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(A) For the first minima to occur at point $P$ (directly in front of $S_1$),the path difference between the waves from $S_1$ and $S_2$ must be equal t

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$\frac{\lambda}{2(\sqrt{5}-2)}$

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(A) For the first minima to occur at point $P$ (directly in front of $S_1$),the path difference between the waves from $S_1$ and $S_2$ must be equal to $\frac{\lambda}{2}$.The distance from $S_1$ to $P$ is $2d$.The distance from $S_2$ to $P$ is $\sqrt{(2d)^2 + d^2} = \sqrt{4d^2 + d^2} = \sqrt{5}d$.The path difference $\Delta x = S_2P - S_1P = \sqrt{5}d - 2d = d(\sqrt{5} - 2)$.Setting the path difference equal to $\frac{\lambda}{2}$ for the first minima:$d(\sqrt{5} - 2) = \frac{\lambda}{2}$Therefore,the slit separation $d$ is:$d = \frac{\lambda}{2(\sqrt{5} - 2)}$

Consider a Young's double slit experiment as shown in the figure. What should be the slit separation $d$ in terms of wavelength $\lambda$ such that the first minima occurs directly in front of the slit $S_1$?

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