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Question: \[100\pi \] phase difference\(=\)_____ path difference \(\begin{aligned} & \text{A}\text{. 10}...

100π100\pi phase difference==_____ path difference
A. 10λ B. 25λ C. 50λ D. 100λ \begin{aligned} & \text{A}\text{. 10}\lambda \\\ & \text{B}\text{. 25}\lambda \\\ & \text{C}\text{. 50}\lambda \\\ & \text{D}\text{. 100}\lambda \\\ \end{aligned}

Explanation

Solution

A wave is represented by sinusoidal form. So at the time of measurement I.e. t=0t=0 it does not pass through the origin and said to have a phase difference or phase shift. The phase difference or phase shift is the angle ϕ\phi in degrees or radians that the waveform has shifted from a certain reference point along the horizontal zero axis. Path difference is the difference between the path travelled of two waves at a certain point.

Formula used:
Relationship between the phase difference(Δϕ)\left( \Delta \phi \right) , path difference(Δx)\left( \Delta x \right) and the wavelength (λ)\left( \lambda \right)of a wave is given by Δϕ=2πλ×Δx\Delta \phi =\dfrac{2\pi }{\lambda }\times \Delta x

Complete answer:
Here the phase difference given is 100π100\pi i.e. Δϕ=100π\Delta \phi =100\pi
But the relationship between the phase difference(Δϕ)\left( \Delta \phi \right) , path difference(Δx)\left( \Delta x \right) and the wavelength (λ)\left( \lambda \right)of a wave is given by
Δϕ=2πλ×Δx, so  100π=2πλ×Δx Δx=50λ \begin{aligned} & \Delta \phi =\dfrac{2\pi }{\lambda }\times \Delta x,\text{ so } \\\ & \Rightarrow 100\pi =\dfrac{2\pi }{\lambda }\times \Delta x \\\ & \Rightarrow \Delta x=50\lambda \\\ \end{aligned}

So, the correct answer is “Option C”.

Additional Information:
The equation of wave motion is given by
y(x,t)=Asin(ωtkx+ϕ0)y\left( x,t \right)=A\sin \left( \omega t-kx+{{\phi }_{0}} \right)
where
y(x,t)= displacement A= amplitude ω= angular frequency t=time k=angular wave number x=position ϕ0= initial phase \begin{aligned} & y\left( x,t \right)=\text{ displacement} \\\ & A=\text{ amplitude} \\\ & \omega =\text{ angular frequency} \\\ & t=\text{time} \\\ & k=\text{angular wave number} \\\ & x=\text{position} \\\ & {{\phi }_{0}}=\text{ initial phase} \\\ \end{aligned}
Phase: The phase of a harmonic wave is a quantity that gives complete information of the wave at any time and any position.
If a wave is represented by y(x,t)=Asin(ωtkx+ϕ0)y\left( x,t \right)=A\sin \left( \omega t-kx+{{\phi }_{0}} \right)
Then phase of the wave at position xx and timett is given by. ϕ=(ωtkx+ϕ0)\phi =\left( \omega t-kx+{{\phi }_{0}} \right)
So clearly the phase of a wave is periodic both in time and space. So at a given point xxthe phase changes with time and at a given time tt the phase changes with position xx
Now, phase ,ϕ=(ωtkx+ϕ0)\phi =\left( \omega t-kx+{{\phi }_{0}} \right)
Taking xxas a constant differentiate ϕ\phi with respect to tt then,
ΔϕΔt=ω\dfrac{\Delta \phi }{\Delta t}=\omega
Thus the phase change at a given position xx in time Δt\Delta t is given by
Δϕ=ωΔt=2πTΔt\Delta \phi =\omega \Delta t=\dfrac{2\pi }{T}\Delta t
Where T= Time period of the wave.
Taking time as constant
ΔϕΔx=k Δϕ=kΔx=2πλΔx \begin{aligned} & \dfrac{\Delta \phi }{\Delta x}=-k \\\ & \Rightarrow \Delta \phi =-k\Delta x=-\dfrac{2\pi }{\lambda }\Delta x \\\ \end{aligned}

Note:
Note that from the equation of wave motion you can calculate the particle velocity particle acceleration, particle energy. Generally waves are of three main types such as.
Mechanical wave-it is the wave which obeys Newton’s law and can only exist within material medium e.g. water wave, sound wave, seismic wave.
Electromagnetic wave- wave associated with a charged objects.e.g x-ray.
Matter waves-waves associated with fundamental particles like electron,proton and neutron.