Diffraction¶

Single-Slit Diffraction¶

The description of the singles-slit diffraction experiment

We considedr the diffraction parrern of plane waves of light of wavelength $\lambda$ that are diffracted by a single long , narrow slit of width $a$ (aperture) in an otherwise opaque screen $B$. Waves from different points within the slit undergo interference and produce a diffraction pattern of bright and dark fringes on screen $C$.

Fraunhofer vs Fresnel Diffraction

At large screen distance $(D>\frac{a^2}{\lambda})$, the shape of the projected pattern is independent of $D$. This is Fraunhofer or far-field diffraction (远场衍射).

At small $D$, however, both the size and shape of the diffraction pattern changes with distance. This phenomenon is known as Fresnel or near-field diffraction (近场衍射).

Locations of the Dark Fringes¶

First Dark Fringes¶

minimg

Paring

To find the dark fringes, we shall use a simplifying strategy that involves pairing up all the rays coming through the slit and then finding what conditions cause the wavelets of the rays in each pair to cancel each other.

We want the wavelets along these two rays to cancel each other when they arrive at $P1$. So they should be out of phase by $\frac{\lambda}{2}$ when they reach $P_1$. Then any similar pairing of rays from the two zones will give cancellation at the same point $P_1$, because $D ≫ a$.

For $D >> a$, we can approximate rays $r_1$ and $r_2$ as being parallel, at angle θ to the central axis. Path Length Difference：

\[\begin{align*} \frac{a}{2}\sin\theta=\frac{\lambda}{2} \end{align*}\]

The angle $\theta$ of the first dark fringe above and (by symmetry) below the central axis is, therefore, determined by

\[\begin{align*} \sin\theta=\frac{\lambda}{a} \end{align*}\]

Narrowing the Slit

If we narrow the slit such that $a=\lambda$ , the angle $\theta$ at which the first dark fringes appear increases to $90^{\circ}$ . This means that bright fringe must then cover the entire viewing screen.

Second Dark Fringes¶

minimg Divide the slit into four zones of equal width $\frac{a}{4}$. To produce the second dark fringe at $P_2$, the path length difference between $r_1$ and $r_2$, that between $r_2$ and $r_3$, and that between $r_3$ and $r_4$ must all be equal to $\frac{\lambda}{2}$.

For $D \gg a$, we can approximately these four rays as being parallel, at angle $\theta$ to the central axis.

\[\begin{align*} &\frac{a}{4}\sin\theta=\frac{\lambda}{2}\\ &\Rightarrow a\sin\theta=2\lambda \end{align*}\]

All Dark Fringes¶

\[\begin{align*} &\frac{a}{2m}\sin\theta=\frac{\lambda}{2}\\ a\sin\theta=&m\lambda \quad \text{for} \quad m=1,2,3,\dots \end{align*}\]

Electric Field and Intensity¶

To find an expression for the intensity at an arbitrary point $P$ on the viewing screen, corresponding to a particular small angle $\theta$, we need to divide the slit into $N$ zones of equal widths $\Delta x=\frac{a}{N}$ small enough that we can assume each zone acts as a source of Huygens wavelets. We then add the phasors for the wavelets, which form a geometric series (notice $r_{i+1}-r_i=\Delta x\sin\theta$)

\[\begin{align*} \tilde{E}_{\theta}=&\frac{E_0}{N} e^{-i\omega t} \sum_{i=1}^N e^{ikr_i}\\ =&\frac{E_0}{N}e^{-i\omega t}e^{ikr_1}\times \left[ 1+ e^{ik(r_2-r_1)}+\cdots+e^{ik(r_N-r_1)} \right] \end{align*}\]

The arc of phasors represents the wavelets that reach an arbitrary point $P$ on the viewing screen, corresponding to a small angle $\theta$. The amplitude $E_{\theta}$ of the resultant wave at $P$ is the vector sum of these phasors. In the limit of $N \rightarrow \infty$, the arc of phasors approaches the arc of a circle.

minimg

We use a graphic construction

\[\begin{align*} &\frac{I(\theta)}{I_{max}}=\frac{E_{\theta}^2}{E_m^2}=\left[ \frac{\left( 2R\sin\alpha \right)^2}{\left( 2R\alpha \right)^2} \right]=\left[ \frac{\sin\alpha}{\alpha} \right]^2\\ &\alpha = \frac{\pi}{\lambda}(a\sin\theta)\\ &\phi = 2\alpha= (\frac{2\pi}{\lambda})(a\sin\theta) \end{align*}\]

Noitce $\phi=2\alpha$ is the phase difference bewteen the rays from the top and bottom of the entire slit.

