Question

Consider two coherent monochromatic sources $S_1$ and $S_2$, each of wavelength $\lambda$ and separated by a distance $d$. The ratio of the intensity of $S_1$ and $S_2$ is $4$. A detector moves on the line perpendicular to $S_1{S}_2$ as shown in the figure. If the resultant intensity at point $P$ is equal to $\frac{9}{4}$ times intensity of $S_1$, then the distance of $P$ from $S_1$ is,
(Given : $d>0$ and $n$ is a positive integer)

• A. $\frac{d^2-n^2{\lambda }^2}{2n\lambda }$
  
• B. $\frac{d^2+n^2{\lambda }^2}{2n\lambda }$
  
• C. $\frac{n\lambda d}{\sqrt{d^2-n^2{\lambda }^2}}$
  
• D. $\frac{2n\lambda d}{\sqrt{d^2-n^2{\lambda }^2}}$
  

Answer 1

The correct option is A $\frac{d^2-n^2{\lambda }^2}{2n\lambda }$


Let the intensity of source $S_1$ and $S_2$ is $I_1$ and $I_2$ respectively.
Let resultant intensity at $P$ is $I_R$
$\Rightarrow I_R=\frac{9}{4}{I}_1$
and $\frac{I_1}{I_2}=4$ (Given)

The resultant intensity is given by,
$I_R=I_1+I_2+2\sqrt{I_1{I}_2}\cos \varphi$,
$(\varphi$=phase difference)
$\Rightarrow \frac{9}{4}{I}_1=I_1+\frac{I_1}{4}+2\sqrt{I_1\times \frac{I_1}{4}}\cos \varphi$

$\Rightarrow \frac{9}{4}{I}_1=\frac{5I_1}{4}+I_1\cos \varphi$

$\Rightarrow \frac{9}{4}-\frac{5}{4}=\cos \varphi$

$\Rightarrow \cos \varphi =1$
$\therefore \varphi =2n\pi$
Where $n=0,1,2,.....$

Therefore, at point $P$, constructive interference takes place.

Let the path difference at $P$ is $\Delta x$.

The relation between path and phase difference is,
$\varphi =2n\pi =\frac{2\pi }{\lambda }\times \Delta x$

Where, $\lambda =$wavelength.
Now, for constructive interference,
$\Delta x=n\lambda ...\left(1\right)$


Also, from the diagram, $\Delta x=S_{2}P-S_{1}P$
From right angled triangle,
$\Delta x=\sqrt{d^2+x^2}-x$
$\Rightarrow \Delta x+x=\sqrt{d^2+x^2}$
$(\Delta x+x{)}^2=d^2+x^2$
$\Delta x^2+x^2+2x\Delta x=d^2+x^2$
$\Delta x^2+2x\Delta x=d^2$

Substituting $\Delta x$ from equation $\left(1\right)$,
$\Rightarrow (n\lambda {)}^2+2n\lambda x=d^2$

$\Rightarrow x=\frac{d^2-n^2{\lambda }^2}{2n\lambda }$

Where $n=0,1,2,3,...$

Hence, $\left(A\right)$ is the correct answer.