Question

Two coherent point sources ${\text{S}}_1\&{\text{S}}_2$ vibrating in phase emits a light of wavelength $\lambda .$ The separation between the sources is $2\lambda .$ The smallest distance from ${\text{S}}_2$ on a line passing through ${\text{S}}_2$ and perpendicular to ${\text{S}}_1{\text{S}}_2$ where a minimum intensity occurs is

• A. $\frac{7\lambda }{12}$
• B. $\frac{15\lambda }{4}$
• C. $\frac{\lambda }{2}$
• D. $\frac{3\lambda }{4}$

Answer 1

The correct option is B $\frac{15\lambda }{4}$

For minima at a point, the path difference is,

$\Delta x=\left(\frac{2n+1}{2}\right)\lambda$

For $n=0,\Delta x=\frac{\lambda }{2}$



For the given situation, from the figure we get,

$\Delta x=S_{1}P-S_{2}P=\sqrt{4{\lambda }^2+x^2}-x$

$\frac{\lambda }{2}=\sqrt{4{\lambda }^2+x^2}-x$

$\sqrt{4{\lambda }^2+x^2}=\frac{\lambda }{2}+x$

On squaring both sides,

$\Rightarrow 4{\lambda }^2+x^2=\frac{{\lambda }^2}{4}+x\lambda +x^2$

$\Rightarrow \frac{15{\lambda }^2}{4}=x\lambda \Rightarrow x=\frac{15\lambda }{4}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.

Why this question ? To understand the relation between minima and maxima and the path difference.