Physics · Book 4 · Bachelor Year 2

University Physics — Year 2

University Physics — Year 2 · Bachelor Year 2

21Multiple-Wave Interference and Gratings

Tilt a compact disc under a lamp and it throws a rainbow: its spiral of pits, a micrometre and a half apart, is a grating — thousands of tiny reflecting strips, each sending back a copy of the wave, all interfering. Two waves give the soft cos⁡2\cos^2 fringes of the last chapters; a thousand give fringes a thousand times narrower, sharp enough to separate two wavelengths that differ by a part in a hundred thousand. That is the instrument with which the composition of the Sun and the speed of galaxies were read, and which sits in every spectrometer. This chapter adds up NN waves, derives the grating equation, its dispersion and its resolving power, and meets the other multiple-wave device, the Fabry–Pérot cavity that will become the laser’s resonator.

A compact disc in sunlight: its spiral of pits, 1.6\, µ m apart, is a reflection grating, and each order spreads the white light into a spectrum.
A compact disc in sunlight: its spiral of pits, 1.6 µm1.6\,\text{µ}\mathrm{m} apart, is a reflection grating, and each order spreads the white light into a spectrum.

21.1 Interference of NN waves

Theorem 21.1 (NN equal waves in arithmetic phase progression)

NN coherent waves of equal amplitude aa, each lagging the previous by the phase φ\varphi, superpose into a wave of amplitude A(φ)A(\varphi) and intensity

A=a sin⁡(Nφ/2)sin⁡(φ/2),I(φ)=I0 sin⁡2(Nφ/2)sin⁡2(φ/2):A = a\,\frac{\sin(N\varphi/2)}{\sin(\varphi/2)} , \qquad I(\varphi) = I_0\,\frac{\sin^2(N\varphi/2)}{\sin^2(\varphi/2)} :

principal maxima of intensity N2I0N^2I_0 where φ=2πp\varphi = 2\pi p (all waves in phase), of angular half-width 2π/N2\pi/N in φ\varphi (the first zeros are at φ=2πp±2π/N\varphi = 2\pi p \pm 2\pi/N), separated by N−2N - 2 secondary maxima no higher than about 4.5%4.5\% of the principal ones for large NN. The more waves, the sharper and brighter the peaks: the width shrinks as 1/N1/N, the height grows as N2N^2, the energy (their product) as NN.

Proof. Complex amplitudes aeikφa\eu^{\iu k\varphi}, k=0,…,N−1k = 0, \dots, N - 1: a geometric series, a(1−eiNφ)/(1−eiφ)=aei(N−1)φ/2sin⁡(Nφ/2)/sin⁡(φ/2)a(1 - \eu^{\iu N\varphi})/(1 - \eu^{\iu\varphi}) = a\eu^{\iu(N - 1)\varphi/2}\sin(N\varphi/2) /\sin(\varphi/2). Zeros where sin⁡(Nφ/2)=0\sin(N\varphi/2) = 0 but sin⁡(φ/2)≠0\sin(\varphi/2) \ne 0: φ=2πm/N\varphi = 2\pi m/N, mm not a multiple of NN. In the phasor picture the NN arrows close into a regular polygon at the zeros and align at the principal maxima. ∎

Left: the phasor sum of five equal waves — a regular fan whose chord is the resultant; it closes into a pentagon, and vanishes, at = 2π/5. Right: the intensity for 2, 5 and 12 waves, normalized to its peak: the principal maxima sharpen as 1/N.
Left: the phasor sum of five equal waves — a regular fan whose chord is the resultant; it closes into a pentagon, and vanishes, at φ=2π/5\varphi = 2\pi/5. Right: the intensity for 22, 55 and 1212 waves, normalized to its peak: the principal maxima sharpen as 1/N1/N.

21.2 The diffraction grating

Proposition 21.2 (Grating equation)

A grating is a set of NN identical, equidistant, parallel slits (or grooves) of period aa. A plane wave incident at the angle θ0\theta_0 from the normal is re-emitted by each slit; in the direction θ\theta the waves from neighbouring slits have the path difference δ=a(sin⁡θ−sin⁡θ0)\delta = a(\sin\theta - \sin\theta_0), and they reinforce in the directions

a (sin⁡θp−sin⁡θ0)=pλ,p=0,±1,±2,…a\,(\sin\theta_p - \sin\theta_0) = p\lambda , \qquad p = 0, \pm1, \pm2, \dots

— the orders of the grating. The zero order is the undeviated beam for every wavelength; each other order spreads the wavelengths, with the angular dispersion

 ⁣dθ ⁣dλ=pacos⁡θ,\frac{\dd\theta}{\dd\lambda} = \frac{p}{a\cos\theta} ,

larger for a finer grating and a higher order; the maximum order is ∣p∣≤a(1+∣sin⁡θ0∣)/λ|p| \le a(1 + |\sin\theta_0|)/\lambda. A reflection grating (grooves ruled on a mirror) obeys the same equation with the reflected angles.

