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  • Two harmonic waves of monochromatic light  and <img alt="y_{2}=a \cos \left (\omega t +\phi \right )" src="

Two harmonic waves of monochromatic light

y_{1}=a \cos \omega t and y_{2}=a \cos \left (\omega t +\phi \right )

are superimposed on each other. Show that maximum intensity in interference pattern is four times the intensity due to each silt. Hence write the conditions for constructive and destructive interference in terms of the phase angle \phi.

 

 
 
 
 
 

Answers (1)

y_{1}=a \cos \omega t , y_{2}=a \cos \left (\omega t +\phi \right )

Resultant displacement 

y = y_{1}+y_{2}

y =A\cos \omega t + A\left ( \cos \omega t + \phi \right )

y =2A\cos \left ( \omega t +\frac{\phi}{2} \right )\cos \frac{\phi}{2}

y =2A\cos \frac{\phi}{2} \cos \left ( \omega t +\frac{\phi}{2} \right )

The amplitude of resultant wave =

 y =2A\cos \frac{\phi}{2}

Intensity I\propto A^{2}

I = 4I_{0}\cos ^{2}\phi

for constructive interference  I is maximum

for \phi = 0,\pm 2\pi,\pm 4\pi; i.e., \phi = 2n\pi

I = 4I_{0}         (constructive interference)

At maxima I = 4I_{0}

For destructive interference I is minimum. i.e., I = 0

\phi = \pm \pi,\pm 3\pi.....

\phi = \left ( 2n\pm 1 \right )\frac{\pi}{2}

Posted by

Safeer PP

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