Question

A source S and a detector D are placed at a distance d apart. A big cardboard is placed at a distance $\sqrt{2}d$ from the source and the detector as shown in figure. The source emits a wave of wavelength = d/2 which is received by the detector after reflection from the cardboard. It is found to be in phase with the direct wave received from the source. By what minimum distance should the cardboard be shifted away so that the reflected wave becomes out of phase with the direct wave?

Answer 1

Given:
Distance between the source and detector = d
Distance of cardboard from the source = $\sqrt{2d}$
Wavelength of the source $\lambda$ = d/2
Path difference between sound waves received by the detector before shifting the cardboard:
$$\begin{gathered} 2\left(\sqrt{{\left(\frac{d}{2}\right)}^2+{\left(\sqrt{2}d\right)}^2}\right)-d \\ \Rightarrow 2\times \frac{3d}{2}-d \\ \Rightarrow 2d \end{gathered}$$
If the cardboard is shifted by a distance x, the path difference will be:

$2\left(\sqrt{{\left(\frac{d}{2}\right)}^2+{\left(\sqrt{2d}+x\right)}^2}\right)-d$
According to the question,

$$\begin{gathered} 2\sqrt{{\left(\frac{d}{2}\right)}^2+{\left(\sqrt{2}d+\mathrm{x}\right)}^2}-\mathrm{d}=2d+\frac{d}{4} \\ \Rightarrow 2\sqrt{{\left(\frac{d}{2}\right)}^2+{\left(\sqrt{2}d+x\right)}^2}-d=\frac{9d}{4} \\ \Rightarrow 2\sqrt{{\left(\frac{d}{2}\right)}^2+{\left(\sqrt{2}d+x\right)}^2}=\frac{9d}{4}+d=\frac{13d}{4} \\ \Rightarrow {\left(\frac{d}{2}\right)}^2+{\left(\sqrt{2}d+x\right)}^2=\frac{169}{64}{d}^2 \\ \Rightarrow {\left(\sqrt{2}d+x\right)}^2=\frac{(169-16)}{64}{d}^2=\frac{153}{64}{d}^2 \\ \Rightarrow \sqrt{2}d+x=1.54d \\ \Rightarrow x=\left(1.54-1.41\right)d=0.13d \end{gathered}$$