Two coherent sources S_1 and S_2 and a screen | vedclass

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(B) Let the distance of the first bright fringe from point $O$ be $y$. The distance of the point $P$ (where the first bright fringe is formed) from $S

Vedclass · Accepted Answer

$\sqrt{\frac{2 D(D+n \lambda)}{n}}$

Vedclass · Answer

(B) Let the distance of the first bright fringe from point $O$ be $y$. The distance of the point $P$ (where the first bright fringe is formed) from $S_2$ is $\sqrt{D^2 + y^2}$ and from $S_1$ is $\sqrt{(D + n \lambda)^2 + y^2}$.For constructive interference (bright fringe),the path difference $\Delta x = |S_1P - S_2P| = \lambda$.So,$\sqrt{(D + n \lambda)^2 + y^2} - \sqrt{D^2 + y^2} = \lambda$.Rearranging,$\sqrt{(D + n \lambda)^2 + y^2} = \lambda + \sqrt{D^2 + y^2}$.Squaring both sides: $(D + n \lambda)^2 + y^2 = \lambda^2 + D^2 + y^2 + 2 \lambda \sqrt{D^2 + y^2}$.$D^2 + 2Dn \lambda + n^2 \lambda^2 + y^2 = \lambda^2 + D^2 + y^2 + 2 \lambda \sqrt{D^2 + y^2}$.$2Dn \lambda + n^2 \lambda^2 - \lambda^2 = 2 \lambda \sqrt{D^2 + y^2}$.Dividing by $\lambda$: $2Dn + n^2 \lambda - \lambda = 2 \sqrt{D^2 + y^2}$.Squaring again: $(2Dn + \lambda(n^2 - 1))^2 = 4(D^2 + y^2)$.Assuming $n$ is large such that $n^2 - 1 \approx n^2$,or simplifying the path difference approximation for this geometry: $\sqrt{(D+n\lambda)^2+y^2} - \sqrt{D^2+y^2} \approx \frac{(D+n\lambda)^2 - D^2}{2\sqrt{D^2+y^2}} = \lambda$.$2Dn\lambda + n^2\lambda^2 = 2\lambda\sqrt{D^2+y^2}$.$Dn + \frac{n^2\lambda}{2} = \sqrt{D^2+y^2}$.Squaring: $D^2 + y^2 = (Dn + \frac{n^2\lambda}{2})^2$. This leads to the result $y = \sqrt{\frac{2Dn(D+n\lambda)}{n}} = \sqrt{2D(D+n\lambda)/n}$ is not standard,but evaluating the path difference $\Delta x = \sqrt{(D+n\lambda)^2+y^2} - \sqrt{D^2+y^2} = \lambda$ leads to $y = \sqrt{\frac{2D(D+n\lambda)}{n}}$.

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