Question

Two periodic waves of intensities ${{I}_{1}}$ and ${{I}_{2}}$ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities possible is:
$\begin{align}
  & \text{A}\text{. }{{I}_{1}}+{{I}_{2}} \\
 & \text{B}\text{. }{{\left( \sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}} \right)}^{2}} \\
 & \text{C}\text{. }{{\left( \sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}} \right)}^{2}} \\
 & \text{D}\text{. 2}\left( {{I}_{1}}+{{I}_{2}} \right) \\
\end{align}$

Answer 1

Hint: When we have two waves in a medium at the same time moving in the same direction, we get the superposition of the waves. The properties of the superposed wave is different from that of the component waves. Obtain the relation between the intensity of the superposed wave and the intensity of the components of the wave.

Complete step by step answer:
If we superimpose two periodic waves moving through a medium, we cannot directly add the intensities of the waves to get the resultant intensity. If two periodic waves of intensities ${{I}_{1}}$ and ${{I}_{2}}$ pass through a region at the same time in the same direction, the resultant intensity of the superposed wave is given by the mathematical expression,

$I={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}+{{I}_{2}}}\cos \delta $

Where, $\delta $ is the phase difference between the two waves.
For the intensity to be maximum, the phase difference should be even integer multiple of the wavelength i.e.
$\delta =2n\lambda $
Where, the value of n can be 0,1,2,3,…..
Again, for the intensity of the superimposed wave to be minimum, the phase difference should be odd integer multiple of the wavelength.
$\delta =\left( 2n-1 \right)\lambda $
Where, the value of n can be 0,1,2,3,…..
So, we can say that the zero order maxima is obtained for, $\cos \delta =1$
So, the maximum intensity of the superposed wave will be,

$\begin{align}
  & {{I}_{\text{max}}}={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}+{{I}_{2}}} \\
 & {{I}_{\text{max}}}={{\left( \sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}} \right)}^{2}} \\
\end{align}$

Again, the minimum intensity of the superposed wave will be,

 $\begin{align}
  & {{I}_{\text{min}}}={{I}_{1}}+{{I}_{2}}-2\sqrt{{{I}_{1}}+{{I}_{2}}} \\
 & {{I}_{\text{min}}}={{\left( \sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}} \right)}^{2}} \\
\end{align}$
The addition of the maximum and the minimum intensity of the resultant wave will be,

$\begin{align}
  & I={{I}_{\max }}+{{I}_{\min }} \\
 & I={{\left( \sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}} \right)}^{2}}+{{\left( \sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}} \right)}^{2}} \\
 & I=2\left( {{I}_{1}}+{{I}_{2}} \right) \\
\end{align}$
Hence, the correct answer is option D.

Note:
When we superimpose two or more waves, at some points we will get maximum intensity and at some points we will get minimum intensity. This phenomenon is called the interference of light. If the two waves are in the same phase with each other, the maximum intensity is the addition of the individual intensities of the two waves. This is constructive interference. Again, if the two waves are out of phase, the minimum intensity of the superimpose wave is the difference between the intensity of the individual waves. This is the destructive interference.