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Sound Waves — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsSound Waves3 Marks
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 $\sum$ at a distance $\text{D}(>>\lambda)$ from the slit $S_1$ as shown in figure. Find the distance $x$ such that the intensity at $P$ is equal to the intensity at $0.$
✓

Answer

Given: $S_1 S_2$ are in the same phase. At $O,$ there will be maximum intensity.
There will be maximum intensity at $P$.
From the $($in questions$)$ : $\triangle\text{S}_1\text{PO}$ and $\triangle\text{S}_2\text{PO}$ are right $-$ angled triangle
So, $(\text{S}_1\text{P})^2-(\text{S}_2\text{P})^2$$=\big[\text{D}^2+\text{x}^2\big]-\Big[(\text{D}-2\lambda)^2+\text{x}^2\Big]^2$
$=4\lambda\text{D}+4\lambda^2=4\lambda\text{D}$
$\Big(\lambda^2$ is small and can be neglected$\Big)$
$\Rightarrow(\text{S}_1\text{P}+\text{S}_2\text{P})(\text{S}_1\text{P}-\text{S}_2\text{P})=4\lambda\text{D}$
$\Rightarrow(\text{S}_1\text{P}-\text{S}_2\text{P})=\frac{4\lambda\text{D}}{(\text{S}_1\text{P}+\text{S}_2\text{P})}$
$\Rightarrow\text{S}_1\text{P}-\text{S}_2\text{P}=\frac{4\lambda\text{D}}{2\sqrt{\text{x}^2+\text{D}^2}}$
For constructive interference, path difference $=\text{n}^\lambda$
So, $\text{S}_1\text{P}-\text{S}_2\text{P}=\frac{4\lambda\text{D}}{2\sqrt{\text{x}^2+\text{D}^2}}=\text{n}\lambda$
$\Rightarrow\frac{2\text{D}}{\sqrt{\text{x}^2+\text{D}^2}}=\text{n}$
$\Rightarrow\text{n}^2\big(\text{x}^2+\text{D}^2\big)=4\text{D}^2$
$\Rightarrow\text{n}^2\text{x}^2+\text{n}^2\text{D}^2=4\text{D}^2$
$\Rightarrow\text{n}^2\text{x}^2=4\text{D}^2-\text{n}^2\text{D}^2$
$\Rightarrow\text{n}^2\text{x}^2=\text{D}^2(4-\text{n}^2)$
$\Rightarrow\text{x}=\frac{\text{D}}{\text{n}}\sqrt{4-\text{n}^2}$
When $n = 1,\text{x}=\sqrt{3}\text{D} \ (1^{st} $ order$)$.
When $ n = 2,\text{x}=0 \ (2^{nd}$ order$).$
So when $\text{x}=\sqrt{3}\text{D},$ the intensity at $P$  is equal to the intensity at $O$.

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