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Chapter 37: Problem 24

Suppose the slit openings in a Young's double-slit experiment have different sizes so that the electric fields and intensities from each slit are different. With \(E_{1}=E_{01} \sin (\omega t)\) and \(E_{2}=E_{02} \sin (\omega t+\phi),\) show that the resultant electric field is \(E=E_{0} \sin (\omega t+\theta),\) where $$ E_{0}=\overline{E_{01}^{2}+E_{02}^{2}+2 E_{01} E_{02} \cos \phi}$$ and $$\sin \theta=\frac{E_{02} \sin \phi}{E_{0}}$$

Short Answer

Expert verified
The resultant electric field is \(E = E_0 \sin(\omega t + \theta)\), where \(E_0 = \sqrt{E_{01}^{2}+E_{02}^{2}+2 E_{01} E_{02} \cos \phi}\) and \(\sin \theta=\frac{E_{02} \sin \phi}{E_{0}}\)

Step by step solution

01

Superposition of Electric Fields

Begin by adding the two electric fields: \[E = E_1 + E_2 = E_{01} \sin(\omega t) + E_{02} \sin (\omega t + \phi)\].
02

Apply Trigonometric Identity

Use sin(A + B) = sinA cosB + cosA sinB identity on \(\sin (\omega t+ \phi)\) in the expression of \(E_2\). This results in: \[E = E_{01} \sin(\omega t) + E_{02}[\sin (\omega t) \cos(\phi) + \cos (\omega t) \sin(\phi)]\]
03

Simplify the Expression

Regrouping gives us \(E = (E_{01}+ E_{02} \cos(\phi)) \sin(\omega t) + E_{02} \sin(\phi) \cos(\omega t)\). This is in the form \(E = a \sin(\omega t) + b \cos(\omega t)\), where \(a\) and \(b\) are constants.
04

Apply Trigonometric Identity Again

Finally, apply the identity \(a \sin(\omega t) + b \cos(\omega t) = r \sin(\omega t + \delta)\). Compared with \(E_0 \sin (\omega t + \theta)\), we equate two equations and get two parts - magnitude and phase. Using Pythagora's theorem and trigonometric identities, calculate \(E_0\) and \(\theta\). We end up with \[E_{0}=\sqrt{E_{01}^{2}+E_{02}^{2}+2 E_{01} E_{02} \cos \phi}\] and \[\sin \theta=\frac{E_{02} \sin \phi}{E_{0}}\]
05

Final Result

This is our final result. The electric field is \(E = E_0 \sin(\omega t + \theta)\) where \(E_0\) and \(\theta\) are as derived.

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Key Concepts

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

Superposition of Electric Fields
The superposition principle is a cornerstone in the study of physics and it is pivotal to understanding the Young's double-slit experiment. It states that when two or more waves overlap, the resultant wave is simply the sum of the individual waves. In more technical terms, superposition refers to the additive nature of wave functions that result in a combined wave.

When dealing with electric fields, like in the given exercise, each slit in the double-slit experiment contributes an electric field described by a sine function. The electric field resulting from one slit is given as \(E_1 = E_{01} \times \text{sin}(\omega t)\), and the other as \(E_2 = E_{02} \times \text{sin}(\omega t + \phi)\). The total electric field \(E\) at any point in space is the vector sum of these two fields. This relationship is essential in predicting the pattern seen on a screen placed to observe the interference pattern formed by the two slits.
Trigonometric Identity
To solve the problem, we harness the power of trigonometric identities, which are equalities involving trigonometric functions that hold true for all values of the variables involved. A commonly employed trigonometric identity in wave interference problems is \(\text{sin}(A + B) = \text{sin}A \times \text{cos}B + \text{cos}A \times \text{sin}B\).

By applying this identity, we are able to break down the composite wave (\(E_2\)) into two components that can more easily be summed with the first wave (\(E_1\)). This makes it feasible to simplify the equation and reformat it into a form that leads to a simple harmonic wave, thus allowing us to extract meaningful physical quantities such as the amplitude \(E_0\) and the phase \(\theta\) of the resultant wave. Understanding how to manipulate these trigonometric identities is crucial for solving complex wave superposition problems.
Constructive and Destructive Interference
The phenomenon of constructive and destructive interference is intrinsically linked to the principle of superposition. Constructive interference occurs when waves add together to produce a wave of greater amplitude, while destructive interference happens when waves combine to produce a reduced amplitude, potentially canceling each other out completely.

In terms of electric fields, if the crest of one wave aligns with the crest of another, they constructively interfere, resulting in a higher intensity spot on the screen. Conversely, if the crest of one wave aligns with the trough of the other, they destructively interfere, leading to a dark spot. These interference effects are responsible for the characteristic bright and dark fringes observed in the Young's double-slit experiment. Understanding how the electric field vectors add together to cause these interference patterns can vastly improve a student's comprehension of wave phenomena and their applications in various fields of physics.

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Most popular questions from this chapter

When illuminated, four equally spaced parallel slits act as multiple coherent sources, each differing in phase from the adjacent one by an angle \(\phi .\) Use a phasor diagram to determine the smallest value of \(\phi\) for which the resultant of the four waves (assumed to be of equal amplitude) is zero.

Young's double-slit experiment is performed with 589 -nm light and a distance of \(2.00 \mathrm{m}\) between the slits and the screen. The tenth interference minimum is observed \(7.26 \mathrm{mm}\) from the central maximum. Determine the spacing of the slits.

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.

A soap film \((n=1.33)\) is contained within a rectangular wire frame. The frame is held vertically so that the film drains downward and forms a wedge with flat faces. The thickness of the film at the top is essentially zero. The film is viewed in reflected white light with near-normal incidence, and the first violet \((\lambda=420 \mathrm{nm})\) interference band is observed \(3.00 \mathrm{cm}\) from the top edge of the film. (a) Locate the first red \((\lambda=680 \mathrm{nm})\) interference band. (b) Determine the film thickness at the positions of the violet and red bands. (c) What is the wedge angle of the film?

Astronomers observe the chromosphere of the Sun with a filter that passes the red hydrogen spectral line of wavelength \(656.3 \mathrm{nm},\) called the \(\mathrm{H}_{\alpha}\) line. The filter consists of a transparent dielectric of thickness \(d\) held between two partially aluminized glass plates. The filter is held at a constant temperature. (a) Find the minimum value of \(d\) that produces maximum transmission of perpendicular \(\mathrm{H}_{\alpha}\) light, if the dielectric has an index of refraction of 1.378 (b) What If? If the temperature of the filter increases above the normal value, what happens to the transmitted wavelength? (Its index of refraction does not change significantly.) (c) The dielectric will also pass what nearvisible wavelength? One of the glass plates is colored red to absorb this light.
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