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# The intensity of two waves is 2 and 3 unit, and then the average intensity of light in the overlapping region will have the value:A). 2.5B). 6C). 5D). 13

Last updated date: 20th Jun 2024
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Hint: In this question we use the expression of maximum intensity and minimum intensity and get two different values. Further, we use the relation between average intensity, maximum intensity, and minimum intensity to get the desired result. We also discuss the basics of intensity and wave propagation.
Formula used:
${I_{\max /\min }} = {I_1} + {I_2} \pm 2\sqrt {{I_1}{I_2}}$
${I_{avg}} = \dfrac{{{I_{\max }} + {I_{\min }}}}{2}$

Complete step by step solution:
As we know that, the intensity of radiant energy is defined as the power transferred per unit area. Here the area is measured on the plane perpendicular to the direction of propagation of the energy or wave. Further, intensity can be calculated as the square of the given amplitude.
In this question, the expression for maximum intensity and minimum intensity is given as:
\eqalign{ & {I_{\max }} = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} ..........(1) \cr & \Rightarrow {I_{\max }} = 2 + 3 + 2\sqrt {2 \times 3} = 5 + 2\sqrt 6 \cr}
\eqalign{ & \because {I_{\min }} = {I_1} + {I_2} - 2\sqrt {{I_1}{I_2}} ..........(2) \cr & \Rightarrow {I_{\max }} = 2 + 3 - 2\sqrt {2 \times 3} = 5 - 2\sqrt 6 \cr}
Now, the average intensity is given by:
${I_{avg}} = \dfrac{{{I_{\max }} + {I_{\min }}}}{2}$
Now, by substituting the values of equation (1) and (2) in above equation, we get:
\eqalign{ & \Rightarrow {I_{avg}} = \dfrac{{5 + 2\sqrt 6 + 5 - 2\sqrt 6 }}{2} \cr & \therefore {I_{avg}} = 5 \cr}
Therefore, we get the required result and the correct option is C) i.e., the average intensity of the two waves is 5.