Proof. The phase between successive slits is φ=2πδ/λ\varphi = 2\pi\delta/\lambda; the principal maxima of Theorem 21.1 are at δ=pλ\delta = p\lambda. Differentiate asin⁡θ=pλ+asin⁡θ0a\sin\theta = p\lambda + a\sin\theta_0 at fixed pp, θ0\theta_0. ∎

Left: the grating — neighbouring slits re-emit with the path difference a at normal incidence. Right: on a screen, the white zero order and the spectra of the orders ±1, ±2, violet nearest the centre, wider (and eventually overlapping) as p grows.
Left: the grating — neighbouring slits re-emit with the path difference asin⁡θa\sin\theta at normal incidence. Right: on a screen, the white zero order and the spectra of the orders ±1\pm1, ±2\pm2, violet nearest the centre, wider (and eventually overlapping) as pp grows.

Example 21.3 (A classroom grating and a CD)

A grating of 600600 lines per millimetre (a=1.67 µma = 1.67\,\text{µ}\mathrm{m}) at normal incidence sends 500 nm500\,\mathrm{nm} light to sin⁡θ=0.3\sin\theta = 0.3, θ=17.5∘\theta = 17.5{}^{\circ} in the first order, 36.9∘36.9{}^{\circ} in the second, and has no third order (3×0.3>13 \times 0.3 > 1); its first-order spectrum runs from 13.9∘13.9{}^{\circ} (400 nm400\,\mathrm{nm}) to 24.8∘24.8{}^{\circ} (700 nm700\,\mathrm{nm}). A compact disc, a=1.6 µma = 1.6\,\text{µ}\mathrm{m}, is such a grating in reflection — hence the rainbow, and the fact that a DVD (a=0.74 µma = 0.74\,\text{µ}\mathrm{m}) spreads it wider. The second-order red (700 nm700\,\mathrm{nm} at sin⁡θ=0.84\sin\theta = 0.84) overlaps the third-order violet: orders overlap beyond the first, a nuisance cured with filters or a prism.

21.3 Resolving power

Proposition 21.4 (Resolving power of a grating)

Two close wavelengths λ\lambda and λ+Δλ\lambda + \Delta\lambda are just resolved in order pp when the principal maximum of one falls on the first zero of the other (Rayleigh’s criterion), which happens for

λΔλ=pN:\frac\lambda{\Delta\lambda} = pN :

the resolving power is the order times the number of lines lit. It can also be written W(sin⁡θ−sin⁡θ0)/λW(\sin\theta - \sin\theta_0)/\lambda with W=NaW = Na the width of the grating: the largest path difference across the grating, in wavelengths — which is why large spectrographs use gratings tens of centimetres wide and work at grazing angles.

Proof. The maximum of order pp for λ+Δλ\lambda + \Delta\lambda is at φ=2πp\varphi = 2\pi p for that wavelength, i.e. at δ=p(λ+Δλ)\delta = p(\lambda + \Delta\lambda); the first zero of the maximum for λ\lambda is at δ=pλ+λ/N\delta = p\lambda + \lambda/N. Equating: pΔλ=λ/Np\Delta\lambda = \lambda/N. Then pN=Na(sin⁡θ−sin⁡θ0)/λpN = Na(\sin\theta - \sin\theta_0)/\lambda. ∎

Example 21.5 (Separating the sodium lines)

The doublet, Δλ=0.6 nm\Delta\lambda = 0.6\,\mathrm{nm} at 589 nm589\,\mathrm{nm}, needs R=1000R = 1000: a thousand lines in the first order, 2 mm2\,\mathrm{mm} of a 600 lines/mm600\,\mathrm{lines}/\mathrm{mm} grating. Its own line width, 0.02 nm0.02\,\mathrm{nm}, needs R=30 000R = 30\,000: a 5 cm5\,\mathrm{cm} grating in the first order, or 2.5 cm2.5\,\mathrm{cm} in the second — and a 10 cm10\,\mathrm{cm} grating of 1200 lines/mm1200\,\mathrm{lines}/\mathrm{mm} in the second order reaches 240 000240\,000, resolving 0.0025 nm0.0025\,\mathrm{nm}: the width of a line broadened by the thermal motion of the atoms, and the Doppler shift of a star moving at 1 km/s1\,\mathrm{km}/\mathrm{s}.

Rayleigh’s criterion: the peak of one wavelength on the first zero of the other; the sum (dashed) shows a dip of about 20\% between two resolved lines.
Rayleigh’s criterion: the peak of one wavelength on the first zero of the other; the sum (dashed) shows a dip of about 20%20\% between two resolved lines.

