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You must have seen light coming out from the laser. Let us carry out a small activity. Take two needles and touch the needles on the surface of the water. Here if both the needles move with the same speed then they are said to be coherent. Let us learn more about coherent waves.
Suppose there is a surface of the water and you take a needle and touch the surface of the water. What will happen? Yes, ripples are formed. Now if you take two needles and you touch the surface of the water with the needles. What do you think will happen?
You will see a pattern. That pattern is the interference pattern. When you touch both the needles at the surface of the water at the same time, both the needles are in the same phase. Needle 1 will produce a wave. Also, needle 2 will produce its own ripples and they will intersect with waves of the first needle.
Now, if both the needles are moving with the same velocity, the wave formed here are coherent. If the velocity of a 1st needle and 2nd needle are not steady they won’t intersect. This is because one is at a steady speed and other is at variable speed.
If the potential difference between two waves is zero or is constant w.r.t time, then the two ways are said to be coherent.
The waves are noncoherent if the potential difference between the two ways keeps on changing. Lightbulb, study lamp are the examples of the coherent waves. They emit waves at random potential difference.
Explanation
Now let us consider there are two needles say S_{1} and S_{2} moving up and down on the surface of the water and are pointing at point P. So the path difference here is given as S_{1}P – S_{2}P. Now the displacement by two needles and S_{1} S_{2} are:
y_{1} = A cos wt ……………… (1)
y_{2} = A cos wt …………….. (2)
So the resultant displacement at point P is, y = y1 + y2. When we substitute the value of y_{1} and y_{2} we write,
y = A cos wt + A cos wt
y = 2A cos wt……………….. (3)
Now, we know the intensity is proportional to the square of the amplitude waves.
I_{0} ∝ A²
Where I0 is the initial intensity and A² is the amplitude of the wave. From equation 3, we say that A = 2A. So,
I_{0} ∝ (2A)² or I0 ∝ 4 A²
I = 4 I0
Now, if two needles that are S_{1 }are S_{2} are in the same phase, the potential difference is,
S_{1}P – S_{2}P = nλ
Where n = 0, 1, 2,3 ……… and λ = the wavelength of the wave. If the two needles S_{1} and S_{2} are vibrating at its destructive interference then, the potential difference is
S_{1}P – S_{2}P = (n + 1/2) λ
Now if the potential difference of the waves is Φ then,
y_{1} = α cos wt
y_{2} = α cos wt
The individual intensity of each wave is I0 , we get,
y = y_{1} + y_{2}
= α cos wt + α cos (wt +Φ)
y = 2 α cos(Φ/2) cos (wt + Φ/2)
Since, the intensity is I0 ∝ A²
I_{0} ∝ 4α² cos² (Φ/2)
I = 4 I_{0} cos² (Φ/2)
Well, the timeaveraged value of cos²(Φ_{t}/2) is 1/2. So, the resultant intensity will be I = 2 I0 at all the points.
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