Introduction To Fourier Optics Third Edition Problem Solutions Direct

Problems focus on 2D Fourier transforms, convolution, and correlation. A typical problem asks: “Find the Fourier transform of a circular aperture of radius (a) and compare it to that of a square aperture.” The solution requires careful handling of Bessel functions and the Fourier slice theorem.

To appreciate the depth required, here is a skeletal structure of a high-quality solution to a third-edition problem (Chapter 6, Problem 6-2):

Problem: Show that the coherent transfer function (CTF) of a diffraction-limited system with an exit pupil function (P(\xi, \eta)) is given by (H_c(f_X, f_Y) = P(\lambda d_i f_X, \lambda d_i f_Y)), where (d_i) is the image distance.

Excerpt from a model solution:

A poor solution omits the delta function step; a great solution also discusses the implications for coherent image formation (e.g., no optical transfer function magnitude decay beyond cutoff).

Problem Statement: An imaging system has a square exit pupil of width $w$. Determine the Coherent Transfer Function (CTF) and the Optical Transfer Function (OTF). Problems focus on 2D Fourier transforms, convolution, and

Solution:

Part A: Coherent Transfer Function (CTF) For a coherent imaging system, the CTF is the scaled pupil function. The pupil function is: $$ P(x,y) = \textrect\left(\fracxw\right) \textrect\left(\fracyw\right) $$

The CTF, $H(f_x, f_y)$, is equal to the pupil function mapped into frequency coordinates. $$ H(f_x, f_y) = P(\lambda d_i f_x, \lambda d_i f_y) $$ Where $d_i$ is the image distance. The cutoff frequency occurs when the argument is $\pm w/2$. $$ \lambda d_i f_cutoff = \fracw2 \implies f_cutoff = \fracw2 \lambda d_i $$

Thus, the CTF is: $$ H(f_x, f_y) = \begincases 1 & |f_x| < f_cutoff \text and |f_y| < f_cutoff \ 0 & \textotherwise \endcases $$

Part B: Optical Transfer Function (OTF) The OTF is the normalized autocorrelation of the CTF (or the pupil function). $$ \textOTF(f_x, f_y) = \fracH(f_x, f_y) \star H(f_x, f_y)\textArea(H) $$ A poor solution omits the delta function step;

Geometrically, the autocorrelation of a square of side $w$ is a triangle function. The area of the pupil is $w^2$. The resulting OTF in one dimension is: $$ \textOTF(f_x) = \Lambda\left(\fracf_x2f_cutoff\right) $$ Where $\Lambda(x)$ is the triangle function ($1-|x|$ for $|x|\le 1$).

Key Insight: The incoherent cutoff frequency ($2f_cutoff$) is twice the coherent cutoff frequency, meaning incoherent imaging passes higher spatial frequencies, but with reduced contrast compared to the "all-or-nothing" pass of the coherent system.

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 $F(f_x) = \int_-\infty^\infty f(x) e^-j 2\pi f_x x dx$. Problem Statement: Calculate the Fourier transform of the

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

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, $e^j\theta - e^-j\theta = 2j\sin(\theta)$: $$ F(f_x) = \frac2j \sin(\pi f_x a)j 2\pi f_x = \frac\sin(\pi f_x a)\pi f_x $$

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.