A clear circular aperture of radius has amplitude transmittance

PHYSICS · Subjective · Fermi University Physics · Optics & Photonics · Interference and diffraction · Fourier optics and image formation

What an Instrument Can See, Ch Scalar diffraction theory, derived

Problem

A clear circular aperture of radius

a

has amplitude transmittance

t(r)=\mathrm{circ}(r/a)

, equal to 1 for

r\le a

and 0 otherwise. Light of vacuum wavelength

\lambda

passes through it and is brought to a focus by a lens of focal length

F

, so that, as the first course states for a lens, the field in the focal plane is the Fraunhofer transform of

t

with spatial frequency

\nu=

(focal-plane radial coordinate)

/(\lambda F)

. (a) Starting from the two-dimensional Fourier transform written in polar coordinates, show that for a circularly symmetric

t(r)

it reduces to the Fourier-Bessel (Hankel) transform

\tilde T(\rho)=2\pi\int_0^\infty t(r)J_0(2\pi r\rho)\,r\,dr.

(b) Using

\dfrac{d}{dx}[xJ_1(x)]=xJ_0(x)

, evaluate this transform for

t(r)=\mathrm{circ}(r/a)

and show that it can be written as

\pi a^2\,\mathrm{jinc}(2\pi a\rho)

with

\mathrm{jinc}(x)\equiv 2J_1(x)/x

. (c) The first zero of

J_1(x)

beyond the origin is at

x=3.8317

. Show that the corresponding angular radius of the first dark ring of the focal-plane pattern is

\theta_0=1.22\lambda/D

, with

D=2a

the aperture diameter. (d) An aperture of diameter

D=250\ \mathrm{mm}

is illuminated at

\lambda=500\ \mathrm{nm}

and focused by a lens of focal length

F=1500\ \mathrm{mm}

. Find the angular radius

\theta_0

of the first dark ring and its linear radius in the focal plane.