/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 66 (a) Consider an arrangement of \... [FREE SOLUTION] | 91Ó°ÊÓ

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(a) Consider an arrangement of \(N\) slits with a distance \(d\) between adjacent slits. The slits emit coherently and in phase at wavelength \(\lambda\). Show that at a time \(t\), the electric field at a distant point \(P\) is $$E_P(t) = E_0 cos(kR - \omega t) + E_0 cos(kR - \omega t + \phi)$$ $$+ E_0 cos(kR - \omega t + 2\phi) + . . .$$ $$+ E_0 cos(kR - \omega t + (N - 1)\phi)$$ where \(E_0\) is the amplitude at \(P\) of the electric field due to an individual slit, \(\phi = (2\pi d\) sin \(\theta)/\lambda\), \(\theta\) is the angle of the rays reaching \(P\) (as measured from the perpendicular bisector of the slit arrangement), and \(R\) is the distance from \(P\) to the most distant slit. In this problem, assume that \(R\) is much larger than \(d\). (b) To carry out the sum in part (a), it is convenient to use the complex-number relationship \(e^{iz} = cos z + i \space sin \space z\), where \(i = \sqrt{-1}\). In this expression, cos \(z\) is the \(real\) \(part\) of the complex number \(e^{iz}\), and sin \(z\) is its \(imaginary\) \(part\). Show that the electric field \(E_P(t)\) is equal to the real part of the complex quantity $$\sum _{n=0} ^{N-1} E_0 e^{i(kR-\omega t+n\phi)}$$ (c) Using the properties of the exponential function that \(e^Ae^B = e^{(A+B)}\) and \((e^A)^n = e^{nA}\), show that the sum in part (b) can be written as $$E_0 ( {e^{iN\phi} - 1 \over e^{i\phi} - 1} )e^{i(kR-\omega t)}$$ $$= E_0 ({e^{iN\phi/2} - e^{-iN\phi/2} \over e^{i\phi/2} - e^{-i\phi/2}} )e^{i[kR-\omega t+(N-1)\phi/2]}$$ Then, using the relationship \(e^{iz}\) = cos \(z\) + \(i\) sin \(z\), show that the (real) electric field at point \(P\) is $$E_P(t) = [E_0 {sin(N\phi/2) \over sin(\phi/2)} ] cos [kR - \omega t + (N - 1)\phi/2]$$ The quantity in the first square brackets in this expression is the amplitude of the electric field at \(P\). (d) Use the result for the electric-field amplitude in part (c) to show that the intensity at an angle \(\theta\) is $$I = I_0 [{ sin(N\phi/2) \over sin(\phi/2) }] ^2$$ where \(I_0\) is the maximum intensity for an individual slit. (e) Check the result in part (d) for the case \(N\) = 2. It will help to recall that sin 2\(A\) = 2 sin \(A\) cos \(A\). Explain why your result differs from Eq. (35.10), the expression for the intensity in two-source interference, by a factor of 4. (\(Hint\): Is I0 defined in the same way in both expressions?)

Short Answer

Expert verified
The intensity at angle \(\theta\) is \(I = I_0 \left( \frac{\sin(N\phi/2)}{\sin(\phi/2)} \right)^2\). For \(N=2\), the difference is due to \(I_0\) being defined per slit.

Step by step solution

01

Understanding the Problem

We need to show how the electric field at a distant point is constructed from the contributions of multiple slits, using trigonometry and complex numbers.
02

Expressing the Electric Field

For part (a), the electric field contributions from each slit are expressed as a sum of cosine terms, each with a phase difference. The field from slit n is \( E_0 \cos(kR - \omega t + n\phi) \). Therefore, the total electric field at point \( P \) due to all slits is:\[ E_P(t) = \sum_{n=0}^{N-1} E_0 \cos(kR - \omega t + n\phi) \]
03

Using Complex Exponential Notation

To simplify the sum in part (a), we apply Euler's formula, expressing \(\cos z\) as the real part of \(e^{iz}\). The electric field becomes:\[ E_P(t) = \text{Re} \left( \sum_{n=0}^{N-1} E_0 e^{i(kR - \omega t + n\phi)} \right) \]This is the solution for part (b).
04

Simplifying the Complex Sum

For part (c), using the properties of exponential functions \((e^Ae^B = e^{A+B})\) and \((e^A)^n = e^{nA})\), the sum can be rewritten as a geometric series:\[ \sum_{n=0}^{N-1} e^{in\phi} = e^{iN\phi} - 1 \over e^{i\phi} - 1 \]Thus, the electric field expression becomes:\[ E_0 \frac{e^{iN\phi} - 1}{e^{i\phi} - 1} e^{i(kR - \omega t)} = E_0 \frac{e^{iN\phi/2} - e^{-iN\phi/2}}{e^{i\phi/2} - e^{-i\phi/2}} e^{i[kR - \omega t + (N-1)\phi/2]} \]
05

Real Part of the Complex Expression

We transform back using Euler's formula, obtaining the real component:\[ E_P(t) = \left[ E_0 \frac{\sin(N\phi/2)}{\sin(\phi/2)} \right] \cos[kR - \omega t + (N-1)\phi/2] \] The term in brackets is the amplitude at point \(P\).
06

Deriving Intensity Formula

Using the known relationship between field amplitude and intensity, \( I \propto E^2 \), we find the intensity:\[ I = I_0 \left( \frac{\sin(N\phi/2)}{\sin(\phi/2)} \right)^2 \] where \( I_0 \) is the intensity from one slit. This result handles part (d).
07

Checking N = 2 Case

For two slits, \(N=2\), we simplify the expression:\[ I = I_0 \left( \frac{\sin(\phi)}{\sin(\phi/2)} \right)^2 = 4I_0 \cos^2(\phi/2) \]This differs by a factor of 4 compared to standard two-source intensity due to square variation, as \(I_0\) is defined for a single slit, doubling amplitude results quadruple intensity.

