Two coherent narrow slits emitting sound of wavelength in the same phase are placed parallel to each other at a small separation of 2 . The sound is detected by moving a detector on the screen at a distance D(>> ) 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.

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): _1PO and _2PO are right-angled triangle So, (S_1P)^2-(S_2P)^2= [D^2+x^2 ]- [(D-2 )^2+x^2 ]^2 =4 +4 ^2=4 ( ^2 is small and can be neglected ) (S_1P+S_2P)(S_1P-S_2P)=4 (S_1P-S_2P)= 4 (S_1P+S_2P) _1P-S_2P= 4 2 x^2+D^2 For constructive interference, path difference =n^ So, S_1P-S_2P= 4 2 x^2+D^2 =n 2D x^2+D^2 =n ^2 (x^2+D^2 )=4D^2 ^2x^2+n^2D^2=4D^2 ^2x^2=4D^2-n^2D^2 ^2x^2=D^2(4-n^2) = D n 4-n^2 When n = 1,x=3D (1^ st order). When n = 2,x=0 (2^ nd order). So when x=3D, the intensity at P is equal to the intensity at O.

Two coherent narrow slits emitting sound of wavelength in the sam…

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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.

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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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