Question

Two coherent narrow slits emitting sound of wavelength λ in the same phase are placed parallel to each other at a small separation of 2λ. The sound is detected by moving a detector on the screen ∑ at a distance D(>>λ) from the slit S₁ as shown in figure. Find the distance x such that the intensity at P is equal to the intensity at O.

Answer 1

Given:
S_1^& S₂ are in the same phase. At O, there will be maximum intensity.
There will be maximum intensity at P.
From the figure (in questions):

$∆S_1\mathrm{PO}$and $∆S_2\mathrm{PO}$ are right-angled triangles.
So,
$$\begin{gathered} {\left(S_1\mathrm{P}\right)}^2-{\left(S_2\mathrm{P}\right)}^2 \\ =\left[D^2+x^2\right]-{\left[{\left(D-2\lambda \right)}^2+x^2\right]}^2 \\ =4\lambda \mathrm{D}+4{\lambda }^2=4\lambda \mathrm{D} \\ ({\mathrm{\lambda }}^2\mathrm{is}\mathrm{small}\mathrm{and}\mathrm{can}be\mathrm{neglected}) \\ \Rightarrow \left(S_{1}P+S_{2}P\right)\left(S_{1}P-S_{2}P\right)=4\lambda D \\ \Rightarrow \left(S_{1}P-S_{2}P\right)=\frac{4\lambda D}{\left(S_{1}P+S_{2}P\right)} \\ \Rightarrow S_1\mathrm{P}-S_2\mathrm{P}=\frac{4\lambda \mathrm{D}}{2\sqrt{x^2+{\mathrm{D}}^2}} \end{gathered}$$
For constructive interference, path difference = n^$\lambda$.
So,
$$\begin{gathered} \Rightarrow S_1\mathrm{P}-S_2\mathrm{P}=\frac{4\lambda \mathrm{D}}{2\sqrt{x^2+{\mathrm{D}}^2}}=n\lambda \\ \Rightarrow \frac{2\mathrm{D}}{\sqrt{x^2+{\mathrm{D}}^2}}=n \\ \Rightarrow n^2(x^2+{\mathrm{D}}^2)=4{\mathrm{D}}^2 \\ \Rightarrow {\mathrm{n}}^2{\mathrm{x}}^2+{\mathrm{n}}^2{\mathrm{D}}^2=4{\mathrm{D}}^2 \\ \Rightarrow {\mathrm{n}}^2{\mathrm{x}}^2=4{\mathrm{D}}^2-{\mathrm{n}}^2{\mathrm{D}}^2 \\ \Rightarrow {\mathrm{n}}^2{\mathrm{x}}^2={\mathrm{D}}^2\left(4-{\mathrm{n}}^2\right) \\ \Rightarrow x=\frac{\mathrm{D}}{n}\sqrt{4-n^2} \\ \mathrm{When}n=1, \\ x=\sqrt{3}\mathrm{D}(1\mathrm{st}\mathrm{order}). \\ \mathrm{When}n=2, \\ x=0(2\mathrm{nd}\mathrm{order}). \end{gathered}$$
So, when x = $\sqrt{3}D$ , the intensity at P is equal to the intensity at O.