Question

If the intensity of two waves is $2$ and $3$ units, then average intensity of light in the overlapping region will be,

• A. $2.5\text{units}$
• B. $6\text{units}$
• C. $5\text{units}$
• D. $13\text{units}$

Answer 1

The correct option is C $5\text{units}$

The resultant intensity of two waves having intensity $I_1$ and $I_2$ is given by,
$I_{res}=I_1+I_2+2\sqrt{I_1{I}_2}\cos \varphi$

The maximum intensity occurs, when $\cos \varphi =1$
$\Rightarrow I_{max}=I_1+I_2+2\sqrt{I_1{I}_2}$
$\Rightarrow I_{max}=2+3+2\sqrt{2\times 3}=5+2\sqrt{6}$

The minimum intensity occurs, when $\cos \varphi =-1$
$\Rightarrow I_{min}=I_1+I_2-2\sqrt{I_1{I}_2}$
$I_{min}=2+3-2\sqrt{2\times 3}=5-2\sqrt{6}$

The average intensity is given by,
$I_{avg}=\frac{I_{max}+I_{min}}{2}$
$I_{avg}=\frac{5+2\sqrt{6}+5-2\sqrt{6}}{2}$
$\Rightarrow I_{avg}=5\text{units}$

Hence, $\left(C\right)$ is the correct answer.

$$\begin{array}{c}\text{Why this question ?}\\ \text{Tip: The resultant intensity due to superposition of two coherent waves is given by,}{I}_R\text{=}{I}_1+I_2+2\sqrt{I_1{I}_2}\cos \varphi \end{array}$$