Huygens–Fresnel Principle

Let us again consider a monochromatic plane light wave, propagating in the $z$-direction, which is normally incident on an opaque screen that occupies the plane $z=0$. Suppose that there is an irregularly shaped aperture cut in the screen, and that the light that passes through this aperture travels to a flat projection screen occupying the plane $z=R$. Let $\psi_0\,\cos(\omega\,t-\phi)$ be the electric component of the wave illuminating the opaque screen from behind (i.e., from $z<0$). We can determine the diffraction pattern that appears on the projection screen by using a more accurate version of Huygen's principle known as the Hugyens–Fresnel principle. According to this principle, each unblocked element of the opaque screen emits a secondary wave of the form (Jackson 1975)

$$\psi(r,\theta,t) = \frac{\psi_0\,K(\theta)\,\cos(\omega\,t-k\,r-\phi+\pi/2)}{\lambda\,r}\,dS,$$ (10.88)

where $r$ is the distance that the wave has propagated, $\theta$ the angle subtended between the direction of propagation and the direction of incidence (i.e., the $z$-axis), $dS$ the area of the element, and $\lambda=2\pi/k$ the wavelength. Here, $K(\theta)$ is known as the obliquity factor. Previously (in Section 10.9), we effectively assumed that $K(\theta)=1$ for $0\leq\theta\leq \pi/2$, and $K(\theta)=0$ for $\pi/2<\theta\leq \pi$. In other words, we assumed that the secondary wave is a half-spherical wave that is emitted isotropically in all forward directions, and has zero amplitude in all backward directions. In fact, the true obliquity factor is (Jackson 1975)

$$K(\theta) = \frac{1}{2}\left(1+\cos\theta\right),$$ (10.89)

which implies that, although the secondary wave propagates predominately in the forward direction, there is some backward propagation. According to Equation (10.88), there is a $-\pi/2$ phase difference between the secondary wave emitted by an element of the aperture, and the light that illuminates the element from behind. We previously (in Section 10.9) assumed that there was no phase difference. However, it is easily demonstrated that none of our previous results would be modified had we taken this phase difference into account.

Let us define a modified aperture function, $f(x,y)$, which is such that $f=1$ if the point ($x$, $y$, 0) on the opaque screen falls within the aperture, and $f=0$ otherwise. It follows from Equation (10.88) that the wave amplitude at the point ($x'$, $y'$, $R$) on the projection screen, which is the resultant of all of the secondary waves that are emitted by the aperture and travel to this particular point, is given by

$$\psi(x',y',t) = \frac{\psi_0}{\lambda} \int_{-\infty}^\infty\int_... ...}^\infty f(x,y)\,K(\theta)\, \frac{\cos(\omega\,t-k\,r-\phi+\pi/2)}{r}\,dx\,dy.$$ (10.90)

Now,

$$r =\left[(x-x')^{\,2}+(y-y')^{\,2}+R^{\,2}\right]^{1/2}$$

$$=R + \frac{(x-x')^{\,2}}{2\,R}+ \frac{(y-y')^{\,2}}{2\,R} +{\cal ... ...-x')^{\,4}}{R^{\,3}}\right] +{\cal O}\left[\frac{(y-y')^{\,4}}{R^{\,3}}\right],$$ (10.91)

and

$$\cos\theta \equiv \frac{R}{r} = 1-\frac{(x-x')^{\,2}}{2\,R^{\,2}}... ...-x')^{\,4}}{R^{\,4}}\right] +{\cal O}\left[\frac{(y-y')^{\,4}}{R^{\,4}}\right].$$ (10.92)

Let us assume that

$$\frac{d}{R}\ll 1,$$ (10.93)

where $d$ is the typical aperture dimension (or, to be more exact, the typical value of $\vert x-x'\vert$ and $\vert y-y'\vert$). In this limit, Equation (10.90) reduces to

$$\psi(x',y',t) \simeq -\frac{\psi_0}{\lambda\,R}\int_{-\infty}^\in... ...ga\,t-k\,R-\phi-\frac{k\,(x-x')^2}{2\,R}-\frac{k\,(y-y')^2}{2\,R}\right]dx\,dy,$$ (10.94)

where use has been made of the trigonometric identity $\cos(x+\pi/2)=-\sin(x)$. The neglect of terms involving $(x-x')^{\,4}$ and $(y-y')^{\,4}$ inside the argument of the sine function in the previous equation is justified provided that

$$\frac{d^{\,4}}{\lambda\,R^{\,3}} \ll 1.$$ (10.95)

Note that this is a far less stringent criterion than the far-field criterion (see Section 10.2)

$$\frac{d^{\,2}}{\lambda\,R} \ll 1.$$ (10.96)

Hence, Equation (10.94) is valid not only in the far-field limit (specified by the previous inequality), but also in the so-called near-field limit specified by the inequality

$$1\lesssim\frac{d^{\,2}}{\lambda\,R} \ll \left(\frac{R}{d}\right)^2.$$ (10.97)

Expression (10.94) is more general than our previous far-field expression, (10.74), because we have retained terms in the argument of the sine function that are quadratic in $x$ and $y$, whereas these terms were previously neglected. Incidentally, far-field diffraction is often referred to as Fraunhofer diffraction, whereas near-field diffraction is termed Fresnel diffraction. Note, finally, that because the ordering $d/R\ll 1$ implies that $\vert\theta\vert\ll 1$, our previous incorrect assumption for the form of the obliquity factor makes no difference, because our previous form and the correct form are both characterized by $K(0)=1$.

