17-Phys-A7 Optics · May 2016

Question 2 of 6: Single-slit and grating diffraction, Airy disk, Rayleigh resolution

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 2: Single-slit and grating diffraction, Airy disk, Rayleigh resolution (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Part (a) — single-slit Fraunhofer minima.

Given. $\lambda=500$ nm, slit width $a=0.015$ cm$=150\ \mu$m, lens focal length $f=60$ cm (minima observed in the focal plane).

Find. (i) $y_1$, central max to first minimum; (ii) $y_2-y_1$, first to second minima.

Approach. Single-slit minima occur at $\sin\theta_m=m\lambda/a$; with $\lambda/a\approx0.0033$ the angles are tiny, so $y_m\approx f\sin\theta_m=fm\lambda/a$ (paraxial, lens maps angle to focal-plane position).

1. First minimum. $y_1=\dfrac{f\lambda}{a}=\dfrac{(0.60)(500\times10^{-9})}{1.5\times10^{-4}}=\boxed{2.00\times10^{-3}\ \text{m}=2.00\ \text{mm}}.$
2. Second minimum, then the spacing. $y_2=\dfrac{f\cdot2\lambda}{a}=4.00$ mm, so $y_2-y_1=\boxed{2.00\ \text{mm}}$ — identical to $y_1$, because in the paraxial regime successive single-slit minima are equally spaced by $f\lambda/a$.

Part (b) — grating: minima flanking the 2nd-order principal maximum.

Given. $N=30$ grooves, groove spacing $d=0.01$ mm$=10\ \mu$m, $\lambda=500$ nm, normal incidence, order $m=2$.

Find. Angular separation of the two minima immediately flanking the $m=2$ principal maximum.

Approach. Principal maxima: $d\sin\theta_m=m\lambda$. For an $N$-slit grating the intensity zero pattern has period $\lambda/(Nd)$ in $\sin\theta$; the first minima adjacent to order $m$ sit at $\sin\theta=\sin\theta_m\pm\lambda/(Nd)$ (the interference maximum is flanked by minima $1/N$ of a fringe-order away on each side).

1. Locate the principal maximum. $\sin\theta_2=\dfrac{2\lambda}{d}=\dfrac{2(500\times10^{-9})}{10^{-5}}=0.1000\ \Rightarrow\ \theta_2=\boxed{5.739^\circ}.$
2. Locate the flanking minima. $\dfrac{\lambda}{Nd}=\dfrac{500\times10^{-9}}{30\times10^{-5}}=1.667\times10^{-3}$, so $\sin\theta_-=0.09833\Rightarrow\theta_-=5.646^\circ$ and $\sin\theta_+=0.10167\Rightarrow\theta_+=5.840^\circ$.
3. Angular separations. $\Delta\theta_-=\theta_2-\theta_-=\boxed{0.094^\circ}$, $\Delta\theta_+=\theta_+-\theta_2=\boxed{0.096^\circ}$ (the tiny asymmetry is the nonlinearity of $\arcsin$; a paraxial estimate $\lambda/(Nd\cos\theta_2)=0.096^\circ$ splits the difference). Total angular width of the principal maximum's central lobe: $\theta_+-\theta_-=\boxed{0.192^\circ}.$

Part (c) — Airy disk of a diffraction-limited telescope objective.

Given. Objective diameter $D=10$ cm, focal length $f=100$ cm, $\lambda=500$ nm, star = distant point source (plane-wave illumination).

Find. Airy-disk diameter in the focal plane.

Approach. The Airy pattern's first dark ring subtends half-angle $\theta_1=1.22\lambda/D$ from the aperture; the focal-plane radius is $f\theta_1$ (paraxial), so the disk diameter is $2f\theta_1=2.44\lambda f/D$.

1. Airy-disk diameter. $d_{\text{Airy}}=\dfrac{2.44\,\lambda f}{D}=\dfrac{2.44(500\times10^{-9})(1.00)}{0.10}=\boxed{1.22\times10^{-5}\ \text{m}=12.2\ \mu\text{m}}.$

Part (d)(i) — Rayleigh's criterion. Two incoherent point sources are just resolved when the central maximum of one source's diffraction pattern falls exactly on the first minimum of the other's. At that separation the summed intensity dips to about $8/\pi^2\approx0.81$ of the peak — a shallow but visible dip — which is why it is taken as the practical resolution limit rather than a hard cutoff.

Part (d)(ii) — resolvable-distance range as the pupil adapts.

Given. Two point sources laterally separated by $s=1$ m, $\lambda=500$ nm, observer pupil diameter adapting over $D_p=2$–$7$ mm.

Find. The distance $L$ at which the two sources are just resolved (Rayleigh criterion, circular-aperture eye pupil), evaluated at each pupil extreme.

Approach. Rayleigh angle $\theta_{\min}=1.22\lambda/D_p$; the sources are just resolved when the subtended angle $s/L$ equals $\theta_{\min}$, so $L=sD_p/(1.22\lambda)$ — a larger (more dilated) pupil gives a smaller diffraction angle and so a larger maximum resolvable distance.

1. Constricted pupil, $D_p=2$ mm. $L=\dfrac{(1.0)(2\times10^{-3})}{1.22(500\times10^{-9})}=\boxed{3.28\times10^{3}\ \text{m}\ (3.28\ \text{km})}.$
2. Dilated pupil, $D_p=7$ mm. $L=\dfrac{(1.0)(7\times10^{-3})}{1.22(500\times10^{-9})}=\boxed{1.148\times10^{4}\ \text{m}\ (11.48\ \text{km})}.$

As the eye adapts from a bright-light 2 mm pupil to a dark-adapted 7 mm pupil, the maximum distance at which the two lights are just resolvable increases from about $3.28$ km to about $11.48$ km — a larger pupil is a better (not worse) resolving aperture, even though it admits more light for the same reason a larger telescope objective resolves finer detail.