Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 2

Show that in the limit of many slits \((N \rightarrow \infty)\), the interference pattern in the vicinity of the principal maxima takes on the form of the single-slit diffraction pattern

Short Answer

Expert verified
Answer: As the number of slits (N) in a diffraction grating approaches infinity, the distance between adjacent slits (a) approaches zero while the total width of the array of slits approaches the width of a single slit (b). In this limit, the intensity of the N-slit interference pattern converges to the intensity of the single-slit diffraction pattern, making them similar near the principal maxima.

Step by step solution

01

Analyze the N-slit interference pattern

(Write the content here) For N slits, the phase difference between adjacent slits is given by the following expression: \[ \delta = \frac{2\pi a\sin\theta}{\lambda} \] where the angle θ is the angle between the incoming light and the plane of the slits, λ is the wavelength of the light, and a is the distance between adjacent slits. The interference pattern is constructive (has a maximum intensity) when N times the phase difference between successive slits is an integer multiple of 2π, that is: \[N\delta = 2\pi m\] where m is an integer. Substitute the expression for the phase difference into this equation: \[ N \cdot \frac{2\pi a\sin\theta}{\lambda} = 2\pi m \]
02

Find the general equation for the intensity of the N-slit interference pattern

We can find the general equation for the intensity (I) of the interference pattern by using the principle of superposition, where the electric fields of each individual slit are added: \[ I = I_0 \left[\frac{\sin{(\frac{N\delta}{2})}}{\sin{(\frac{\delta}{2})}}\right]^2 \] where I₀ is the maximum intensity of the interference pattern.
03

Calculate the intensity of the diffraction pattern from a single slit

For a single slit, the diffraction pattern is given by the equation: \[I = I_0 \left[\frac{\sin{( \frac{\pi b \sin\theta}{\lambda})}}{(\frac{\pi b \sin\theta}{\lambda})}\right]^2\] where b is the width of the single slit.
04

Show that the intensity patterns become similar as N approaches infinity

As N approaches infinity, the distance a between adjacent slits vanishes, and the total width of the array of slits approaches b, the width of the single slit. Thus, we can set a = b/N, and as N goes to infinity, a approaches 0. In this limit: \[ I_\infty = I_0 \left[\frac{\sin{(\frac{\pi b \sin\theta}{\lambda})}}{(\frac{\pi b \sin\theta}{\lambda})}\right]^2 = I_{single} \] Hence, the interference pattern approaches that of a single-slit diffraction pattern as N approaches infinity, thus proving the statement.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Show that the Fraunhofer diffraction pattern of an elliptical aperture, of semiaxes \(a\) and \(b\), is identical to that of a circular aperture except for a linear expansion by the ratio \(a / b\) in the direction parallel to the minor axis \(b\).

A particular transparent diffraction grating is ruled with 5,500 lines per centimeter, and \(2.5 \mathrm{~cm}\) of it is used in a simple spectrometer having a source slit, collimating lens, and telescope. At what angles are the sodium D lines \((\lambda=589.0\) and \(589.6 \mathrm{~nm})\) found if the grating is illuminated at normal incidence? What is the dispersion in minutes of arc per nanometer and the theoretical resolution at each of these angular positions?

Show that when a circular aperture is illuminated by a plane wave whose amplitude varies with radius as \(T(\rho)\), from (9.4.19), the diffraction pattern is given by $$ \psi(u)=2 \pi \breve{C}^{\prime} \int_{0}^{a} T(\rho) J_{c}\left(\frac{u \rho}{a}\right) \rho d \rho, $$ which reduces to \((10.5 .4)\) when \(T(\rho)=1\).

The largest fully steerable radio telescope at the National Radio Astronomy Observatory (Green Bank, W.Va.) has a parabolic mirror \(140 \mathrm{ft}\) in diameter. Approximately what angular resolution does it have for the 1,420-M Hz \((21-\mathrm{cm})\) line radiated by interstellar atomic hydrogen? (In practice, the sensitivity of the detector at the parabola focus is not isotropic. A careful analysis would include a nonuniform aperture-illumination function analogous to Probs. \(10.4 .8\) and 10.4.9.) The mirror was constructed to conform to an ideal paraboloid within a tolerance of \(0.030 \mathrm{in}\). What minimum wavelength can be used with this instrument if the criterion is that the construction errors not exceed \(\lambda / 8\) ? What is the approximate resolution for the minimum wavelength? Answer: 21 minutes of arc; \(\lambda=6 \mathrm{~mm}, 0.6\) minute of arc.

A radio-frequency antenna (such as a TV antenna) may be thought of semiquantitatively as an aperture with dimensions comparable to the physical dimensions (usually of the order of a wavelength). How far away from the antenna, expressed in wavelengths, must one be for the Fraunhofer case to be valid? Compare with the case for a pinhole of diameter \(0.1 \mathrm{~mm}\) for visible wavelengths.
See all solutions

Recommended explanations on Physics Textbooks

Particle Model of Matter

Read Explanation

Electricity

Read Explanation

Circular Motion and Gravitation

Read Explanation

Fields in Physics

Read Explanation

Torque and Rotational Motion

Read Explanation

Thermodynamics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free