21.4 The Fabry–Pérot cavity

Proposition 21.6 (Multiple-wave interference by division of amplitude)

Two parallel partially reflecting mirrors (intensity reflectance RR) a distance LL apart transmit a wave at normal incidence only in so far as the waves that have bounced 0,1,2,…0, 1, 2, \dots times between them add in phase: with φ=4πL/λ\varphi = 4\pi L/\lambda the round-trip phase, the transmitted intensity is (Airy’s function)

ItI0=11+Fsin⁡2(φ/2),F=4R(1−R)2:\frac{I_t}{I_0} = \frac1{1 + F\sin^2(\varphi/2)} , \qquad F = \frac{4R}{(1 - R)^2} :

sharp peaks of full transmission at 2L=pλ2L = p\lambda (the resonances of the cavity, spaced in frequency by the free spectral range c/2Lc/2L), of relative width 1/F1/\mathcal F with the finesse F=πR/(1−R)\mathcal F = \pi \sqrt R/(1 - R) — 3030 for R=0.9R = 0.9, 300300 for R=0.99R = 0.99. It is the grating’s equal in resolving power (pFp\mathcal F with p=2L/λ∼105p = 2L/\lambda \sim 10^5, hence R∼107R \sim 10^7) and, closed on itself, the resonator of the laser (Chapter 23).

Proof. Each round trip multiplies the amplitude by ReiφR\eu^{\iu\varphi} (up to a constant phase); the transmitted amplitude is t2a∑(Reiφ)k=t2a/(1−Reiφ)t^2a\sum(R\eu^{\iu\varphi})^k = t^2a/(1 - R\eu^{\iu\varphi}) with t2=1−Rt^2 = 1 - R; its squared modulus is (1−R)2/(1−2Rcos⁡φ+R2)=(1−R)2/[(1−R)2+4Rsin⁡2(φ/2)](1 - R)^2/ (1 - 2R\cos\varphi + R^2) = (1 - R)^2/[(1 - R)^2 + 4R\sin^2(\varphi/2)]. Peaks at φ=2πp\varphi = 2\pi p; half-maximum where Fsin⁡2(φ/2)=1F\sin^2(\varphi/2) = 1, i.e. Δφ≈4/F=2(1−R)/R\Delta\varphi \approx 4/\sqrt F = 2(1 - R)/\sqrt R, against the 2π2\pi between peaks. ∎

The Airy transmission of a Fabry–Pérot cavity: full transmission at the resonances 2L = p, peaks narrowing as the mirrors’ reflectance grows.
The Airy transmission of a Fabry–Pérot cavity: full transmission at the resonances 2L=pλ2L = p\lambda, peaks narrowing as the mirrors’ reflectance grows.

Method 21.7 (Grating calculations)

(1) Period a=1/(lines per mm)a = 1/(\text{lines per mm}); grating equation with the signs of the angles. (2) List the orders that exist (∣sin⁡θ∣≤1|\sin\theta| \le 1) and check their overlap (pλmax⁡p\lambda_{\max} against (p+1)λmin⁡(p + 1)\lambda_{\min}). (3) Dispersion p/acos⁡θp/a\cos\theta, converted to a position on the detector with the focal length:  ⁣dx=f  ⁣dθ\dd x = f\,\dd\theta. (4) Resolving power pNpN with NN the lines actually illuminated; compare λ/R\lambda/R with the structure to be resolved. (5) For real instruments add the entrance slit’s width (it blurs the lines) and the detector’s pixels.

21.5 Exercises

Exercise 21.1 ★

A grating of 500500 lines per millimetre at normal incidence, λ=550 nm\lambda = 550\,\mathrm{nm}: angles of the orders; highest order; angular width of the first-order visible spectrum (400400–700 nm700\,\mathrm{nm}); same grating at 30∘30{}^{\circ} incidence: orders on each side.

Solution

Solution of Exercise 21.1.

a=2 µma = 2\,\text{µ}\mathrm{m}, sin⁡θp=0.275p\sin\theta_p = 0.275p: 16.0∘16.0{}^{\circ}, 33.4∘33.4{}^{\circ}, 55.6∘55.6{}^{\circ}; no fourth order. First-order visible from 11.5∘11.5{}^{\circ} to 20.5∘20.5{}^{\circ}: 9∘9{}^{\circ}. At 30∘30{}^{\circ}: sin⁡θ=0.5+0.275p\sin\theta = 0.5 + 0.275p: p=+1p = +1 (51∘51{}^{\circ}) on one side, p=−1p = -1 to −5-5 (13∘13^\circ, −3∘-3^\circ, −19∘-19^\circ, −37∘-37^\circ, −61∘-61^\circ) on the other.

Exercise 21.2 ★

The sodium doublet with a 600600-lines/mm grating and a lens of f=50 cmf = 50\,\mathrm{cm}: angular and linear separation of the two lines on the detector in the first and second orders; pixel size needed to see them apart.

Solution

Solution of Exercise 21.2.

a=1.67 µma = 1.67\,\text{µ}\mathrm{m}. Order 1: θ=20.7∘\theta = 20.7{}^{\circ}, Δθ=Δλ/acos⁡θ=3.9×10−4 rad\Delta\theta = \Delta\lambda/a\cos\theta = 3.9 \times 10^{-4}\,\mathrm{rad}, 0.19 mm0.19\,\mathrm{mm}; order 2: θ=45∘\theta = 45{}^{\circ}, 1.0×10−3 rad1.0 \times 10^{-3}\,\mathrm{rad}, 0.51 mm0.51\,\mathrm{mm}. Pixels of 50 µm50\,\text{µ}\mathrm{m} or less.

