lambda D}{2} \Rightarrow d = \sqrt{\frac{\lambda D}{2}}$.

List-$I$	List-$II$
$(i)$ The equation of the polar of $(-5, 1)$ with respect to $C$	$(A)$ $y = 0$
$(ii)$ The equation of the tangent at $(8, 0)$ to $C$	$(B)$ $y = 6$
$(iii)$ The equation of the normal at $(2, 6)$ to $C$	$(C)$ $x + y = 7$
$(iv)$ The equation of the diameter of $C$ through $(8, 12)$	$(D)$ $13x + 5y = 98$
	$(E)$ $x = 8$

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