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Chapter 10: Problem 4

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 \(b\).

Short Answer

Expert verified
Answer: The Fraunhofer diffraction pattern for an elliptical aperture with 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.

Step by step solution

01

Fraunhofer diffraction definition

Fraunhofer diffraction is the far-field diffraction pattern that occurs when a plane wave is incident on an aperture. In this case, we have an elliptical aperture and a circular aperture. The intensity of light (diffraction pattern) at any distance from the aperture can be calculated using the Fourier transform of the aperture's transmission function.
02

Calculate the transmission functions for elliptical and circular apertures

The transmission function of an aperture is a function that describes the amplitude and phase of the transmitted electric field in the aperture plane. For an elliptical aperture: \(T_e(x, y) = \begin{cases} 1 & \frac{x^2}{a^2}+\frac{y^2}{b^2} \leq 1 \\ 0 & \text{otherwise} \end{cases}\) For a circular aperture: \(T_c(x, y) = \begin{cases} 1 & x^2 + y^2 \leq b^2 \\ 0 & \text{otherwise} \end{cases}\)
03

Calculate the Fraunhofer diffraction patterns for both apertures

Using the Fourier transform, we can calculate the Fraunhofer diffraction patterns: For the elliptical aperture, the diffraction pattern is given by: \(I_e(\xi, \eta) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} T_e(x, y) e^{-i2\pi (\xi x + \eta y)} dx dy\) For the circular aperture, the diffraction pattern is given by: \(I_c(\xi, \eta) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} T_c(x, y) e^{-i2\pi (\xi x + \eta y)} dx dy\)
04

Simplify the integral expressions

For the elliptical aperture, the integral becomes the double integral over the limits of the ellipse: \(I_e(\xi, \eta) = \int_{-a}^{a} \int_{-\frac{b}{a}\sqrt{a^2 - x^2}}^{\frac{b}{a}\sqrt{a^2 - x^2}} e^{-i2\pi (\xi x + \eta y)} dy dx\) For the circular aperture, the integral becomes the double integral over the limits of the circle: \(I_c(\xi, \eta) = \int_{-b}^{b} \int_{-\sqrt{b^2 - y^2}}^{\sqrt{b^2 - y^2}} e^{-i2\pi (\xi x + \eta y)} dx dy\)
05

Perform a variable substitution on the elliptical diffraction pattern

To compare the diffraction patterns, we will perform a variable substitution (\(u = \frac{x}{a}\)) on the elliptical aperture diffraction pattern: \(I_e(\xi, \eta) = a^2 \int_{-1}^{1} \int_{-b\sqrt{1-u^2}}^{b\sqrt{1-u^2}} e^{-i2\pi a(\xi u + \frac{\eta y}{a})} dy du\)
06

Compare the elliptical and circular aperture diffraction patterns

Upon examination of the integrals, we can notice that the elliptical aperture diffraction pattern can be expressed as a linear expansion of the circular aperture diffraction pattern: \(I_e(\xi, \eta) = a^2 I_c(\frac{\xi}{a}, \frac{\eta}{a})\) This shows that the Fraunhofer diffraction pattern for an elliptical aperture is indeed identical to that of a circular aperture, with a linear expansion by the ratio a/b in the direction parallel to the minor axis b.

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Most popular questions from this chapter

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 \(\frac{1}{2}\) ?

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 \(10.2\).

The Bragg equation (10.10.13) is similar in form to the grating equation (10.8.10) with \(\theta_{\phi}=\theta_{2}\) However, for the grating the angles \(\theta_{\alpha}, \theta_{\theta}\) are measured from the normal, whereas in the Bragg case the angle \(\theta\) is measured, by convention, from the diffracting planes in question (rather than from the normal to the planes). What is the relation between the two situations? Why not use the cosine in place of the sine in one equation, since the angles appear to be complements?

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

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 \(T(\rho)=1\).
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