Exercise 21.3 ★

Resolving power of a 2 cm2\,\mathrm{cm} grating of 600600 lines/mm in orders 11, 22, 33; smallest Δλ\Delta\lambda resolved at 550 nm550\,\mathrm{nm}; does it separate the sodium doublet? The hydrogen Hα\alpha line from its deuterium twin (0.18 nm0.18\,\mathrm{nm} apart)?

Solution

Solution of Exercise 21.3.

N=12 000N = 12\,000: R=12 000R = 12\,000, 24 00024\,000, 36 00036\,000; Δλ=0.046\Delta\lambda = 0.046, 0.0230.023, 0.015 nm0.015\,\mathrm{nm}; the doublet (0.6 nm0.6\,\mathrm{nm}) and the H/D pair (0.18 nm0.18\,\mathrm{nm}, needing R=3600R = 3600) are easily separated.

Exercise 21.4 ★

A compact disc (a=1.6 µma = 1.6\,\text{µ}\mathrm{m}) and a DVD (a=0.74 µma = 0.74\,\text{µ}\mathrm{m}) lit at normal incidence: angles of first-order violet and red; which orders exist for each; why does the CD show two rainbows and the DVD one?

Solution

Solution of Exercise 21.4.

CD: violet 14.5∘14.5{}^{\circ}, red 25.9∘25.9{}^{\circ}; red exists to order 2, violet to order 4: two (partly overlapping) rainbows. DVD: violet 32.7∘32.7{}^{\circ}, red 71∘71{}^{\circ}; red has only the first order: one rainbow, wider.

Exercise 21.5 ★★

The NN-slit pattern. (a) Show that between two principal maxima there are N−1N - 1 zeros and N−2N - 2 secondary maxima. (b) For large NN, intensity of the first secondary maximum relative to the principal one (at Nφ/2≈3π/2N\varphi/2 \approx 3\pi/2): about 1/221/22. (c) Sketch the pattern for N=3N = 3 and N=6N = 6 and give the heights of the secondary maxima for N=3N = 3. (d) Fraction of the total energy in the principal maxima for large NN (compare the integrals of the peak, width ∝1/N\propto 1/N, height ∝N2\propto N^2, with the rest).

Solution

Solution of Exercise 21.5.

(a) Zeros at φ=2πm/N\varphi = 2\pi m/N, m=1,…,N−1m = 1, \dots, N - 1: N−1N - 1 zeros, N−2N - 2 maxima between them. (b) At Nφ/2=3π/2N\varphi/2 = 3\pi/2, I≈I0/sin⁡2(3π/2N)≈I0(2N/3π)2I \approx I_0/\sin^2(3\pi/2N) \approx I_0(2N/3\pi)^2: 4/9π2=0.0454/9\pi^2 = 0.045 of N2I0N^2I_0. (c) N=3N = 3: a single secondary maximum at φ=π\varphi = \pi of height I0I_0, one ninth of the principal 9I09I_0; N=6N = 6: four secondary maxima of a few percent. (d) The central lobe of sin⁡2x/x2\sin^2x/x^2 holds 90%90\% of the energy.

Exercise 21.6 ★★

Overlapping orders. (a) Show that order pp of λmax⁡\lambda_{\max} overlaps order p+1p + 1 of λmin⁡\lambda_{\min} when pλmax⁡>(p+1)λmin⁡p\lambda_{\max} > (p + 1)\lambda_{\min}; for 400400–700 nm700\,\mathrm{nm}, from which order? (b) Free spectral range in order pp: ΔλFSR=λ/p\Delta\lambda_{\text{FSR}} = \lambda/p. (c) A spectrograph works in order 2020 near 500 nm500\,\mathrm{nm} (an echelle grating): free spectral range; how is the overlap removed (cross-disperser)? (d) The Fabry–Pérot’s free spectral range is c/2Lc/2L: write it in wavelength and compare with the grating’s.

Solution

Solution of Exercise 21.6.

(a) p>λmin⁡/(λmax⁡−λmin⁡)=1.33p > \lambda_{\min}/(\lambda_{\max} - \lambda_{\min}) = 1.33: from p=2p = 2. (b) λ/p\lambda/p. (c) 25 nm25\,\mathrm{nm}; a prism crossed with the grating stacks the orders one above the other. (d) Δλ=λ2/2L\Delta\lambda = \lambda^2/2L, a fraction of a nanometre — hundreds of times smaller than a grating’s.

