17-Phys-A7 Optics · May 2015

Question 6 of 6: Diffraction gratings, double slit, and resolution

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

Question 6: Diffraction gratings, double slit, and resolution (20 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).

Given. $\lambda=500$ nm, slit spacing $d=5\ \mu$m, slit width $a=1\ \mu$m, lens focal length $f=2$ m directly behind the apertures ($N=2,10,1000$ slits).

Find. (i) $y_1$ (0th→1st order separation); (ii) count of principal maxima inside the single-slit envelope's first zero; (iii) full width of a principal maximum, for each $N$.

Approach. A lens placed directly at the aperture forms the exact Fraunhofer pattern in its back focal plane, with position related to diffraction angle by $y=f\sin\theta$ (the Fourier-transform property of a thin lens). Principal (grating) maxima occur at $d\sin\theta=m\lambda$; the single-slit envelope's zeros occur at $a\sin\theta=k\lambda$; an $N$-slit principal maximum has zeros on either side at $\Delta(\sin\theta)=\pm\lambda/(Nd)$.

1. (i) 0th→1st order separation — same for all three apertures (position depends on $d$ alone, not $N$): $$y_1=f\sin\theta_1=f\cdot\frac{\lambda}{d}=2\times\frac{500\times10^{-9}}{5\times10^{-6}}=\boxed{0.200\ \text{m}=20.0\ \text{cm}}.$$
2. (ii) Orders inside the first envelope zero. Envelope zeros: $\sin\theta=\pm\lambda/a=\pm0.500$. Grating orders: $\sin\theta_m=m\lambda/d=0.100\,m$. Since $d/a=5$ exactly, order $m=\pm5$ lands exactly on an envelope zero (intensity $=0$ there — a "missing order"), so the orders with non-zero intensity strictly between the envelope zeros are $m=-4,\dots,-1,0,1,\dots,4$: $$\boxed{9\ \text{principal maxima}}\quad (m=0,\pm1,\pm2,\pm3,\pm4)$$

   — the same count for $N=2,10,1000$, since it is set by $d/a$ alone, independent of $N$.

   For $N=2$ these are the broad, low-contrast fringes of ordinary two-beam (Young's) interference rather than sharp grating lines, but their positions and the missing-order count are identical — both are governed by $d$ and $a$ alone, independent of $N$.
3. (iii) Width of a principal maximum. The $N$-slit intensity $\propto[\sin(N\beta)/\sin\beta]^2$ ($\beta=\pi d\sin\theta/\lambda$) has zeros adjacent to each principal maximum at $\Delta(\sin\theta)=\pm\lambda/(Nd)$, so the full width (zero-to-zero) is $\Delta(\sin\theta)_{\text{full}}=2\lambda/(Nd)$; mapping through $y=f\sin\theta$ gives $\Delta y_{\text{full}}=2f\lambda/(Nd)$: $$\begin{gathered} N=2:\ \Delta y=\dfrac{2(2)(500\text{nm})}{2(5\,\mu\text{m})}=\boxed{200\ \text{mm}} \\ N=10:\ \Delta y=\boxed{40.0\ \text{mm}} \\ N=1000:\ \Delta y=\boxed{0.400\ \text{mm}} \end{gathered}$$ As $N$ grows the maxima sharpen dramatically (by a factor of $500$ from $N=2$ to $N=1000$) while staying at the same $9$ envelope-limited positions — exactly the transition from a two-beam interference pattern to a sharp diffraction-grating spectrum.

Part (b).

Given. Two-slit spacing $d=0.2$ mm; 5th-order maximum at $y_5=15$ cm from centre.

Find. Wavelength $\lambda$.

Approach. Young's double-slit maxima: $d\sin\theta_m=m\lambda$, and for small angles $\sin\theta_m\approx y_m/L$, so $\lambda=y_md/(mL)$. Report the general relation, then illustrate the physically required screen distance for the answer to be visible light.

1. General relation. $$\lambda=\frac{y_5\,d}{5L}=\frac{(0.15\ \text{m})(2\times10^{-4}\ \text{m})}{5L}=\frac{6\times10^{-6}}{L}\ \text{m}.$$
2. Illustrative value. Requiring $\lambda$ to fall in the visible band — e.g. an unremarkable orange-red $\lambda=600$ nm — fixes $L=6\times10^{-6}/(600\times10^{-9})=\boxed{10.0\ \text{m}}$, a perfectly ordinary optical-bench/lecture-hall distance. So the data are consistent with $\boxed{\lambda\approx600\ \text{nm}}$ if (and only if) the screen sits about $10$ m from the slits; the general boxed relation above is the answer that does not depend on this assumption.

Part (c).

Given. Circular aperture (hole) diameter $D=2$ mm, $\lambda=623.8$ nm, plane-wave illumination, target beam diameter $4$ cm.

Find. Propagation distance $z$.

Approach. A circular aperture produces an Airy diffraction pattern whose first dark ring sits at half-angle $\theta_1=1.22\lambda/D$; far from the aperture the beam's diameter grows as $2z\theta_1$.

1. Airy divergence half-angle. $\theta_1=1.22\lambda/D=1.22(623.8\times10^{-9})/(2\times10^{-3})=3.805\times10^{-4}$ rad.
2. Solve for $z$. $2z\theta_1=0.04\ \text{m}\ \Rightarrow\ z=\dfrac{0.04}{2\times3.805\times10^{-4}}=\boxed{52.6\ \text{m}}$.
3. Regime check. Fresnel number $N_F=(D/2)^2/(\lambda z)=(1\times10^{-3})^2/(623.8\times10^{-9}\times52.6)=0.030\ll1$, confirming the Fraunhofer (far-field) approximation used is self-consistent at this distance.

Part (d).

Given. Headlight separation $s=2$ m, observer pupil diameter $D_p=5$ mm, $\lambda=500$ nm.

Find. Maximum distance $L$ at which the headlights are just resolvable (Rayleigh criterion).

Approach. Two points are just resolved when their angular separation equals the Airy first-minimum half-angle of the (circular) pupil aperture, $\theta_{\min}=1.22\lambda/D_p$; the angular separation of the headlights is $s/L$.

1. Rayleigh angle. $\theta_{\min}=1.22(500\times10^{-9})/(5\times10^{-3})=1.22\times10^{-4}$ rad.
2. Solve for $L$. $s/L=\theta_{\min}\ \Rightarrow\ L=s/\theta_{\min}=2/(1.22\times10^{-4})=\boxed{1.64\times10^{4}\ \text{m}\approx16.4\ \text{km}}$.