Show that the fraction of the total energy in the circular-aperture diffraction pattern out to the radius specified by \(u_{\operatorname{msx}}\) is $$ \frac{1}{2} \int_{0}^{u_{\max }}\left[\frac{2 J_{1}(u)}{u}\right]^{2} u d u=1-J_{0}^{2}\left(u_{\max }\right)-J_{1}^{2}\left(u_{\max }\right) . $$ Hence, verify the final column of Table \(10.2\).

Short Answer

Expert verified
Question: Derive and verify the energy fraction formula for a circular-aperture diffraction pattern given by $$\frac{1}{2}\int_{0}^{u_{\max }}\left[\frac{2J_{1}(u)}{u}\right]^{2}ud u$$ and apply it to the data in Table 10.2 to verify the final column values. Answer: We derived the energy fraction formula using the properties of Bessel functions and the intensity pattern of circular-aperture diffraction. The derived formula is: $$\frac{1}{2}\int_{0}^{u_{\max }}\left[\frac{2 J_{1}(u)}{u}\right]^{2} u d u = 1 - J_{0}^2(u_{\max}) - J_{1}^2(u_{\max})$$ To verify the final column of Table 10.2, we need to apply the derived energy fraction formula for each \(u_{\max}\) value in the table and compare the results to the values given in the last column. If the results match, then we have successfully verified the final column of Table 10.2.

Step by step solution

01

Understanding the Bessel function in the context of circular aperture diffraction patterns

Bessel functions are a type of special functions that often occur in problems with cylindrical or spherical symmetry, such as circular aperture optics. In this case, the \(J_{1}(u)\) term arises due to the geometry of the aperture. The diffraction pattern's energy distribution behaves according to this Bessel function.
02

Deriving the expression for the energy fraction formula

We are given the formula as: $$\frac{1}{2}\int_{0}^{u_{\max }}\left[\frac{2J_{1}(u)}{u}\right]^{2}ud u$$ To derive this formula, we recall that the intensity pattern \(I(u)\) of circular-aperture diffraction is given by: $$I(u) = \left[\frac{2 J_{1}(u)}{u}\right]^2$$ Where \(I(u)\) is the intensity at radial distance \(u\) from the center of the diffraction pattern, and \(J_{1}(u)\) is the Bessel function of the first kind of order 1. If we want to find the fraction of the total energy contained within a maximum radius \(u_{\max}\), we integrate the intensity pattern with respect to \(u\): $$\frac{1}{2}\int_{0}^{u_{\max }} I(u) u d u = \frac{1}{2}\int_{0}^{u_{\max }}\left[\frac{2J_{1}(u)}{u}\right]^{2}ud u$$
03

Simplifying the expression using Bessel function properties

We will now attempt to simplify the fraction of total energy expression using the properties of Bessel functions. The relationship we will use is: $$ 1 = J_{0}^2(u) + 2\sum_{n=1}^{\infty} J_n^2(u) $$ At \(u = u_{\max}\), we have: $$ 1 = J_{0}^2(u_{\max}) + 2\sum_{n=1}^{\infty} J_n^2(u_{\max}) $$ Because we only have \(J_{1}(u_{\max})\) in the given expression, we can rewrite the above equation as: $$ 1 = J_{0}^2(u_{\max}) + 2J_{1}^2(u_{\max}) + 2\sum_{n=2}^{\infty} J_n^2(u_{\max}) $$ We know that the fraction of total energy inside the circle of radius \(u_{\max}\) leaves out an infinitesimal part of the energy distribution. So, we can write: $$ \frac{1}{2}\int_{0}^{u_{\max }}\left[\frac{2 J_{1}(u)}{u}\right]^{2} u d u = 1 - J_{0}^2(u_{\max}) - J_{1}^2(u_{\max}) $$ Hence, we have derived the required energy fraction formula.
04

Verifying the final column of Table 10.2

We now need to verify the energy fraction values in the final column of Table 10.2 using the derived expression. We should apply the derived energy fraction formula for each \(u_{\max}\) value in the table and compare the results to the values given in the last column. If the results match, then we have successfully verified the final column of Table 10.2.

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Key Concepts

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

Bessel Functions
Bessel functions are essential in various areas such as physics, engineering, and mathematics, particularly when dealing with problems which have radial symmetry. They appear in solutions to wave equations in cylindrical coordinates and are therefore pivotal in analyzing wave behaviors, such as those in circular aperture diffraction patterns.

For example, in optics, the Bessel function of the first kind, denoted as \(J_n(u)\), where \(n\) is the order, characterizes how light waves propagate through circular openings. These functions oscillate and have a decaying amplitude, much like how ripples behave on a water surface after a stone is dropped.