Exercise 21.7 ★★

Blazed grating. A reflection grating has its grooves tilted by the blaze angle β\beta so that each facet reflects specularly into the direction where the order pp is wanted. (a) For the Littrow mount (θ=θ0=β\theta = \theta_0 = \beta): show 2asin⁡β=pλ2a\sin\beta = p\lambda; blaze angle for first-order 500 nm500\,\mathrm{nm} with 12001200 lines/mm. (b) Why does the blaze send most of the light into one order (think of the envelope of a single facet’s diffraction, Chapter 22)? (c) The same grating is blazed for 1 µm1\,\text{µ}\mathrm{m}: in which order is it efficient at 500 nm500\,\mathrm{nm}? (d) Resolving power of a 10 cm10\,\mathrm{cm} grating in Littrow at β=64∘\beta = 64{}^{\circ}, λ=500 nm\lambda = 500\,\mathrm{nm} (use R=2Wsin⁡β/λR = 2W\sin\beta/\lambda).

Solution

Solution of Exercise 21.7.

(a) sin⁡β=λ/2a=0.3\sin\beta = \lambda/2a = 0.3, β=17.5∘\beta = 17.5{}^{\circ}. (b) The single facet diffracts into a broad lobe centred on its specular direction; the orders that fall inside it get the light. (c) Order 2 (2×500 nm=1 µm2 \times 500\,\mathrm{nm} = 1\,\text{µ}\mathrm{m} at the same angle). (d) R=2Wsin⁡β/λ=3.6×105R = 2W\sin\beta/\lambda = 3.6 \times 10^5.

Exercise 21.8 ★★

The slit. A spectrometer’s entrance slit of width ss is imaged on the detector with magnification 11; the grating (600600 lines/mm, 5 cm5\,\mathrm{cm}, order 11, f=50 cmf = 50\,\mathrm{cm}) disperses. (a) Width of a line on the detector due to the slit alone for s=50 µms = 50\,\text{µ}\mathrm{m}, in nm. (b) Width due to the grating’s resolving power. (c) Which dominates; slit width at which the two are equal. (d) Why not close the slit further?

Solution

Solution of Exercise 21.8.

(a)  ⁣dλ/ ⁣dx=acos⁡θ/pf=3.1 nm/mm\dd\lambda/\dd x = a\cos\theta/pf = 3.1\,\mathrm{nm}/\mathrm{mm}: 50 µm50\,\text{µ}\mathrm{m} is 0.16 nm0.16\,\mathrm{nm}. (b) R=30 000R = 30\,000: 0.018 nm0.018\,\mathrm{nm}. (c) The slit, by nine; equal at s=6 µms = 6\,\text{µ}\mathrm{m}. (d) Diffraction at the slit and a starved detector.

Exercise 21.9 ★★

Fabry–Pérot. Mirrors R=0.95R = 0.95, L=5 mmL = 5\,\mathrm{mm}, λ=600 nm\lambda = 600\,\mathrm{nm}. (a) Order pp, free spectral range in frequency and in wavelength. (b) Finesse; width of a transmission peak in frequency and wavelength. (c) Resolving power pFp\mathcal F; compare with the 10 cm10\,\mathrm{cm} grating. (d) Why is a Fabry–Pérot always used with a prefilter or a grating?

Solution

Solution of Exercise 21.9.

(a) p=2L/λ=16 700p = 2L/\lambda = 16\,700; FSR c/2L=30 GHzc/2L = 30\,\mathrm{GHz}, λ2/2L=0.036 nm\lambda^2/2L = 0.036\,\mathrm{nm}. (b) F=π0.95/0.05=61\mathcal F = \pi\sqrt{0.95}/0.05 = 61; 0.5 GHz0.5\,\mathrm{GHz}, 0.6 pm0.6\,\mathrm{pm}. (c) pF=106p\mathcal F = 10^6, four times the 10 cm10\,\mathrm{cm} grating’s. (d) Its free spectral range is a fraction of a nanometre: the orders overlap unless the light is prefiltered to one of them.

Exercise 21.10 ★★★

The limit of a grating. (a) Show that R=pN=W(sin⁡θ−sin⁡θ0)/λ≤2W/λR = pN = W(\sin\theta - \sin\theta_0) /\lambda \le 2W/\lambda: the resolving power cannot exceed the width of the grating in wavelengths, times two. (b) Interpret: the largest path difference between the extreme rays. (c) A 30 cm30\,\mathrm{cm} grating at 500 nm500\,\mathrm{nm}: ultimate RR; smallest resolvable velocity by Doppler shift (Δλ/λ=v/c\Delta\lambda/\lambda = v/c). (d) Why do astronomers nevertheless reach 1 m/s1\,\mathrm{m}/\mathrm{s} (think of measuring the position of a line to a fraction of its width, with many lines)?

Solution

Solution of Exercise 21.10.

