A Young's double slit interference arrangemen | vedclass

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(C) Let the distance of a point on the water surface from the point midway between the slits be $x$. The distance of this point from each slit $S_1$ a

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$3$

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(C) Let the distance of a point on the water surface from the point midway between the slits be $x$. The distance of this point from each slit $S_1$ and $S_2$ is $\sqrt{d^2 + x^2}$.The path difference $\Delta$ at this point is given by the difference in optical paths. The light travels through water to reach the point on the surface.Optical path from $S_2$ to the point is $\mu_w \sqrt{d^2 + x^2}$.Optical path from $S_1$ to the point is $\mu_w \sqrt{d^2 + x^2}$.Wait,looking at the figure,the point $x$ is on the surface of water. The path from $S_2$ is entirely in water,while the path from $S_1$ is in air and then enters water. However,the standard interpretation for this specific problem setup is that the path difference is $\Delta = \mu_w \sqrt{d^2+x^2} - \sqrt{d^2+x^2} = m\lambda$.$\Delta = (\mu_w - 1) \sqrt{d^2 + x^2} = m\lambda$Substituting $\mu_w = 4/3$:$(\frac{4}{3} - 1) \sqrt{d^2 + x^2} = m\lambda$$\frac{1}{3} \sqrt{d^2 + x^2} = m\lambda$$\sqrt{d^2 + x^2} = 3m\lambda$Squaring both sides:$d^2 + x^2 = 9m^2\lambda^2$$x^2 = 9m^2\lambda^2 - d^2$Comparing this with $x^2 = p^2 m^2 \lambda^2 - d^2$,we get $p^2 = 9$,so $p = 3$.

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