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

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(C) For darkness at point $O$,the path difference between the light waves reaching $O$ from $S_1$ and $S_2$ must be an odd multiple of $\frac{\lambda}

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) For darkness at point $O$,the path difference between the light waves reaching $O$ from $S_1$ and $S_2$ must be an odd multiple of $\frac{\lambda}{2}$.The path taken by light from $S$ to $O$ via $S_2$ is $SO + OS_2 = D + D = 2D$.The path taken by light from $S$ to $O$ via $S_1$ is $SS_1 + S_1O$. Here,$SS_1 = \sqrt{D^2 + d^2}$ and $S_1O = \sqrt{D^2 + d^2}$.Thus,the total path via $S_1$ is $2\sqrt{D^2 + d^2}$.The path difference $\Delta x = 2\sqrt{D^2 + d^2} - 2D$.For minimum darkness,$\Delta x = \frac{\lambda}{2}$.$2\sqrt{D^2 + d^2} - 2D = \frac{\lambda}{2} \implies \sqrt{D^2 + d^2} = D + \frac{\lambda}{4}$.Squaring both sides: $D^2 + d^2 = D^2 + \frac{\lambda D}{2} + \frac{\lambda^2}{16}$.Since $d \ll D$,we can neglect the $\lambda^2$ term: $d^2 \approx \frac{\lambda D}{2}$.Therefore,$d = \sqrt{\frac{\lambda D}{2}}$.

Similar Questions

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