Question

Figure shown two coherent sources S1 and S2 which emit sound of wavelength λ in phase. The separation between the sources is 3λ. A circular wire of large radius is placed in such way that S1,S2 is at the centre of the wire. Find the angular positions θ on the wire for which constructive interference takes place.

Answer 1

Explanation:

Step 1:

Let the circle radius= R. P is a point on the wire when it creates an angle equal to the distance it connects with S1S2.

The interference at P will be positive if the path difference PS1-PS2 is an integral multiple of the wavelength (i.e. the phase difference at P between the two waves is 2nπ).

Here $\mathrm{S}_{1} \mathrm{S}_{2}=3$

$\mathrm{S}_{1} \mathrm{O}=\mathrm{S}_{2} \mathrm{O}=\frac{3}{2}$

Step 2:

Draw a circle perpendicular PQ

$\mathrm{PQ}=\mathrm{R} \sin \theta, \text { and } \mathrm{OQ}=\mathrm{R} \cos \theta$

Hence $\mathrm{S}_{2} \mathrm{Q}=\mathrm{OQ}-\mathrm{OS}_{2}=\mathrm{R} \cos \theta-\frac{3}{2^{\prime}}$

and  $\text { and } S_{1} Q=3+R \cos \theta-\frac{3}{2}=R \cos \theta+\frac{3}{2}$

Now $\mathrm{PS}_{1}^{2}=\mathrm{PQ}^{2}+\mathrm{S}_{1} \mathrm{Q}^{2}$

and  

$\mathrm{PS}_{2}^{2}=\mathrm{PQ}^{2}+\mathrm{S}_{2} \mathrm{Q}^{2}$

So, $\mathrm{PS}_{1}^{2}-\mathrm{PS}_{2}^{2}=\mathrm{S}_{1} \mathrm{Q}^{2}-\mathrm{S}_{2} \mathrm{Q}^{2}$

$\left(P S_{1}-P S_{2}\right)\left(P S_{1}+P S_{2}\right)=\left(R \cos \theta+\frac{3}{2}\right)^{2}-\left(R \cos \theta-\frac{3}{2}\right)^{2}$

$\left(P S_{1}-P S_{2}\right)(2 R)=6 R \cos \theta$  {Since R is large assuming $\mathrm{PS}_{1} \approx \mathrm{PS}_{2}$}  

$\mathrm{PS}_{1}-\mathrm{PS}_{2}=3 \cos \theta$

Step 3:

This should be equal to an integral multiple of the wavelength, for constructive interference.

Hence $3 \cos \theta=n$

$\cos \theta=\frac{n}{2}$  

$\text { For } n=0, \cos \theta=0$

$\theta=90^{\circ} \text { for } n=1, \cos \theta=\frac{1}{3}$

$\theta=70.5^{\circ} \text { For } n=2, \cos \theta=\frac{2}{3}=0.667$

$\theta=48.2^{\circ} \text { for } n=3, \cos \theta=1$

$\theta=0^{\circ}$

Therefore the positive interference in other quadrants will be = 0 °, 48.2 °, 70.5 °, 90 ° and similar points.