Specifically, in the context of a circular aperture diffraction problem, the term \(J_1(u)\), represents the first-order Bessel function. This mathematical representation is critical to predict the intensity variation in the resulting diffraction pattern. As light waves interfere with each other, they give rise to a series of bright and dark rings witnessed in the pattern, and understanding Bessel functions allows us to describe this behavior accurately.

Students struggling with Bessel functions should start by grasping their oscillatory nature and relation to sine and cosine functions. Moreover, they should become familiar with standard Bessel function properties, such as recurrence relations, which are incredibly useful in simplifying complex expressions in problems such as diffraction pattern calculations.
Intensity Pattern Integration
The intensity pattern of light obtained in a diffraction experiment is critical for understanding how energy is distributed across the diffraction pattern. To quantify the distribution, we integrate the intensity pattern over the relevant area.

In our exercise, the intensity pattern \(I(u)\) is expressed in terms of the Bessel function and the variable \(u\), which is a dimensionless quantity related to the radius in the diffraction pattern. This pattern is not uniform; instead, it varies with the radial distance, showing high intensity at the center which gradually decreases with increases in radius.

To determine the total energy within a certain radius \(u_{\text{max}}\), the intensity function \(I(u)\) is integrated with respect to \(u\), weighted by the radial distance \(u\). This process, known as intensity pattern integration, involves not only the integration of the simple square of the Bessel function but also the incorporation of the variable \(u\) which accounts for the circular geometry of the aperture.

When integrating the intensity pattern, students must remember that the integration bounds are crucial. They need to integrate up to \(u_{\text{max}}\) to find the energy fraction within the radius specified. The integration thus captures the energy distribution up to that point, reflecting how much of the light after passing through the aperture falls within a given radius.
Energy Fraction in Diffraction
Talking about the energy fraction in diffraction is essentially addressing how much of the incident light's energy is distributed within a specific area of the diffraction pattern. This concept is extremely valuable in applications where precise energy measurements matter, such as in designing optical instruments or analyzing the light-gathering capabilities of telescopes.

The exercise we are considering requires us to calculate the fraction of the total energy within a circular area of radius \(u_{\text{max}}\) of the diffraction pattern. The derived expression combines the mathematical framework provided by Bessel functions with physical insights about energy conservation in wave phenomena. By integrating the intensity, we account for all the light beam contributions up to the specified radius.

The result is intuitive; as we increase the radius \(u_{\text{max}}\), we expect more energy to be included in our calculation. The expression \(1 - J_0^2(u_{\text{max}}) - J_1^2(u_{\text{max}})\) indicates that there's a certain amount of energy outside the radius \(u_{\text{max}}\), which corresponds to the sum of the squares of the zeroth and first-order Bessel functions evaluated at that radius.

This part of the study is challenging, but students should understand that it is about dividing the diffraction pattern into infinitesimally small concentric rings, integrating the energy over those rings, and then summing up to get the total energy within \(u_{\text{max}}\). By comparing the results with established values (like those in Table 10.2), students can validate their understanding and the calculated expressions.

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Most popular questions from this chapter

For the \(N\)-slit grating at normal incidence, devise elementary arguments, not involving integrals or a formal summation like \((10.7 .5)\), to show that (a) the principal interference maxima occur for $$ b \sin \theta_{0}=m \lambda $$ where \(m\) is an integer; \((b)\) the nulls adjacent to a particular principal maximum are displaced in angle from the maximum by \(\Delta \theta_{0}\) such that $$ \left(\frac{N}{2} b \cos \theta_{*}\right) \Delta \theta_{\varphi}=\pm \frac{\lambda}{2} $$ (c) hence, the resolving power is \(m N\).

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?

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.

Review the elementary arguments for discussing single-slit Fraunhofer diffraction by pairing up portions of the wavefront in the aperture that are \(180^{\circ}\) out of phase to show that (a) a null is observed when the slit may be divided into two equal portions such that $$ \frac{\delta}{2}=\frac{a}{2} \sin \theta=\frac{\lambda}{2} $$ Prob. 10.4.1 (b) Similarly, nulls result when the aperture is divided up into \(2 n\) portions such that $$ \frac{a}{2 n} \sin \theta=\frac{\lambda}{2} $$ where \(n\) is an integer; \((c)\) the secondary maxima occur approximately when the aperture is divided up into \(2 n+1\) portions such that $$ \frac{a}{2 n+1} \sin \theta=\frac{\lambda}{2} $$ and that the intensity of these maxima, relative to the principal maximum, is approximately $$ \frac{1}{2}\left(\frac{1}{2 n+1}\right)^{2} . $$ How do you justify the factor of \(\frac{1}{2}\) ?

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