Introduction To Fourier Optics Third Edition Problem Solutions May 2026

Unlike many engineering texts, Goodman’s publisher (McGraw-Hill) does not release an official solutions manual to the public. This is intentional: the problems are designed for graduate courses where the instructor guides discovery.

Legitimate resources for solutions and hints:

Warning: Avoid generic online “solution manuals” – they are often for earlier editions, contain critical sign errors in the Fresnel integrals, or omit the all-important step of justifying the paraxial approximation.

Problem Statement: A slit of width $w$ is illuminated by a unit-amplitude plane wave normal to the aperture. Find the field distribution a distance $z$ away under the Fresnel approximation.

Solution: Let the aperture function be $t(x) = \textrect(x/w)$. The Fresnel diffraction integral for the field $U(x, z)$ is given by:

$$ U(x, z) = \frace^jkzj\lambda z e^j \frack2zx^2 \int_-\infty^\infty t(\xi) e^j \frack2z\xi^2 e^-j \frac2\pi\lambda z x \xi d\xi $$

Substituting $t(\xi) = \textrect(\xi/w)$, the limits of integration become $-w/2$ to $w/2$. The integral represents the Fourier transform of the product of the aperture and a quadratic phase factor. Subject: Fourier Optics & Wave Phenomena Reference: Goodman,

While this integral cannot be solved in closed form using elementary functions, the standard method involves expanding the term $e^j \frack2z\xi^2$ inside the slit or utilizing the Fresnel Integrals.

Let us perform a coordinate transformation. The field is proportional to: $$ U(x, z) \propto \int_-w/2^w/2 e^j \frac\pi\lambda z (x-\xi)^2 d\xi $$ (Note: This simplifies the algebra by completing the square).

Let $u = \sqrt\frac2\lambda z (x - \xi)$. The limits become: Upper limit: $u_2 = \sqrt\frac2\lambda z (x + w/2)$ Lower limit: $u_1 = \sqrt\frac2\lambda z (x - w/2)$

The solution is expressed in terms of the Fresnel Integrals $C(u)$ and $S(u)$: $$ U(x, z) = \frac12 \left( \frac1+j2 \right) \left[ [C(u_2) + jS(u_2)] - [C(u_1) + jS(u_1)] \right] $$

Key Insight: Fresnel diffraction requires numerical evaluation of Fresnel integrals unless the distance $z$ is very large (Fraunhofer regime) or very small (Rayleigh-Sommerfeld regime).

Subject: Fourier Optics & Wave Phenomena Reference: Goodman, J. W. Introduction to Fourier Optics, 3rd Edition. Purpose: To demonstrate the methodology for solving characteristic problems involving Fourier transforms, Fresnel diffraction, and lens imaging. J. W. Introduction to Fourier Optics

Problem Statement: Calculate the Fourier transform of the function $f(x) = \textrect(x/a)$ where $a > 0$.

Solution: Recall the definition of the rectangular function: $$ \textrect\left(\fracxa\right) = \begincases 1 & |x| < a/2 \ 0 & \textotherwise \endcases $$

The Fourier transform $\mathcalFf(x)$ is defined as $F(f_x) = \int_-\infty^\infty f(x) e^-j 2\pi f_x x dx$.

$$ F(f_x) = \int_-a/2^a/2 (1) e^-j 2\pi f_x x dx $$

Integrating: $$ F(f_x) = \left[ \frace^-j 2\pi f_x x-j 2\pi f_x \right]_-a/2^a/2 $$ $$ F(f_x) = \frac1-j 2\pi f_x \left( e^-j \pi f_x a - e^j \pi f_x a \right) $$

Using Euler's formula, $e^j\theta - e^-j\theta = 2j\sin(\theta)$: $$ F(f_x) = \frac2j \sin(\pi f_x a)j 2\pi f_x = \frac\sin(\pi f_x a)\pi f_x $$ and lens imaging.

Using the definition of the sinc function, $\textsinc(z) = \frac\sin(\pi z)\pi z$: $$ F(f_x) = a \cdot \textsinc(a f_x) $$

Key Insight: The width of the function in the space domain ($a$) is inversely proportional to the width of the spectrum in the frequency domain.

The third edition contains approximately 130 problems across 10 chapters. They fall into four major categories:

Typical question: Derive the conditions to avoid overlap between the twin images and the dc term in an off-axis hologram.

Solution strategy:

This level of detail turns a simple answer into a pedagogical tool.