Interference of two waves - class-XI

For incoherent sources (random phases), the intensities add linearly. If there are 10 sources each with intensity I0, the total intensity is 10 * I0.

$\begin{array}{l} \lambda =\dfrac { v }{ n } =\dfrac { { 350 } }{ { 350 } } =1m=100cm \ Path, difference, , \Delta x=AP-BP=25cm \ Phase, difference, , \Delta \varphi =\dfrac { { 2\pi  } }{ \lambda  } \Delta x=\dfrac { { 2\pi  } }{ 1 } \times \left( { \dfrac { { 25 } }{ { 100 } }  } \right) =\dfrac { \pi  }{ 2 }  \ Amplitude, A=\sqrt { { { \left( { { a _{ 1 } } } \right)  }^{ 2 } }+{ { \left( { { a _{ 2 } } } \right)  }^{ 2 } } } =\sqrt { { { \left( { 0.3 } \right)  }^{ 2 } }+{ { \left( { 0.4 } \right)  }^{ 2 } } } =0.5mm \end{array}$

The beat frequency is the difference between the two frequencies. f1 = 100/2 = 50 Hz, f2 = 92/2 = 46 Hz. Beat frequency = 50 - 46 = 4 Hz.

The resultant amplitude of two waves with amplitude A and phase difference phi is sqrt(A^2 + A^2 + 2AAcos(phi)). For phi = pi/2, this is sqrt(2A^2) = A*sqrt(2). The frequency remains the same.

Let the intensity of the two waves be $I _1$  and  $I _2$

The resultant intensity of two interfering waves is proportional to the square of the resultant amplitude. Using phasor addition for waves with identical amplitude a and phase difference delta, the resultant amplitude is 2a cos(delta/2). Squaring this gives an intensity proportional to 4a^2 cos^2(delta/2).

Resultant amplitude is $A=\sqrt{A^2 _1+A^2 _2+2A _1A _2\cos(\phi)}$, where 

$\displaystyle \Delta x = 0.4m$
 $\displaystyle\Phi = k\Delta x = \frac{1\pi}{3} = \sqrt{5^2+5^2+2\times 5 cos  (2\pi/3)}$
$\displaystyle= 5.$

Given :   $I _1 = I _2  = I _o$

$y _1=A _0 \sin{\omega t}$, 

$\dfrac { { I } _{ max }-{ I } _{ min } }{ { I } _{ max }+{ I } _{ min } } =\dfrac { { \left( { a } _{ 1 }+{ a } _{ 2 } \right)  }^{ 2 }-{ \left( { a } _{ 1 }-{ a } _{ 2 } \right)  }^{ 2 } }{ { \left( { a } _{ 1 }+{ a } _{ 2 } \right)  }^{ 2 }+{ \left( { a } _{ 1 }-{ a } _{ 2 } \right)  }^{ 2 } }$
                                       [$\because { I } _{ max }={ \left( { a } _{ 1 }+{ a } _{ 2 } \right)  }^{ 2 },{ I } _{ min }={ \left( { a } _{ 1 }-{ a } _{ 2 } \right)  }^{ 2 }$  where $a =$ amplitude]
                $=\dfrac { 4{ a } _{ 1 }{ a } _{ 2 } }{ 2\left( { a } _{ 1 }^{ 2 }+{ a } _{ 2 }^{ 2 } \right)  } =\dfrac { 2{ a } _{ 1 }{ a } _{ 2 } }{ { a } _{ 1 }^{ 2 }+{ a } _{ 2 }^{ 2 } } $
Now, dividing the numerator and denominator by ${a} _{1}{a} _{2}$, we get
$\dfrac { { I } _{ max }-{ I } _{ min } }{ { I } _{ max }+{ I } _{ min } } =\dfrac { 2 }{ \left[ \dfrac { { a } _{ 1 } }{ { a } _{ 2 } } +\dfrac { { a } _{ 2 } }{ { a } _{ 1 } }  \right]  } =\dfrac { 2 }{ \left[ \sqrt { \alpha  } +\dfrac { 1 }{ \sqrt { \alpha  }  }  \right]  } =\dfrac { 2\sqrt { \alpha  }  }{ \left( \alpha +1 \right)  } $

Beat $=\delta _{1}-\delta _{2}$$=\dfrac{\omega _{1}}{2\pi}-\dfrac{\omega _{2}}{2\pi}$$=\dfrac{1026\pi}{2\pi}-\dfrac{1014\pi}{2\pi}$$=6$