Plot the Relative Intensity- $\theta$ with different a/$\lambda$

minimg As the slit width increases (relative to the wavelength), - the width of the central diffraction maximum increases;
- the relative height of the secondary maxima increases, but the width of the secondary maxima decreases;

The geometric series of phasors from sum to integral is

\[\begin{align*} \tilde{E}_{\theta}=&\frac{E_0}{N} e^{-i\omega t} \sum_{i=1}^N e^{ikr_i}\\ =&\frac{E_0 \Delta x}{a} e^{-i\omega t} \sum_{i=1}^N e^{ikr_i}\\ =&\lim_{N\rightarrow \infty} E_0 e^{-i\omega t}\frac{1}{a}\int_0^a e^{ik(r_1+x\sin\theta)}\,\mathrm{d}x\\ =&\frac{E_0}ae^{i(kr_1-\omega t)}\int_0^ae^{ik_xx}dx\\ =&e^{i(kr_1-\omega t+k_xa/2)}\frac{E_0}a\int_{-a/2}^{a/2}e^{ik_xx}dx, \qquad\text{where } k_x=k\sin\theta \end{align*}\]

This leads to the result we have derived

\[\begin{align*} \boxed{I_{sq}(\theta)=I_{max}\left[ \frac{\sin\alpha}{\alpha} \right]^2} \end{align*}\]

Fourier Methods in Diffraction Theory¶

Review Fourier Transform

The Fourier transform of a one-dimensional function $f(x)$ is defined as

\[\begin{align*} F(k_x)=\int_{\infty}^{\infty} f(x)e^{-ik_x x}\,\mathrm{d}x \end{align*}\]

whose inverse transform is

\[\begin{align*} f(x)=\frac{1}{2\pi}\int_{\infty}^{\infty}F(k_x)e^{ik_x x}\,\mathrm{d}k_x \end{align*}\]

The Fourier Method for the Single Slit¶

Consider the long slit in the y-direction, illuminated by a plane wave. Assuming that there are no phase or amplitude variations across the aperture, the one-dimensional aperture function has the form of a square pulse(方波):

\[\begin{align*} E_{sq}(x)=\left\{ \begin{array}{cc} E_0 & |x|\le \frac{a}{2}\\ 0 & |x|>\frac{a}{2} \end{array} \right. \end{align*}\]

The Fourier transform of $E_{sq}(x)$ is

\[\begin{align*} \tilde{E}_{sq}(k_x)&=\frac{1}{a}\int_{-\infty}^{\infty}E_{sq}(x)e^{-ik_x x}\,\mathrm{d}x\\ &=\frac{E_0}{a}\int_{-\frac{a}{2}}^{\frac{a}{2}} e^{-ik_x x}\,\mathrm{d}x\\ &=-\frac{E_0}{a}\frac{1}{ik_x}\left( e^{-ik_x \frac{a}{2}} - e^{ik_x \frac{a}{2}}\right)\\ &\overset{\alpha = \frac{a}{2}k_x}{=\!=\!=\!=}\frac{E_0}{a} \frac{a}{2}\frac{2i\sin\alpha}{i\alpha}\\ &=E_0 \frac{\sin\alpha}{\alpha} \end{align*}\]

where $\alpha=\frac{a}{2}k_x=\frac{a}{2}k\sin\theta$.

This leads to the result we have derived

\[\begin{align*} \boxed{I_{sq}(\theta)=I_{max}\left[ \frac{\sin\alpha}{\alpha} \right]^2} \end{align*}\]

The key message is that the field distribution in the Fraunhofer diffraction pattern is the Fourier transform of the field distribution across the aperture (打到屏幕的光分布是过小孔的光分布的傅里叶变换).

The Fourier Method for the Double Silt¶

Consider the double slit with its long side in the y -direction, illuminated by a plane wave. Assuming the one-dimensional aperture function along the x direction has the following form

\[\begin{align*} E_{ds}(x)=\left\{\begin{array}{cc} 0 & |x|<\frac{d-a}{2}\\ E_0 & \frac{d-a}{2}\le |x| \le \frac{d+a}{2} \\ 0 & |x|>\frac{d+a}{2} \end{array}\right. \end{align*}\]

The Fourier transform of $E_{ds}(x)$ is

\[\begin{align*} \tilde{E}_{ds}(k_x)&=\int_{\infty}^{\infty}E_{ds}(x)e^{-ik_x x}\,\mathrm{d}x\\ &=\int_{-\frac{d+a}{2}}^{-\frac{d-a}{2}} E_0 e^{-ik_x x}\,\mathrm{d} x + \int_{\frac{d-a}{2}}^{\frac{d+a}{2}} E_0 e^{-ik_x x}\,\mathrm{d} x\\ &\overset{x^{\prime}=x+\frac{d}{2}, x^{\prime\prime}=x-\frac{d}{2}}{=\!=\!=\!=\!=\!=\!}\int_{-\frac{a}{2}}^{\frac{a}{2}} E_0 e^{-ik_x (-\frac{d}{2}+x^{\prime})}\,\mathrm{d}x^{\prime} + \int_{-\frac{a}{2}}^{\frac{a}{2}} E_0 e^{-ik_x (\frac{d}{2}+x^{\prime\prime})}\,\mathrm{d}x^{\prime\prime}\\ &=\left( e^{ik_x \frac{d}{2}} + e^{-ik_x \frac{d}{2}}\right)\int_{-\frac{a}{2}}^{\frac{a}{2}} E_0 e^{-ik_x x^{\prime}}\,\mathrm{d} x^{\prime} \end{align*}\]

