17-Phys-A7 Optics · December 2016

Question 2 of 6: Single-slit and grating diffraction, aperture spreading, Rayleigh resolution

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

Question 2: Single-slit and grating diffraction, aperture spreading, 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 overlap gives the unknown wavelength.

Given. $\lambda_1=500$ nm; single-slit Fraunhofer minima at $\sin\theta_m=m\lambda/a$; the 5th minimum of the unknown $\lambda_x$ coincides (same angle, same slit) with the 4th minimum of $\lambda_1$.

Find. $\lambda_x$.

Approach. "Overlap" means the two minima occur at the same diffraction angle, so $5\lambda_x/a=4\lambda_1/a$ — the slit width $a$ cancels, leaving a pure ratio.

1. Equate the minimum conditions. $5\lambda_x=4\lambda_1=4(500\ \text{nm})=2000\ \text{nm}$, so $\lambda_x=\dfrac{2000}{5}=\boxed{400\ \text{nm}}$ (violet light).

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

Given. $N=50$ grooves, groove spacing $d=0.01$ mm$=10\ \mu$m, $\lambda=500$ nm, normal incidence, lens focal length $f=100$ cm, order $m=2$.

Find. The separation, in the focal plane, between the $m=2$ principal maximum and each of its two neighbouring minima.

Approach. Principal maxima: $d\sin\theta_m=m\lambda$. For an $N$-slit grating the $N$-line interference pattern has intensity zeros spaced by $\Delta(\sin\theta)=\lambda/(Nd)$ in $\sin\theta$; the minima immediately flanking order $m$ sit exactly one such spacing away, at $\sin\theta=\sin\theta_m\pm\lambda/(Nd)$. Because $\theta_2\approx5.7^\circ$ is not vanishingly small, the focal-plane position is mapped with $y=f\tan\theta$ (not the small-angle $y\approx f\theta$) for full precision.

1. Principal maximum position. $\sin\theta_2=\dfrac{2\lambda}{d}=\dfrac{2(500\times10^{-9})}{10\times10^{-6}}=0.1000\Rightarrow\theta_2=5.7392^\circ$, so $y_2=f\tan\theta_2=(100\ \text{cm})\tan(5.7392^\circ)=100.504\ \text{cm}$.
2. Flanking minima. $\Delta(\sin\theta)=\dfrac{\lambda}{Nd}=\dfrac{500\times10^{-9}}{50(10\times10^{-6})}=1.000\times10^{-3}$, so $\sin\theta_{\pm}=0.1000\pm0.0010$. Outer: $\theta_+=5.7994^\circ$, $y_+=f\tan\theta_+=101.519\ \text{cm}$. Inner: $\theta_-=5.6790^\circ$, $y_-=f\tan\theta_-=99.489\ \text{cm}$.
3. Separations. Outer side: $y_+-y_2=\boxed{1.015\ \text{mm}}$. Inner side: $y_2-y_-=\boxed{1.015\ \text{mm}}$ — the two separations agree to 3 figures because $\tan\theta$ is still nearly linear in $\theta$ over this small a range, even though the absolute angle ($5.7^\circ$) is not itself paraxial.

Part (c) — distance for the diffracted beam to reach twice the aperture diameter.

Given. A monochromatic plane wave of wavelength $\lambda$ normally illuminates an aperture of diameter $D$ (symbolic — no numeric $\lambda$ or $D$ is supplied for this part).

Find. The propagation distance $z$ at which the transmitted beam's diameter has grown to $\approx2D$; state and justify the criterion used.

Approach. Diffraction from an aperture of size $D$ spreads with a characteristic half-angle $\theta\sim\lambda/D$ — the same reciprocal aperture–angle relation used for the single-slit and grating patterns above (a more precise circular-aperture criterion would use the Rayleigh half-angle $1.22\lambda/D$ from part (d); either is defensible since the question explicitly asks the solver to choose and justify one). At distance $z$ the beam radius grows from $D/2$ to $D/2+z\theta$; "essentially twice the diameter" means the new radius is $\approx D$ (full diameter $2D$).

1. State the criterion. Take the characteristic diffraction half-angle as $\theta=\lambda/D$ (order-of-magnitude spreading angle of an aperture-limited beam, consistent with the single-slit first-minimum angle $\lambda/a$ used in part (a)).
2. Solve for $z$. $\dfrac{D}{2}+z\theta=D\ \Rightarrow\ z\theta=\dfrac{D}{2}\ \Rightarrow\ z=\dfrac{D}{2\theta}=\boxed{\dfrac{D^2}{2\lambda}}.$

Part (d)(i) — Rayleigh's criterion. Two incoherent point sources, each producing an Airy diffraction pattern through a circular aperture of diameter $D_p$, are said to be just resolved when the central maximum of one source's Airy pattern falls exactly on the first dark ring (first minimum) of the other's. This occurs when the sources subtend an angle $\theta_{\min}=1.22\lambda/D_p$ at the aperture; the combined intensity then shows a shallow but distinguishable dip (about 74% of the peak) between the two maxima.

Part (d)(ii) — resolvable distance over the adapting pupil range.

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 sources are just resolved (Rayleigh criterion) 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 $L=sD_p/(1.22\lambda)$.

1. Constricted pupil (bright light), $D_p=2$ mm. $\theta_{\min}=\dfrac{1.22(500\times10^{-9})}{2\times10^{-3}}=3.050\times10^{-4}\ \text{rad}$, so $L=\dfrac{s}{\theta_{\min}}=\dfrac{1}{3.050\times10^{-4}}=\boxed{3.279\times10^{3}\ \text{m}=3.28\ \text{km}}.$
2. Dilated pupil (dark-adapted), $D_p=7$ mm. $\theta_{\min}=\dfrac{1.22(500\times10^{-9})}{7\times10^{-3}}=8.714\times10^{-5}\ \text{rad}$, so $L=\dfrac{1}{8.714\times10^{-5}}=\boxed{1.148\times10^{4}\ \text{m}=11.48\ \text{km}}.$