The Timpson Lyceum

1. (2 pts ••) Monochromatic light of wavelength $\lambda=633\,\text{nm}$ from a helium–neon laser passes through a single slit of width $a=0.12\,\text{mm}$ and forms a diffraction pattern on a screen $L=2.5\,\text{m}$ away. The angles are small enough that $\sin\theta\approx\tan\theta$ .

(a) Find the angle to the first dark fringe ( $m=1$ ).

(b) Find the distance on the screen from the center of the pattern to the first dark fringe.

(c) Find the full width of the central bright band (the distance between the first dark fringes on either side).

Solution

1.

(a) First dark fringe: $a\sin\theta=\lambda$ , so $\sin\theta=\dfrac{\lambda}{a}=\dfrac{633\times10^{-9}}{0.12\times10^{-3}}=5.28\times10^{-3}$ , giving $\theta\approx0.30^\circ$ ( $5.3\times10^{-3}\,\text{rad}$ ).

(b) With small angles, $y_1=L\tan\theta\approx L\sin\theta=(2.5)(5.28\times10^{-3})\approx1.3\times10^{-2}\,\text{m}=13\,\text{mm}$ .

(c) The central band runs from the $m=-1$ to the $m=+1$ dark fringe, so its full width is $2y_1\approx2(13\,\text{mm})\approx26\,\text{mm}$ .

2. (2 pts ••) A single slit of width $a=0.15\,\text{mm}$ is illuminated by light of wavelength $\lambda=600\,\text{nm}$ . The intensity as a function of the diffraction angle is

I=I_0\left[\frac{\sin(\beta/2)}{\beta/2}\right]^2,\qquad \beta=\frac{2\pi a}{\lambda}\sin\theta .

(a) Find the angle at which $\beta/2=\pi/2$ (i.e. halfway, in the phase variable, to the first minimum).

(b) Find the intensity at that angle as a fraction of the central intensity $I_0$ .

(c) The first secondary maximum occurs near $\beta/2=1.43\pi$ . Find its intensity as a fraction of $I_0$ , and comment on how much dimmer it is than the central peak.

Solution

2.

(a) $\beta/2=\dfrac{\pi a}{\lambda}\sin\theta=\dfrac{\pi}{2}$ requires $\sin\theta=\dfrac{\lambda}{2a}=\dfrac{600\times10^{-9}}{2(0.15\times10^{-3})}=2.0\times10^{-3}$ , so $\theta\approx0.11^\circ$ .

(b) $\dfrac{I}{I_0}=\left[\dfrac{\sin(\pi/2)}{\pi/2}\right]^2=\left(\dfrac{1}{\pi/2}\right)^2=\left(\dfrac{2}{\pi}\right)^2\approx0.41$ . The intensity has already dropped to about $41\%$ of the peak here.

(c) At $\beta/2=1.43\pi$ : $\dfrac{I}{I_0}=\left[\dfrac{\sin(1.43\pi)}{1.43\pi}\right]^2\approx\left(\dfrac{-0.968}{4.49}\right)^2\approx0.047$ . The first side lobe carries only about $4.7\%$ of the central intensity — the central peak dominates by more than a factor of $20$ .

3. (2 pts ••) A diffraction grating has $600$ lines per millimetre and is illuminated at normal incidence by sodium light, $\lambda=589\,\text{nm}$ .

(a) Find the slit spacing $d$ .

(b) Find the diffraction angles of the first- and second-order bright fringes.

(c) What is the highest order that can be observed with this grating and wavelength? Explain.

Solution

3.

(a) $d=\dfrac{1\,\text{mm}}{600}=1.667\times10^{-3}\,\text{mm}=1.67\times10^{-6}\,\text{m}$ ( $1667\,\text{nm}$ ).

(b) $\sin\theta_m=\dfrac{m\lambda}{d}$ .

$m=1$ : $\sin\theta_1=\dfrac{589\times10^{-9}}{1.667\times10^{-6}}=0.3534$ , so $\theta_1\approx20.7^\circ$ .

$m=2$ : $\sin\theta_2=0.7068$ , so $\theta_2\approx45.0^\circ$ .

(c) The largest possible order has $\sin\theta\le1$ : $m_{\max}=\dfrac{d}{\lambda}=\dfrac{1.667\times10^{-6}}{589\times10^{-9}}=2.83$ . Since $m$ must be an integer, the highest observable order is $m=2$ . (Order $3$ would need $\sin\theta=1.06>1$ , which is impossible.)

4. (2 pts ••) The sodium doublet consists of two closely spaced lines at $\lambda_1=589.00\,\text{nm}$ and $\lambda_2=589.59\,\text{nm}$ . The resolving power of a grating is $R=Nm$ , where $N$ is the number of illuminated slits and $m$ is the order.

(a) What resolving power is needed to just separate the two lines?

