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.

\[∆  S_1 PO\] and \[∆  S_2 PO\] are right-angled triangles.
So,

\[\left( S_1 P \right)^2  -  \left( S_2 P \right)^2 \] 

\[ = \left[ D^2 + x^2 \right] -  \left[ \left( D - 2\lambda \right)^2 + x^2 \right]^2 \] 

\[ = 4\lambda D + 4 \lambda^2  = 4\lambda D\] 

\[( \lambda^2 \text { is  small  and  can  be  neglected })\] 

\[ \Rightarrow \left( S_1 P + S_2 P \right)\left( S_1 P - S_2 P \right) = 4\lambda D\] 

\[ \Rightarrow \left( S_1 P - S_2 P \right) = \frac{4\lambda D}{\left( S_1 P + S_2 P \right)}\] 

\[ \Rightarrow    S_1 P -  S_2 P = \frac{4\lambda D}{2\sqrt{x^2 + D^2}}\]

For constructive interference, path difference = n \[\lambda\]

So,

\[\Rightarrow    S_1 P -  S_2 P = \frac{4\lambda D}{2\sqrt{x^2 + D^2}} = n\lambda\] 

\[ \Rightarrow   \frac{2D}{\sqrt{x^2 + D^2}} = n\] 

\[ \Rightarrow    n^2 ( x^2  +  D^2 ) = 4 D^2 \] 

\[ \Rightarrow    n^2  x^2  +  n^2  D^2  = 4 D^2 \] 

\[ \Rightarrow  n^2  x^2  = 4 D^2  -  n^2  D^2 \] 

\[ \Rightarrow  n^2  x^2  =  D^2 \left( 4 - n^2 \right)\] 

\[ \Rightarrow   x = \frac{D}{n}\sqrt{4 - n^2}\] 

\[\text { When  n = 1 },   \] 

\[x = \sqrt{3}D  (\text { 1st  order }) . \] 

\[\text { When  n } = 2, \] 

\[x = 0 \  (\text { 2nd  order }) . \]

So, when x = \[\sqrt{3}D\] , the intensity at P is equal to the intensity at O.