
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). 2.5
B). 6
C). 5
D). 13
Answer
457.5k+ views
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.
Additional information:
As we know, a wave involves the transfer of energy without the transfer of the matter. So, we can say that a wave can be defined as a disturbance that travels through a medium, transporting energy from one point to another point without transfer of matter.
Further, the frequency is defined as the number of waves that pass a fixed point in unit time. The S.I unit of frequency is Hertz or we can represent by Hz. The unit of wavelength is meters or we can represent by m. Furthermore, we also know that the S.I unit of time is given by second or we can represent by s.
Also, we know that two waves are said to be coherent if they are moving with the same frequency and have constant phase difference.
When the two or more waves are propagating, the summation or adding or subtraction of all these waves travelling in a particular medium, will give us the superposition of waves. If the direction or amplitude of these waves are opposite then the superposition of these waves are calculated by subtracting the waves. If these two or more waves are travelling in the same direction or they have the same amplitude the resultant is given by adding up the two or more waves.
Also, the phase of a wave gives us the location of a point within a wave cycle of a repetitive waveform.
Note: We should remember that, depending on the direction of propagation. Also, a sine wave starts from zero point, whereas the cosine wave starts from one. A wave which has the same amplitude but opposite orientation will cancel out each other and thereby give zero output.
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.
Additional information:
As we know, a wave involves the transfer of energy without the transfer of the matter. So, we can say that a wave can be defined as a disturbance that travels through a medium, transporting energy from one point to another point without transfer of matter.
Further, the frequency is defined as the number of waves that pass a fixed point in unit time. The S.I unit of frequency is Hertz or we can represent by Hz. The unit of wavelength is meters or we can represent by m. Furthermore, we also know that the S.I unit of time is given by second or we can represent by s.
Also, we know that two waves are said to be coherent if they are moving with the same frequency and have constant phase difference.
When the two or more waves are propagating, the summation or adding or subtraction of all these waves travelling in a particular medium, will give us the superposition of waves. If the direction or amplitude of these waves are opposite then the superposition of these waves are calculated by subtracting the waves. If these two or more waves are travelling in the same direction or they have the same amplitude the resultant is given by adding up the two or more waves.
Also, the phase of a wave gives us the location of a point within a wave cycle of a repetitive waveform.
Note: We should remember that, depending on the direction of propagation. Also, a sine wave starts from zero point, whereas the cosine wave starts from one. A wave which has the same amplitude but opposite orientation will cancel out each other and thereby give zero output.
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