\[\begin{align*} &\left( e^{ik_x \frac{d}{2}} + e^{-ik_x \frac{d}{2}}\right) \text{ is double-slit interference with $a << \lambda$}\\ &\int_{-\frac{a}{2}}^{\frac{a}{2}} E_0 e^{-ik_x x^{\prime}}\,\mathrm{d} x^{\prime} \text{ is single-slit diffraction} \end{align*}\]

The interference pattern can be understood by a convolution theorem(卷积) for Fourier transformation: The transform of the convolution of two function is the product of their transforms. ($\mathcal{F} [f\circ g]=\mathcal{F}[f]\times \mathcal{F}[g]$) .

The Young's double-slit interference result (with negligible slit width that is $a << \lambda$) as the Fourier transform of an aperture function

\[\begin{align*} h(x)=\delta\left(x+\frac{d}{2}\right)+\delta\left(x-\frac{d}{2}\right) \end{align*}\]

\[\begin{align*} \mathcal{F}[h(x)]=&\int_{-\infty}^{\infty} h(x)e^{-ikx}\,\mathrm{d}x\\ =&e^{ik\frac{d}{2}}+e^{-ik\frac{d}{2}}\\ =&2\cos\left(k\frac{d}{2}\right) \end{align*}\]

Notice that

\[\begin{align*} \delta (x) =&\left\{ \begin{array}{cc} 1 & x=0\\ 0 & otherwise \end{array} \right.\\ f(a)=&\int_{-\infty}^{\infty}f(x)\delta (x-a)\,\mathrm{d}x \end{align*}\]

Therefore, we can express the double-slit aperture function and its Fourier transform as

\[\begin{align*} E_{ds}(x)=&\int_{-\infty}^{\infty}E_{sq}(x^{\prime})h(x-x^{\prime})\,\mathrm{d}x^{\prime}\\ \mathcal{F}\left[E_{ds}(x)\right]=&\mathcal{F}\left[E_{sq}(x)\right]\mathcal{F}\left[h(x)\right] \end{align*}\]

Formally, with diffraction effects taken into account, the intensity of a double-slit interference pattern is

\[\begin{align*} &\boxed{I(\theta)=I_0\left(\frac{\sin\alpha}{\alpha}\right)^2 \cos^2\beta}\\ &\beta=\frac{\delta_2}{2}=\frac{\pi}{\lambda}d\sin\theta\\ &\alpha=\frac{\pi}{\lambda}a\sin\theta \end{align*}\]

The cause of two factors in the intensity

(1) The interference factor $\cos2\beta$ is due to the interference between two slits with slit separation d.
(2) The diffraction factor $[(\sin \alpha)/\alpha]2$ is due to diffraction by a single slit of width a.

The Double-Slit Interference with width > wavelength

In the double-slit interference experiments , we implicitly $a << \lambda$ ($a$ is slit width and $\lambda$ is wavelength) . For such narrow slits, the central maximum of the diffraction pattern of either slit covers the entire viewing screen.
For relatively wide slits, the intensities of the fringes produced by double-slit interference are modified by diffraction of the light passing through each slit.

Plot the Relative intensity- $\theta$

(a) The intensity plot to be expected in a double-slit bright fringes.double-slit interference experiment with vanishinglynarrow slits.
(b) The intensity plot for diffraction by a typical slit of width a (not vanishingly narrow).

The curve of (b) acts as an envelope, limiting the intensity of the double-slit fringes in (a). Note that the first minima of the diffraction pattern of (b) eliminate the double-slit fringes that would occur near 12° in ©.

The analysis of the minimum intensity

The first minimum occurs where the phase differencebetween the two slits $(N = 2)$ is

\[\begin{align*} \delta_2=\frac{2\pi}{\lambda}d\sin\theta=\pi \end{align*}\]

The first minimum of the envelope occurs where the phase difference between one edge and the center of a single slit is

\[\begin{align*} \alpha=\frac{2\pi}{\lambda}\sin\theta=\pi \end{align*}\]

Therefore, one can determine $\frac{d}{a}$ by counting fringes. In both cases, the larger the length ($d$ or $a$) is, the smaller the $\theta$ ($k_x=k\in\theta$) is.

Diffraction by a Circular Aperture¶

The first minimum for the diffraction pattern of a circular aperture of diameter $a$ is located :

\[\begin{align*} \boxed{\sin\theta=1.22\frac{\lambda}{a}} \end{align*}\]