(b) Working in second order ( $m=2$ ), how many slits must be illuminated?

(c) If the grating has $600$ lines per millimetre, what width of the grating must the beam cover to achieve this?

Solution

4.

(a) $R=\dfrac{\lambda}{\Delta\lambda}=\dfrac{589.00}{0.59}\approx1.0\times10^{3}$ (about $998$ ).

(b) $N=\dfrac{R}{m}=\dfrac{998}{2}\approx5.0\times10^{2}$ slits (about $499$ ).

(c) At $600$ lines per mm, $N$ slits span $\dfrac{N}{600}\,\text{mm}=\dfrac{499}{600}\approx0.83\,\text{mm}$ . A beam covering roughly $0.83\,\text{mm}$ of the grating suffices.

5. (2 pts ••) A camera lens and a telescope are limited by the diffraction of a circular aperture; the Rayleigh criterion gives the smallest resolvable angular separation as $\theta=1.22\,\lambda/D$ . Take $\lambda=550\,\text{nm}$ throughout.

(a) A space telescope has a primary mirror of diameter $D=2.4\,\text{m}$ . Find its diffraction-limited angular resolution, in radians and in arcseconds.

(b) Aimed at the Moon ( $3.84\times10^{8}\,\text{m}$ away), what is the smallest surface separation it could resolve?

(c) The dark-adapted human eye has a pupil of diameter $D=3.0\,\text{mm}$ . How far away can two headlights separated by $1.5\,\text{m}$ be and still be resolved as two sources?

Solution

5.

(a) $\theta=\dfrac{1.22\lambda}{D}=\dfrac{1.22(550\times10^{-9})}{2.4}=2.80\times10^{-7}\,\text{rad}$ . In arcseconds: $(2.80\times10^{-7})\left(\dfrac{180}{\pi}\right)(3600)\approx0.058''$ .

(b) $s=\theta\,L=(2.80\times10^{-7})(3.84\times10^{8})\approx1.1\times10^{2}\,\text{m}$ (about $107\,\text{m}$ ).

(c) $\theta=\dfrac{1.22(550\times10^{-9})}{3.0\times10^{-3}}=2.24\times10^{-4}\,\text{rad}$ . The headlights subtend this angle at distance $L=\dfrac{s}{\theta}=\dfrac{1.5}{2.24\times10^{-4}}\approx6.7\times10^{3}\,\text{m}\approx6.7\,\text{km}$ . (Real eyes do far worse; diffraction is only the theoretical limit.)

6. (2 pts ••) X-rays are directed at a crystal of table salt (NaCl), whose atomic planes are separated by $d=0.282\,\text{nm}$ . A strong first-order reflection ( $m=1$ ) is observed when the beam makes a grazing angle of $\theta=15.9^\circ$ with the planes. Bragg's law is $2d\sin\theta=m\lambda$ .

(a) Find the wavelength of the X-rays.

(b) Find the grazing angle at which the second-order reflection ( $m=2$ ) appears.

(c) What is the highest order that can be observed with these X-rays and this spacing?

Solution

6.

(a) $2d\sin\theta=m\lambda$ with $m=1$ : $\lambda=2d\sin\theta=2(0.282\times10^{-9})\sin15.9^\circ=2(0.282\times10^{-9})(0.2740)\approx1.55\times10^{-10}\,\text{m}=0.154\,\text{nm}$ . (This is essentially the copper $K_\alpha$ line.)

(b) $m=2$ : $\sin\theta=\dfrac{2\lambda}{2d}=\dfrac{2(0.1545\times10^{-9})}{2(0.282\times10^{-9})}=\dfrac{0.1545}{0.282}=0.5480$ , so $\theta\approx33.2^\circ$ .

(c) $m_{\max}=\dfrac{2d}{\lambda}=\dfrac{2(0.282\times10^{-9})}{0.1545\times10^{-9}}=3.65$ , so the highest observable order is $m=3$ (at $\sin\theta=0.822$ , $\theta\approx55.3^\circ$ ).

7. (2 pts ••) A double-slit apparatus has slits of width $a=0.050\,\text{mm}$ with their centers separated by $d=0.25\,\text{mm}$ , illuminated by light of wavelength $\lambda=550\,\text{nm}$ . The observed pattern is the two-slit interference fringes modulated by the single-slit diffraction envelope.

(a) At what interference orders $m$ are bright fringes missing because they fall on a zero of the single-slit envelope? Identify the lowest such order.

(b) How many bright interference fringes appear within the central diffraction maximum (counting the central fringe and those on both sides, but excluding any that land exactly on the first diffraction minimum)?

Solution

7.

