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Chapter 37: Problem 29

Consider \(N\) coherent sources described as follows: \(E_{1}=\) \(E_{0} \sin (\omega t+\phi), E_{2}=E_{0} \sin (\omega t+2 \phi), E_{3}=E_{0} \sin (\omega t+3 \phi)\) \(, E_{N}=E_{0} \sin (\omega t+N \phi) .\) Find the minimum value of \(\phi\) for which \(E_{R}=E_{1}+E_{2}+E_{3}+\cdots+E_{N}\) is zero.

Short Answer

Expert verified
The minimum value of \(\phi\) for which \(E_{R}\) is zero is \(\phi = \frac{2\pi}{N}\).

Step by step solution

01

Expression of General Term

Let's start by expressing the \(nth\) term in the series in sinusoidal form. It can be given as: \(E_{n}=E_{0} \sin (\omega t+n \phi)\)
02

Expression for Resultant Displacement

The resultant displacement can be written as the sum of all the individual terms. So, we will apply the superposition principle here. \(E_{R}=E_{1}+E_{2}+\ldots+E_{N}\) which equals to the sum of all the individual terms. Writing these with values, it becomes: \(E_R = E_{0} [ \sin(\omega t) + \sin(\omega t + \phi) + \sin(\omega t +2\phi) + \ldots + \sin(\omega t+ (N-1)\phi) ]\)
03

Application of the Trigonometric Identity

We can simplify the above expression by using the trigonometric identity that sin(x+y) + sin(x-y) = 2sinx cosy. Adding the terms in pairs, and using the identity, the expression for \(E_R\) can be simplified further.
04

Simplified Resultant Displacement

Using the appropriate trigonometric identities, the expression for \(E_R\) will become : \(E_R = [2E_{0} \sin(\omega t + (N-1) \phi /2 ) \sin(N\phi /2 )]\). Since we are looking for the condition when the resultant displacement becomes zero, set \(E_R = 0\).
05

Final Calculation

Setting \(E_R = 0\), we get two conditions: 1. \(\sin(\omega t + (N-1) \phi /2 ) = 0\) which suggests \(\phi = \frac{2m\pi}{N-1}\) and 2. \(\sin(N\phi /2 ) = 0\) which suggests \(\phi = \frac{2m\pi}{N}\), where \(m\) is an integer. The minimum non-zero positive solution which satisfies both conditions will be \(\phi = \frac{2\pi}{N}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coherent Sources Interference
When we talk about coherent sources interference, we refer to the phenomenon that occurs when two or more wave sources of the same frequency and constant phase difference interact with each other. This interaction causes the waves to add up (constructively) or cancel out (destructively) at different points in space.

Imagine two stones thrown into a pond at different locations but creating ripples of the same wavelength. Where the crests or troughs of these ripples meet, they reinforce each other, leading to higher waves. This is constructive interference. On the contrary, where a crest meets a trough, they cancel out, creating a flat surface, known as destructive interference. In the exercise, we look for the minimum phase difference, equal to equals (phi), that would cause complete destructive interference, meaning the waves perfectly cancel out, leaving no resultant displacement.
Resultant Displacement in Waves
The concept of resultant displacement in waves is key to understanding interference patterns. Displacement in this context refers to the distance a point on the wave has moved from its rest position. The superposition principle in physics states that the resultant displacement of multiple waves occupying the same space is the vector sum of their individual displacements.

In simpler terms, if several waves meet, the displacement at any point is the sum of the displacements due to each wave at that point. If they are in phase, they add up to a greater amplitude; if they are out of phase, they can reduce or even cancel out each other entirely, resulting in a resultant displacement of zero which is the case in the exercise's solution that we analyzed to determine the required phase difference for complete destructive interference.
Trigonometric Identities in Physics
In physics, trigonometric identities are not just for solving triangle problems; they are crucial for wave motion analysis and other phenomena. The identity used in the solution, equal to equals equals equals equals equals equals equals equals equals (x+y) + equal to equals equals equals equals equals equals equals equals equals equals (x-y)) = 2equal to equals equals equals equals equals equals equals equals equals equals (x) equal to equals equals equals equals equals equals equals equals equals equals equals (y), is a powerful tool in this context. It helps us simplify the sum of sinusoidal functions, which would be complex otherwise.

This identity transforms the sum of waves in the problem into a more manageable form that helps reveal the points of constructive and destructive interference, such as determining at what phase difference (equal to equals (phi)) the resultant displacement of a set of coherent waves would be zero. These identities are not mere mathematical quirks but the backbone of solving complex wave interference problems in physics.

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Most popular questions from this chapter

Monochromatic light is beamed into a Michelson interferometer. The movable mirror is displaced \(0.382 \mathrm{mm}\) causing the interferometer pattern to reproduce itself 1700 times. Determine the wavelength of the light. What color is it?

An oil film \((n=1.45)\) floating on water is illuminated by white light at normal incidence. The film is 280 nm thick. Find (a) the color of the light in the visible spectrum most strongly reflected and (b) the color of the light in the spectrum most strongly transmitted. Explain your reasoning.

Sketch a phasor diagram to illustrate the resultant of \(E_{1}=\) \(E_{01} \sin \omega t\) and \(E_{2}=E_{02} \sin (\omega t+\phi),\) where \(E_{02}=1.50 E_{01}\) and \(\pi / 6 \leq \phi \leq \pi / 3 .\) Use the sketch and the law of cosines to show that, for two coherent waves, the resultant intensity can be written in the form \(I_{R}=I_{1}+I_{2}+2 I_{1} I_{2} \cos \phi.\)

The waves from a radio station can reach a home receiver by two paths. One is a straight-line path from transmitter to home, a distance of \(30.0 \mathrm{km}\). The second path is by reflection from the ionosphere (a layer of ionized air molecules high in the atmosphere). Assume this reflection takes place at a point midway between receiver and transmitter and that the wavelength broadcast by the radio station is \(350 \mathrm{m}\). Find the minimum height of the ionospheric layer that could produce destructive interference between the direct and reflected beams. (Assume that no phase change occurs on reflection.)

A Young's interference experiment is performed with monochromatic light. The separation between the slits is \(0.500 \mathrm{mm},\) and the interference pattern on a screen \(3.30 \mathrm{m}\) away shows the first side maximum \(3.40 \mathrm{mm}\) from the center of the pattern. What is the wavelength?
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