(a) ∣sin⁡θ−sin⁡θ0∣≤2|\sin\theta - \sin\theta_0| \le 2. (b) The extreme rays differ in path by W(sin⁡θ−sin⁡θ0)W(\sin\theta - \sin\theta_0) at most 2W2W. (c) 2×0.3/5×10−7=1.2×1062 \times 0.3/5 \times 10^{-7} = 1.2 \times 10^6; c/R=250 m/sc/R = 250\,\mathrm{m}/\mathrm{s}. (d) A line’s centre is located to a hundredth of its width, and a thousand lines average down by 1000\sqrt{1000}: metres per second, with heroic stability.

Exercise 21.11 ★★★

Bragg. A crystal is a three-dimensional grating: X-rays reflected by successive atomic planes a distance dd apart, at the glancing angle θ\theta, interfere constructively when 2dsin⁡θ=pλ2d\sin\theta = p\lambda (Bragg). (a) Derive it from the path difference between two planes. (b) Copper Kα\alpha radiation (0.154 nm0.154\,\mathrm{nm}) on rock salt (d=0.282 nmd = 0.282\,\mathrm{nm}): the Bragg angles. (c) Why do visible wavelengths give no Bragg reflection from crystals, and what does the converse (no X-ray diffraction from glass) tell about glass? (d) Why must λ<2d\lambda < 2d?

Solution

Solution of Exercise 21.11.

(a) The wave reflected by the lower plane travels 2dsin⁡θ2d\sin\theta more. (b) sin⁡θ=0.273p\sin\theta = 0.273p: 15.8∘15.8{}^{\circ}, 33.1∘33.1{}^{\circ}, 55.0∘55.0{}^{\circ}. (c) λ≫2d\lambda \gg 2d: no angle satisfies the equation; glass has no regular planes, hence only diffuse rings — it is amorphous. (d) sin⁡θ≤1\sin\theta \le 1.

Exercise 21.12 ★★★

Phased array. NN identical antennas in a line, spaced aa, fed with the same amplitude and a phase step ψ\psi between neighbours. (a) Show that the radiated field in the direction θ\theta is that of the NN-wave sum with φ=2πasin⁡θ/λ−ψ\varphi = 2\pi a\sin\theta/\lambda - \psi: the main beam points where sin⁡θ=λψ/2πa\sin\theta = \lambda\psi/2\pi a. (b) Half-width of the main beam, Δθ≈λ/Nacos⁡θ\Delta\theta \approx \lambda/Na\cos\theta. (c) Why must a≤λ/2a \le \lambda/2 to avoid a second main beam (a grating lobe) when steering? (d) A radar with 6464 elements at λ/2\lambda/2, λ=3 cm\lambda = 3\,\mathrm{cm}: beam width; time to scan 90∘90^\circ in steps of one beam width if the phases switch in 1 µs1\,\text{µ}\mathrm{s}.

Solution

Solution of Exercise 21.12.

(a) Neighbours differ in phase by 2πasin⁡θ/λ2\pi a\sin\theta/\lambda from the path and −ψ-\psi from the feed; the principal maximum at φ=0\varphi = 0. (b) First zero at Δφ=2π/N\Delta\varphi = 2\pi/N: Δθ=λ/Nacos⁡θ\Delta\theta = \lambda/Na\cos\theta. (c) A second principal maximum at φ=±2π\varphi = \pm2\pi needs ∣sin⁡θ±λ/a∣≤1|\sin\theta \pm \lambda/a| \le 1: impossible for every steering angle only if a≤λ/2a \le \lambda/2. (d) Na=0.96 mNa = 0.96\,\mathrm{m}: 1.8∘1.8{}^{\circ}; 5050 steps, 50 µs50\,\text{µ}\mathrm{s}.

21.6 Problem: A spectrograph for a star

Problem 21.1

Weekend problem — designing the spectrograph that reads a star’s light: the grating, the detector, the lines of hydrogen, and the wobble of a planet

A spectrograph: entrance slit, collimator of focal length f1=1.0 mf_1 = 1.0\,\mathrm{m}, a reflection grating of 12001200 lines/mm and width W=12 cmW = 12\,\mathrm{cm}, a camera of f2=1.0 mf_2 = 1.0\,\mathrm{m}, a detector with 15 µm15\,\text{µ}\mathrm{m} pixels. It works in the first order, around 550 nm550\,\mathrm{nm}, with the grating at 20∘20{}^{\circ} incidence.

Part I — The grating.

  1. Period of the grating; number of lines lit if the beam covers the whole width.
  2. Diffraction angle of 550 nm550\,\mathrm{nm} in the first order; of 400 nm400\,\mathrm{nm} and 700 nm700\,\mathrm{nm}; angular width of the visible spectrum.
  3. Angular dispersion at 550 nm550\,\mathrm{nm}; linear dispersion on the detector in nm per mm and nm per pixel.
  4. Resolving power; smallest Δλ\Delta\lambda resolved at 550 nm550\,\mathrm{nm}; in pixels.
  5. Does the second-order violet fall onto the first-order visible? Which filter removes it?
  6. Length of detector needed for the whole visible spectrum in the first order.
  7. The grating is blazed at 17∘17{}^{\circ}: for which wavelength is it most efficient in the first order at this incidence (use the Littrow estimate 2asin⁡β=λ2a\sin\beta = \lambda)?
  8. The grating returns 80%80\% of the light at the blaze wavelength and the telescope plus optics 40%40\%: fraction of the star’s photons that reach the detector.

