Question

Figure shows two coherent microwave source $s_1$ and $s_2$ emitting waves of wavelength and separated by a distance $3\lambda$. For $\lambda \ll D$ and $y\ne 0$, the minimum value of $y$ for point $p$ to be an intensity maximum is $D\frac{\sqrt{m}}{n}$. Find $(m+n)$

Answer 1

$\cos \theta =\frac{BQ}{QA}$

$\cos \theta =\frac{\Delta x}{3\lambda }$

$\Delta x=3\lambda \cos \theta$

Path difference between two waves at point $p$,

$\Delta x=3\lambda \cos \theta$
For $n^{th}$ maxima the path difference should be $n\lambda$.

For minimum $y$, we have to take nearest maxima to $O$.

At centre of screen $O$,

Path difference $\Delta x=S_1-S_2=3\lambda$
$\Rightarrow$ Third maxima is formed

At centre, third maxima is formed.
So, nearest maxima to $O$ is second.
For, second maxima at $P$
$3\lambda \cos \theta =2\lambda \Rightarrow \cos \theta =\frac{2}{3}$

In $\Delta OAP$,

$AO=2,PA=3$

$\Rightarrow PO=\sqrt{5}$

$tan\theta =\frac{\sqrt{5}}{2}=\frac{y}{D}$

$y=\frac{D\sqrt{5}}{2}$

Comparing with
$\frac{D\sqrt{5}}{2}$

$m=5,n=2$

$m+n=7$

FINAL ANSWER: $7$