Question

Two coherent point sources S₁ and S_2, vibrating in phase, emit light of wavelength $\lambda$. The separation between the sources is $2\lambda$. Consider a line passing through S₂ and perpendicular to the line S₁ S₂. What is the smallest distance from S₂ where a minimum intensity occurs?

Answer 1

Let P be the point of minimum intensity.
For minimum intensity at point P,

$S_1\mathrm{P}-S_2\mathrm{P}=x=\left(2n+1\right)\frac{\lambda }{2}$

Thus, we get:

$$\begin{gathered} \sqrt{{\mathrm{Z}}^2+{\left(2\lambda \right)}^2}-\mathrm{Z}=\left(2n+1\right)\frac{\lambda }{2} \\ \Rightarrow {\mathrm{Z}}^2+4{\lambda }^2={\mathrm{Z}}^2{\left(2n+1\right)}^2\frac{{\lambda }^2}{4}+2\mathrm{Z}\left(2n+1\right)\frac{\lambda }{2} \\ \Rightarrow \mathrm{Z}=\frac{4{\lambda }^2-\left(2n+1\right)2{\lambda }^2/4}{\left(2n+1\right)\lambda } \\ =\frac{16{\lambda }^2-\left(2n+1\right)}{4\left(2n+1\right)\lambda }...\left(i\right) \end{gathered}$$

$$\begin{gathered} \text{When}n=0,\text{Z}=\frac{15\lambda }{4} \\ n=-1,\mathrm{Z}=\frac{-15\lambda }{4} \\ n=1,\mathrm{Z}=\frac{7\lambda }{12} \\ n=2,\text{Z}=\frac{-9\lambda }{20} \end{gathered}$$

Thus, $\mathrm{Z}=\frac{7\lambda }{12}$ is the smallest distance for which there will be minimum intensity.