Question

Two coherent narrow slits emitting sound of wavelength $\lambda$ in the same phase are placed parallel to each other at a small separation of $2\lambda$. The sound is detected by moving a detector on the screen E at a distance $D(>>\lambda )$ from the slit S, as shown in figure (16-E6). Find the distance x such that the intensity at P is equal to the intensity at O.

Answer 1

Since $S_1,S_2$ are in same phase, at O there will be maximum intensity.

Given that there will be a maximum intensity at P.

$\Rightarrow$ Path difference $=\Delta x=nl.$

From the figure (in questions)

$(S_{1}P{)}^2-(S_{2}P{)}^2$

$=({\sqrt{D}}^2+x_2{)}^2-[(\sqrt{d}-2\lambda {)}^2+x^2{]}^2$

$=4\lambda D=4{\lambda }^2=4\lambda D$

$({\lambda }^2$ is so small and can neglected)

$\Rightarrow S_{1}P-S_{2}P=\frac{4\lambda D}{2\sqrt{x^2+D^2}}=n\lambda$

$\Rightarrow \frac{2D}{\sqrt{x^2+D^2}}=n$

$\Rightarrow n^2(x^2+D^2)=4D^2$

$\Rightarrow x=\frac{D}{n}\sqrt{4}-n^2$

When n =1, $x=\sqrt{3}$ D (1st order)

n=2,

x =0 (2nd order)

$\therefore$ When $x=\sqrt{3}$ D,at P there will be maximum intensity.