Let

$$u = \left(\frac{2}{\lambda\,R}\right)^{1/2}x,$$ (10.98)

$$u' = \left(\frac{2}{\lambda\,R}\right)^{1/2}x',$$ (10.99)

$$v = \left(\frac{2}{\lambda\,R}\right)^{1/2}y,$$ (10.100)

$$v' = \left(\frac{2}{\lambda\,R}\right)^{1/2}y'.$$ (10.101)

Thus, $u$ and $v$ are dimensionless coordinates that locate a point within the aperture, whereas $u'$ and $v'$ are corresponding coordinates that locate a point on the projection screen. Equation (10.94) transforms to give

$$\psi(u',v',t)\simeq -\psi_0\,\sin(\omega\,t-k\,R-\phi)\,f_c(u',v') + \psi_0\,\cos(\omega\,t-k\,R-\phi)\,f_s(u',v'),$$ (10.102)

where

$$f_c(u',v') =\frac{1}{2}\int_{-\infty}^\infty \int_{-\infty}^\infty f(u,v)\,\cos\left[\frac{\pi}{2}\,(u-u')^{\,2} +\frac{\pi}{2}\,(v-v')^{\,2}\right] du\,dv,$$ (10.103)

$$f_s(u',v') =\frac{1}{2}\int_{-\infty}^\infty \int_{-\infty}^\infty f(u,v)\,\sin\left[\frac{\pi}{2}\,(u-u')^{\,2} +\frac{\pi}{2}\,(v-v')^{\,2}\right] du\,dv.$$ (10.104)

Suppose that the aperture is completely transparent, so that $f(u,v)=1$ for all $u$ and $v$. In this case, the two-dimensional integrals in the previous two equations become separable. Making use of some standard trigonometric identities (see Section B.3), we obtain

$$f_c(u',v') = \frac{1}{2}\left[\int_{-\infty}^\infty \cos\left(\frac{\pi}{2}\... ...t[\int_{-\infty}^\infty \sin\left(\frac{\pi}{2}\,t^{\,2}\right)dt\right]^{\,2},$$ (10.105)

$$f_s(u',v') =\left[\int_{-\infty}^\infty \cos\left(\frac{\pi}{2}\,t^{\,2}\rig... ...t]\left[\int_{-\infty}^\infty \sin\left(\frac{\pi}{2}\,t^{\,2}\right)dt\right].$$ (10.106)

However, (Abramowitz and Stegun 1965)

$$\int_{-\infty}^\infty \cos\left(\frac{\pi}{2}\,t^{\,2}\right)dt= \int_{-\infty}^\infty \sin\left(\frac{\pi}{2}\,t^{\,2}\right)dt=1.$$ (10.107)

Hence, we deduce that $f_c(u',v')=0$ and $f_s(u',v')=1$, which, from Equation (10.102), implies that

$$\psi(u',v',t)= \psi_0\,\cos(\omega\,t-k\,R-\phi).$$ (10.108)

Of course, this is the correct answer; if the aperture is completely transparent (i.e., if $f=1$ everywhere) then the incident wave, $\psi_0\,\cos(\omega\,t-\phi)$, illuminating the aperture from behind, propagates in the $z$-direction without changing amplitude, and acquires a phase shift $k\,R$ by the time it reaches the projection screen. The previous result is the ultimate justification for the Hugyens–Fresnel formula, (10.88).

The intensity at which a given point on the projection screen is illuminated is (see Section 10.2)

$${\cal I}(u',v') \propto \langle \psi^{\,2}(u',v',t)\rangle,$$ (10.109)

where $\langle\cdots\rangle$ denotes an average over a wave period. It follows from Equation (10.102) that

$$\frac{{\cal I}(u',v')}{{\cal I}_0}=f_c^{\,2}(u',v') + f_s^{\,2}(u',v'),$$ (10.110)

where ${\cal I}_0$ is the illumination intensity when the aperture is completely transparent. Here, use has been made of the standard results $\langle \cos^2(\omega\,t-k\,R-\phi)\rangle =\langle \sin^2(\omega\,t-k\,R-\phi)\rangle = 1/2$ and $\langle \cos(\omega\,t-k\,R-\phi)\,\sin(\omega\,t-k\,R-\phi)\rangle=0$.