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Key Concepts

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

Electric Field
In the context of interference and diffraction, understanding the electric field is crucial. The electric field at a distant point caused by multiple slits can be represented as a sum of contributions from each slit. Each slit emits light coherently and in phase, meaning their emissions influence the resultant electric field at any given point. For multiple slits separated by a distance \(d\), with light emitted at a wavelength \(\lambda\), this is expressed mathematically for a point \(P\) as:
  • \(E_P(t) = E_0 \cos(kR - \omega t) + E_0 \cos(kR - \omega t + \phi) + \ldots + E_0 \cos(kR - \omega t + (N-1)\phi)\)
Here, \(E_0\) represents the amplitude due to one slit, \(\phi = \frac{2\pi d \sin \theta}{\lambda}\) denotes the phase difference between emissions from adjacent slits, and \(R\) is the distance from the slits to point \(P\). This equation forms the basis for deriving more complex interference patterns seen in wave physics.
Phase Difference
The phase difference is a key concept in understanding interference and diffraction patterns. It represents the difference in phase angle that each slit contributes to the electrical field at a point due to differences in path length. This can cause constructive or destructive interference, influencing resultant intensity patterns.In our exercise, the phase difference \(\phi\) is given by
  • \(\phi = \frac{2\pi d \sin \theta}{\lambda}\)
This formula calculates \(\phi\) based on the geometry of the setup, where \(d\) is the distance between slits, \(\lambda\) is the wavelength, and \(\theta\) is the angle relative to the normal. Understanding phase differences helps explain interference phenomena like bright and dark fringes, which occur when waves are in-phase (constructive interference) or out-of-phase (destructive interference).
Intensity Formula
When studying wave phenomena, especially interference and diffraction, the intensity formula is fundamental. It expresses how the amplitude of the electric field relates to the light intensity at a particular location. The exercise derives an expression for intensity at an angle \(\theta\) as:
  • \(I = I_0 \left( \frac{\sin(N\phi/2)}{\sin(\phi/2)} \right)^2\)
Here, \(I_0\) is the maximum intensity for a single slit, and \(N\) represents the number of slits. This formula is pivotal in calculating the resultant light intensity pattern generated by multiple slits, accounting for phase differences and multiplying effects as \(N\) increases.The sine term inside the squared fraction modulates the intensity, creating patterns corresponding to maxima and minima of light, illustrating the effects of constructive and destructive interference.
Complex Numbers in Physics
Utilizing complex numbers greatly simplifies calculations in physics involving wave phenomena, such as interference and diffraction. By representing waves as complex exponentials, often using Euler's formula \(e^{iz} = \cos z + i\sin z\), complex numbers allow us to streamline the addition of phase differences and amplitudes in calculations. In the exercise:
  • Electric field contributions become simpler: \(E_P(t) = \text{Re} \left( \sum_{n=0}^{N-1} E_0 e^{i(kR - \omega t + n\phi)} \right)\)
This expression shows that the sum of phases for the electric field can be handled elegantly with complex algebra, where \(\text{Re}\) indicates taking the real part of the complex sum. Properties such as \(e^Ae^B = e^{(A+B)}\) and complex exponentiation simplify these computations.Using complex numbers is not just a mathematical trick but a powerful tool in physics to understand and predict interference patterns, making complex wave computations, like our given exercise, more intuitive and manageable.

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Most popular questions from this chapter

If the planes of a crystal are 3.50 \(\AA\) (1 \(\AA\) = 10\(^{-10}\) m = 1 \(\AA\)ngstrom unit) apart, (a) what wavelength of electromagnetic waves is needed so that the first strong interference maximum in the Bragg reflection occurs when the waves strike the planes at an angle of 22.0\(^\circ\), and in what part of the electromagnetic spectrum do these waves lie? (See Fig. 32.4.) (b) At what other angles will strong interference maxima occur?

A laser beam of wavelength \(\lambda\) = 632.8 nm shines at normal incidence on the reflective side of a compact disc. (a) The tracks of tiny pits in which information is coded onto the CD are 1.60 \(\mu\)m apart. For what angles of reflection (measured from the normal) will the intensity of light be maximum? (b) On a DVD, the tracks are only 0.740 \(\mu\)m apart. Repeat the calculation of part (a) for the DVD.

The light from an iron arc includes many different wavelengths. Two of these are at \(\lambda\) = 587.9782 nm and \(\lambda\) = 587.8002 nm. You wish to resolve these spectral lines in first order using a grating 1.20 cm in length. What minimum number of slits per centimeter must the grating have?

If a diffraction grating produces a third-order bright spot for red light (of wavelength 700 nm) at 65.0\(^\circ\) from the central maximum, at what angle will the second-order bright spot be for violet light (of wavelength 400 nm)?

Monochromatic x rays are incident on a crystal for which the spacing of the atomic planes is 0.440 nm. The first-order maximum in the Bragg reflection occurs when the incident and reflected x rays make an angle of 39.4\(^\circ\) with the crystal planes. What is the wavelength of the x rays?

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