17-Phys-A7 Optics · December 2013

Question 6 of 6: Diffraction-limited focal spot, a two-wavelength double slit, an anti-reflection coating, and a moving mirror

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

Question 6: Diffraction-limited focal spot, a two-wavelength double slit, an anti-reflection coating, and a moving mirror (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. Circular lens, f/#=5, f=10 cm (so D=2 cm), point source at 20 cm in front of the lens.

1. Huygens–Fresnel principle. Every point of the illuminated circular aperture is treated as a source of a secondary spherical wavelet; the field at a point in the focal plane is the coherent (phase-correct) sum of all these wavelets, i.e. the Fraunhofer diffraction integral of the circular aperture. For a circular aperture this integral is the Airy function $I(v)=I_0\left[2J_1(v)/v\right]^2$.
2. Working f-number. The source is not at infinity, so the relevant cone angle is set by the actual image conjugate: $\dfrac1{d_i}=\dfrac1f-\dfrac1{d_o}=\dfrac1{10}-\dfrac1{20}=\dfrac1{20}\Rightarrow d_i=20$ cm ($m=-1$), so the working f-number is $N_{eff}=d_i/D=20/2=\boxed{10}$ (twice the nominal f/5, as expected for unit magnification, $N_{eff}=(f/\#)(1+|m|)$).
3. Airy-disc dimensions. The first dark ring sits at $r_1=1.22\,\lambda\,N_{eff}=12.2\,\lambda$ and the second dark ring at $r_2=2.23\,\lambda\,N_{eff}=22.3\,\lambda$ (no source wavelength is given, so these are left proportional to $\lambda$); illustratively, at $\lambda=550$ nm, $\boxed{r_1\approx6.7\ \mu\text{m}}$, $r_2\approx12.3\ \mu$m, with the first bright ring only about 1.75% as intense as the central peak.

Part (b). Given. $\lambda_1=500$ nm; minima at $d\sin\theta=(k-\tfrac12)\lambda$, $k=1,2,\dots$; maxima at $d\sin\theta=n\lambda$, $n=0,1,2,\dots$ (central fringe not counted as the “1st maximum”).

1. Coincidence condition. 4th minimum of 500 nm: path difference $\Delta=(4-\tfrac12)(500\text{ nm})=1750$ nm. 3rd maximum of the unknown line: $\Delta=3\lambda\prime$. Equating, $$\lambda\prime=\frac{1750}{3}=\boxed{583.3\text{ nm}}$$

Part (c). Given. $n_{sub}=1.50$, $\lambda=600$ nm, normal incidence.

1. Coating index. Zero reflection requires the two surface reflections (air/coating and coating/substrate) to have EQUAL magnitude, which for normal incidence needs $$n_{coat}=\sqrt{n_{air}\,n_{sub}}=\sqrt{(1.00)(1.50)}=\boxed{1.2247}$$
2. Coating thickness. The two reflections must also be 180° out of phase, which for the thinnest (quarter-wave) coating gives $$t=\frac{\lambda}{4\,n_{coat}}=\frac{600\text{ nm}}{4(1.2247)}=\boxed{122.5\text{ nm}}$$

Part (d). Given. Normal incidence, mirror speed $v\ll c$ along the beam direction.

1. Double Doppler shift. Reflection from a moving mirror is equivalent to two successive Doppler shifts — once as the mirror “receives” the wave, once as it “re-emits” it. For $\beta=v/c\ll1$ the two first-order shifts add: $$\boxed{\frac{\Delta f}{f}\approx\pm\frac{2v}{c}}\ ,\qquad \boxed{\frac{\Delta\lambda}{\lambda}\approx\mp\frac{2v}{c}}$$ (upper sign: mirror approaching the source → blueshift, wavelength decreases; lower sign: mirror receding → redshift), exactly twice the single-source Doppler shift because the path length itself is changing at rate $2v$.