Part II — Lines.

  1. The hydrogen Balmer lines of the star: Hα\alpha 656.3 nm656.3\,\mathrm{nm}, Hβ\beta 486.1 nm486.1\,\mathrm{nm}, Hγ\gamma 434.0 nm434.0\,\mathrm{nm}: their positions on the detector relative to 550 nm550\,\mathrm{nm}.
  2. Hα\alpha from a deuterium atom is 0.18 nm0.18\,\mathrm{nm} shorter: separated? By how many pixels?
  3. The sodium doublet in the star’s spectrum: separation in pixels; and the width of each line if the star’s atmosphere broadens them to 0.05 nm0.05\,\mathrm{nm}.
  4. The entrance slit is 100 µm100\,\text{µ}\mathrm{m} wide, imaged at magnification f2/f1f_2/f_1: its width on the detector in pixels and in nm; is the spectrograph slit-limited or grating-limited?
  5. The light from the telescope comes as a disc of 1′′1'' on the sky, which the telescope (f=20 mf = 20\,\mathrm{m}) makes 100 µm100\,\text{µ}\mathrm{m} wide at the slit: fraction of light lost if the slit is narrowed to 50 µm50\,\text{µ}\mathrm{m} (take the disc as uniform).
  6. Why does a narrower slit give sharper lines but noisier spectra?

Part III — The star moves.

  1. A star recedes at 30 km/s30\,\mathrm{km}/\mathrm{s}: Doppler shift of Hα\alpha in nm and in pixels.
  2. The Earth’s orbital motion adds ±30 km/s\pm30\,\mathrm{km}/\mathrm{s} over the year: why must it be subtracted, and to what precision if one wants 100 m/s100\,\mathrm{m}/\mathrm{s}?
  3. A planet makes its star wobble at 10 m/s10\,\mathrm{m}/\mathrm{s}: shift in nm and in pixels; is it below the resolution? Explain how measuring the centre of a line to a hundredth of its width, over a thousand lines, recovers it (statistical gain N\sqrt{N}).
  4. The temperature of the instrument changes by 1 K1\,\mathrm{K}, and the grating (glass, expansion 1×10−5 K−11 \times 10^{-5}\,\mathrm{K}^{-1}) dilates: shift of the lines in pixels; what stability is needed for 10 m/s10\,\mathrm{m}/\mathrm{s}?
  5. A calibration lamp (thorium–argon) puts hundreds of known lines on the same detector: explain its role.

Part IV — Alternatives.

  1. The same width of grating ruled at 300300 lines/mm and used in the fourth order (with a cross-disperser to sort the orders): diffraction angle, resolving power, dispersion and free spectral range at 550 nm550\,\mathrm{nm}; what has changed, what has not?
  2. A Fabry–Pérot etalon (L=1 mmL = 1\,\mathrm{mm}, R=0.9R = 0.9) placed before the slit: free spectral range in nm, finesse, peak width; what does its comb of transmission peaks look like on the detector, and what is it used for?
  3. A prism of 60∘60{}^{\circ} in glass with  ⁣dn/ ⁣dλ=−1×10−4 nm−1\dd n/\dd\lambda = -1 \times 10^{-4}\,\mathrm{nm}^{-1} has the angular dispersion  ⁣dD/ ⁣dλ≈2  ⁣dn/ ⁣dλ\dd D/\dd\lambda \approx 2\,\dd n/\dd\lambda near minimum deviation: compare with the grating’s; why did gratings replace prisms?
  4. Why are the largest astronomical spectrographs built with gratings a metre wide and used at large angles (the limit 2W/λ2W/\lambda)?
  5. A magnitude-8 star delivers 1×104 1 \times 10^{4}\, photons per second and per nanometre at the telescope: photons per pixel per second on the detector, and the signal-to-noise ratio of a 100 s100\,\mathrm{s} exposure (photon noise only).
  6. Summarize: the three numbers that describe a spectrograph (dispersion, resolving power, free spectral range) and what sets each.
Solution

Solution of Problem 21.1.

1. a=0.833 µma = 0.833\,\text{µ}\mathrm{m}; N=144 000N = 144\,000.

2. sin⁡θ=sin⁡20∘−λ/a\sin\theta = \sin20^\circ - \lambda/a (the order lies on the other side of the normal): 550 nm550\,\mathrm{nm} at 18.5∘18.5{}^{\circ}, 400 nm400\,\mathrm{nm} at 7.9∘7.9{}^{\circ}, 700 nm700\,\mathrm{nm} at 29.9∘29.9{}^{\circ}: 22∘22{}^{\circ} of spectrum.