(a) An interference maximum at $d\sin\theta=m\lambda$ is missing when it coincides with a diffraction minimum $a\sin\theta=p\lambda$ . Dividing, the missing orders satisfy $\dfrac{m}{p}=\dfrac{d}{a}=\dfrac{0.25}{0.050}=5$ . So $m=5p$ : orders $m=\pm5,\pm10,\dots$ are missing. The lowest is $m=5$ .

(b) The central diffraction maximum extends out to the first diffraction minimum at $\sin\theta=\lambda/a$ . Interference maxima lie at $\sin\theta=m\lambda/d$ , so they fall inside the central envelope when $\dfrac{m\lambda}{d}<\dfrac{\lambda}{a}$ , i.e. $|m|<\dfrac{d}{a}=5$ . Thus $m=-4,-3,\dots,3,4$ : that is $9$ bright fringes. (The $m=\pm5$ fringes land exactly on the diffraction minimum and are suppressed.)

Extra Credit (3 pts •••)

Derive the single-slit intensity pattern from scratch using phasor (amplitude) integration.

Consider a slit of width $a$ illuminated by a plane wave. Divide the slit into infinitesimal strips of width $dy$ at position $y$ (with $-a/2\le y\le a/2$ ). Each strip radiates a wavelet toward a distant screen at angle $\theta$ ; relative to the strip at $y=0$ , the wavelet from the strip at $y$ has an extra path length $y\sin\theta$ , hence a phase $\phi(y)=ky\sin\theta$ with $k=2\pi/\lambda$ . Each strip contributes an amplitude $dE_0=(E_0/a)\,dy$ , where $E_0$ is the total amplitude that would arrive with all strips in phase.

(a) Write the total (complex) field at the screen as the integral $E=\displaystyle\int_{-a/2}^{a/2}\frac{E_0}{a}\,e^{i\,ky\sin\theta}\,dy$ , and carry out the integration.

(b) Show that the magnitude reduces to

|E|=E_0\,\frac{\sin(\beta/2)}{\beta/2},\qquad \beta=\frac{2\pi a}{\lambda}\sin\theta ,

and hence that the intensity (proportional to $|E|^2$ ) is $I=I_0\left[\dfrac{\sin(\beta/2)}{\beta/2}\right]^2$ .

(c) From this result, show that the minima occur at $a\sin\theta=m\lambda$ for nonzero integers $m$ , and use a limit to verify that $I\to I_0$ as $\theta\to0$ .

Solution

Extra Credit

(a) The total field is the sum (integral) of the strip contributions:

E=\int_{-a/2}^{a/2}\frac{E_0}{a}\,e^{i\,ky\sin\theta}\,dy .

Let $\alpha=k\sin\theta=\dfrac{2\pi}{\lambda}\sin\theta$ (a constant for a given angle). Then

E=\frac{E_0}{a}\int_{-a/2}^{a/2} e^{i\alpha y}\,dy=\frac{E_0}{a}\cdot\frac{e^{i\alpha y}}{i\alpha}\Bigg|_{-a/2}^{a/2}=\frac{E_0}{a}\cdot\frac{e^{i\alpha a/2}-e^{-i\alpha a/2}}{i\alpha}.

(b) Using $e^{ix}-e^{-ix}=2i\sin x$ ,

E=\frac{E_0}{a}\cdot\frac{2i\sin(\alpha a/2)}{i\alpha}=\frac{E_0}{a}\cdot\frac{2\sin(\alpha a/2)}{\alpha}=E_0\,\frac{\sin(\alpha a/2)}{\alpha a/2}.

Define $\beta=\alpha a=ka\sin\theta=\dfrac{2\pi a}{\lambda}\sin\theta$ , so $\alpha a/2=\beta/2$ and

E=E_0\,\frac{\sin(\beta/2)}{\beta/2}.

This is real for our phase reference, so $|E|=E_0\,\dfrac{\sin(\beta/2)}{\beta/2}$ (up to sign). Since intensity $\propto|E|^2$ , dividing by the central value $I_0$ (where the ratio is $1$ ),

I=I_0\left[\frac{\sin(\beta/2)}{\beta/2}\right]^2 .

(c) Minima occur where $E=0$ with the peak not also vanishing — that is where $\sin(\beta/2)=0$ but $\beta/2\ne0$ , i.e. $\dfrac{\beta}{2}=m\pi$ for nonzero integer $m$ . Then

\frac{\pi a}{\lambda}\sin\theta=m\pi\quad\Longrightarrow\quad a\sin\theta=m\lambda,

the standard single-slit dark-fringe condition.

As $\theta\to0$ , $\beta/2\to0$ , and using $\displaystyle\lim_{x\to0}\frac{\sin x}{x}=1$ ,

\lim_{\theta\to0}I=I_0(1)^2=I_0,

so the pattern peaks at the center, as expected. $\blacksquare$

Chapter 36 — Diffraction

Chapter 36 — Diffraction

Extra Credit (3 pts •••)

Extra Credit