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 is

• A. $I_1+I_2$
• B. ${\left(\sqrt{I_1}+\sqrt{I_2}\right)}^2$
• C. ${\left(\sqrt{I_1}-\sqrt{I_2}\right)}^2$
• D. $2\left(I{}_1+I_2\right)$

Answer 1

The correct option is D

$2\left(I{}_1+I_2\right)$

Step 1: Given

Intensity of the first wave is $I_1$

Intensity of the second wave is $I_2$

Phase difference between waves is $\varphi$

Step 2: Formula Used

If two periodic waves of intensities $I_1$ and $I_2$ pass through a region at the same time in the same direction, then the resultant intensity of the superposed wave is given by the mathematical expression,

$I=I_1+I_2+2\sqrt{I_1{I}_2}\cos \varphi$

For the intensity to be maximum,

$\varphi =2n\lambda$, where $\lambda$ is wavelength and $n=0,1,2,3,\dots$

For the intensity to be minimum,

$\varphi =\left(2n-1\right)\lambda$, where $\lambda$ is wavelength and $n=0,1,2,3,\dots$

Therefore, zero-order maxima are obtained for $\cos \varphi =1$

Step 3: Solution

The maximum intensity of the superposed wave will be,

$\begin{array}{rcl}{I}_{max}& =& I_1+I_2+2\sqrt{I_1{I}_2}\\ & =& {(\sqrt{I_1}+\sqrt{I_2})}^2\end{array}$

The minimum intensity of the superposed wave will be,

$\begin{array}{rcl}{I}_{min}& =& I_1+I_2-2\sqrt{I_1{I}_2}\\ & =& {(\sqrt{I_1}-\sqrt{I_2})}^2\end{array}$

The sum of the maximum and the minimum intensity of the resultant wave will be,
$$\begin{gathered} I=I_{max}+I_{min} \\ ={\left(\sqrt{I_1}+\sqrt{I_2}\right)}^2+{\left(\sqrt{I_1}-\sqrt{I_2}\right)}^2 \\ =2\left(I_1^{}+I_2^{}\right) \end{gathered}$$

Hence, the sum is $2\left(I_1+I_2\right)$ and option D is correct.