3. 1/acos⁡θ=1.27×10−3 rad/nm1/a\cos\theta = 1.27 \times 10^{-3}\,\mathrm{rad}/\mathrm{nm}; f2×f_2\times: 1.27 mm/nm1.27\,\mathrm{mm}/\mathrm{nm}, i.e. 0.79 nm/mm0.79\,\mathrm{nm}/\mathrm{mm}, 0.012 nm0.012\,\mathrm{nm} per pixel.

4. R=N=144 000R = N = 144\,000: 0.0038 nm0.0038\,\mathrm{nm}, a third of a pixel.

5. Second-order 400 nm400\,\mathrm{nm} lands where first-order 800 nm800\,\mathrm{nm} would, beyond the red: no overlap in the visible; the overlap would start at 350 nm350\,\mathrm{nm}, which a glass filter removes.

6. 22∘22{}^{\circ} ×\times 1 m1\,\mathrm{m} =38 cm= 38\,\mathrm{cm}: a detector covers a slice; the grating is turned to choose it.

7. λ=2asin⁡17∘=490 nm\lambda = 2a\sin17^\circ = 490\,\mathrm{nm}.

8. 0.8×0.4=32%0.8 \times 0.4 = 32\%.

9. Angles 26.5∘26.5{}^{\circ}, 14.0∘14.0{}^{\circ}, 10.3∘10.3{}^{\circ}: +14+14, −8-8 and −14 cm-14\,\mathrm{cm} from the 550 nm550\,\mathrm{nm} position.

10. 0.18 nm0.18\,\mathrm{nm} is 1515 pixels: separated.

11. 5050 pixels apart, each 44 pixels wide.

12. 100 µm100\,\text{µ}\mathrm{m}, 6.76.7 pixels, 0.08 nm0.08\,\mathrm{nm} — twenty times the grating’s limit: slit-limited.

13. The image is 97 µm97\,\text{µ}\mathrm{m} wide: a 50 µm50\,\text{µ}\mathrm{m} slit passes half the light.

14. Less light for the same sharpness: the photon noise grows.

15. Δλ=λv/c=0.066 nm\Delta\lambda = \lambda v/c = 0.066\,\mathrm{nm}, 5.55.5 pixels.

16. The Earth’s velocity shifts every line by up to ±5.5\pm5.5 pixels through the year; it is known from the ephemerides to far better than the 0.3%0.3\% needed for 100 m/s100\,\mathrm{m}/\mathrm{s}.

17. 2.2×10−5 nm2.2 \times 10^{-5}\,\mathrm{nm}, 0.0020.002 pixel — far below the line width. A centroid to a hundredth of a 7 px7\,\mathrm{px} line is 0.07 px0.07\,\mathrm{px}; a thousand lines give 1000\sqrt{1000}: 0.0020.002 pixel — the signal, just. Real instruments add resolving power and stability.

18. Δsin⁡θ=(λ/a)×10−5=6.6×10−6\Delta\sin\theta = (\lambda/a) \times 10^{-5} = 6.6 \times 10^{-6}, Δθ=7×10−6 rad\Delta\theta = 7 \times 10^{-6}\,\mathrm{rad}, 7 µm7\,\text{µ}\mathrm{m}: half a pixel per kelvin — millikelvins for 10 m/s10\,\mathrm{m}/\mathrm{s}.

19. Its lines, recorded at the same instant on the same pixels, give the wavelength scale and track every drift.

20. a=3.33 µma = 3.33\,\text{µ}\mathrm{m}, N=36 000N = 36\,000, same angle (18.5∘18.5{}^{\circ}), R=4N=144 000R = 4N = 144\,000, same dispersion, FSR λ/4=137 nm\lambda/4 = 137\,\mathrm{nm}: resolving power and dispersion depend on WW and the angle only; the free spectral range shrank.

21. FSR λ2/2L=0.15 nm\lambda^2/2L = 0.15\,\mathrm{nm} (1212 pixels); F=30\mathcal F = 30; peaks 0.005 nm0.005\,\mathrm{nm} wide: a comb of sharp lines across the detector, a wavelength ruler.

22. 2×10−42 \times 10^{-4} against 1.3×10−3 rad/nm1.3 \times 10^{-3}\,\mathrm{rad}/\mathrm{nm}: six times less, and a prism’s resolving power (b  ⁣dn/ ⁣dλ∼104b\,\dd n/\dd\lambda \sim 10^4) and non-linear scale lose too; gratings also work in the ultraviolet.

23. R≤2W/λR \le 2W/\lambda: 4×1064 \times 10^6 for a metre at 500 nm500\,\mathrm{nm} — width and angle are the only ways up.

24. 104×0.012×0.32=3810^4 \times 0.012 \times 0.32 = 38 photons per pixel per second; 38003800 in 100 s100\,\mathrm{s}: SNR ≈60\approx 60.

25. Dispersion, set by p/acos⁡θp/a\cos\theta and the camera’s focal length; resolving power, by the illuminated width and the angles (pNpN); free spectral range, by the order (λ/p\lambda/p).