In a Young's double slit experiment,two slits | vedclass

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(C) The condition for the first minimum at point $P$ is given by the path difference $\Delta x = A_2 P - A_1 P = \frac{\lambda}{2}$.From the geometry

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$0.2$

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(C) The condition for the first minimum at point $P$ is given by the path difference $\Delta x = A_2 P - A_1 P = \frac{\lambda}{2}$.From the geometry of the figure,$A_1 P = D$ and $A_2 P = \sqrt{D^2 + a^2}$.Thus,$\sqrt{D^2 + a^2} - D = \frac{\lambda}{2}$.Using the binomial approximation for $a \ll D$,we have $\sqrt{D^2 + a^2} = D(1 + \frac{a^2}{D^2})^{1/2} \approx D(1 + \frac{a^2}{2D^2}) = D + \frac{a^2}{2D}$.Substituting this into the path difference equation: $(D + \frac{a^2}{2D}) - D = \frac{\lambda}{2}$.This simplifies to $\frac{a^2}{2D} = \frac{\lambda}{2}$,which gives $a = \sqrt{\lambda D}$.Given $\lambda = 800 \, nm = 800 	imes 10^{-9} \, m = 8 	imes 10^{-7} \, m$ and $D = 5 \, cm = 0.05 \, m$.$a = \sqrt{8 	imes 10^{-7} 	imes 0.05} = \sqrt{40 	imes 10^{-9}} = \sqrt{400 	imes 10^{-10}} = 20 	imes 10^{-5} \, m = 0.2 	imes 10^{-3} \, m = 0.2 \, mm$.

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