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Two coherent point sources $S_1$ and $S_2$ vibrating in phase emit light of wavelength $\lambda$. The separation between them is $2 \lambda$ as shown in figure. The first bright fringe is formed at $P$ due to interference on a screen placed at a distance $D$ from $S_1(D>>\lambda)$, then $O P$ is
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Verified Answer
The correct answer is:
$\sqrt{3} D$
From the figure,
$\begin{aligned} & \left(S_1 P\right)^2=D^2+x^2 \\ & \text { and } \\ & \left(S_2 P\right)^2=(D-2 \lambda)^2+x^2 \\ & \text { So, }\left(S_1 P\right)^2-\left(S_2 P\right)^2=D^2+x^2 \\ & -\left[D^2+4 \lambda^2-4 D \lambda+x^2\right] \\ & =4 D \lambda-4 \lambda^2 \\ & \left(S_1 P+S_2 P\right)\left(S_1 P-S_2 P\right)=4 D \lambda-4 \lambda^2 \\ & {[D>>\lambda]} \\ & \text { Since, }[D>>\lambda] \\ & \therefore \quad S_1 P=S_2 P \\ & \end{aligned}$
and neglecting higher power of $\lambda$
$\begin{aligned} & \therefore \quad 2 S_1 P\left(S_1 P-S_2 P\right)=4 D \lambda \\ & 2\left(S_1 P\right)(\lambda)=4 D \lambda \\ & \text { or } \quad S_1 P=2 D \\ & \sqrt{D^2+x^2}=2 D \\ & x^2=3 D^2 \\ & x=\sqrt{3} D \\ & \end{aligned}$
Alternative
$\begin{gathered}2 \lambda \cos \theta=\lambda \\ \Rightarrow \quad \cos \theta=\frac{1}{2} \Rightarrow \theta=60^{\circ}, \tan \theta=\frac{x}{D} \\ \sqrt{3}=\frac{x}{D} \Rightarrow x=\sqrt{3} D\end{gathered}$
$\begin{aligned} & \left(S_1 P\right)^2=D^2+x^2 \\ & \text { and } \\ & \left(S_2 P\right)^2=(D-2 \lambda)^2+x^2 \\ & \text { So, }\left(S_1 P\right)^2-\left(S_2 P\right)^2=D^2+x^2 \\ & -\left[D^2+4 \lambda^2-4 D \lambda+x^2\right] \\ & =4 D \lambda-4 \lambda^2 \\ & \left(S_1 P+S_2 P\right)\left(S_1 P-S_2 P\right)=4 D \lambda-4 \lambda^2 \\ & {[D>>\lambda]} \\ & \text { Since, }[D>>\lambda] \\ & \therefore \quad S_1 P=S_2 P \\ & \end{aligned}$
and neglecting higher power of $\lambda$
$\begin{aligned} & \therefore \quad 2 S_1 P\left(S_1 P-S_2 P\right)=4 D \lambda \\ & 2\left(S_1 P\right)(\lambda)=4 D \lambda \\ & \text { or } \quad S_1 P=2 D \\ & \sqrt{D^2+x^2}=2 D \\ & x^2=3 D^2 \\ & x=\sqrt{3} D \\ & \end{aligned}$
Alternative
$\begin{gathered}2 \lambda \cos \theta=\lambda \\ \Rightarrow \quad \cos \theta=\frac{1}{2} \Rightarrow \theta=60^{\circ}, \tan \theta=\frac{x}{D} \\ \sqrt{3}=\frac{x}{D} \Rightarrow x=\sqrt{3} D\end{gathered}$
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