17-Phys-A7 Optics · Undated paper
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
17-Phys-A7, Optics — National Exams, May 2019. 3 hours; closed book (approved Casio/Sharp calculator only). Each question value is as indicated; exam is out of 67. Questions 1–6 are mandatory; the paper then offers a choice of Question 7 or 8, and a choice of Question 9 or 10. Every question is solved in full below as a complete study resource, including both members of each either/or pair.
Reference texts. Hecht, Optics, 5th ed.; Pedrotti, Pedrotti & Pedrotti, Introduction to Optics, 3rd ed.; Griffiths, Introduction to Electrodynamics, 4th ed. (Ch. 7–9, Maxwell’s equations and EM waves).
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.
a) Geometrical optics, in layman terms. The branch of optics that treats light as travelling in straight-line rays that bend only at interfaces (reflection, refraction) or are blocked by opaque bodies — the wave nature of light (diffraction, interference) is ignored because the objects and apertures involved are all much larger than the wavelength. It is the ray-tracing approximation used to design lenses, mirrors, and imaging systems.
b) Anti-reflection coating. A thin dielectric film (thickness ≈ λ/4, index between air and the glass) is deposited on the surface. Two reflected rays are now produced — one from the air–film interface, one from the film–glass interface — and for the design wavelength they emerge exactly out of phase (a net path difference of λ/2 once the film’s own ½-wave thickness and the reflection phase shifts at each interface are accounted for). The two reflected waves undergo destructive interference and cancel, so the reflected ray is removed and (by energy conservation) more light is transmitted.
c) Elliptical polarization. The general state of polarization: the tip of the electric-field vector, viewed head-on along the propagation direction, traces an ellipse once per optical cycle rather than a line (linear polarization, a degenerate ellipse) or a circle (equal-amplitude, 90°-out-of-phase special case). It arises whenever the orthogonal field components $E_x=E_{0x}\cos(kz-\omega t)$ and $E_y=E_{0y}\cos(kz-\omega t+\delta)$ have unequal amplitudes and/or a phase difference $\delta$ that is neither $0$, $\pi$, nor $\pm\pi/2$ with equal amplitudes.
d) Ampère’s and Faraday’s laws (two of Maxwell’s four equations, differential form):
$$\nabla\times\mathbf E=-\dfrac{\partial\mathbf B}{\partial t}\qquad\text{(Faraday's law)}\qquad\qquad \nabla\times\mathbf H=\mathbf J_f+\dfrac{\partial\mathbf D}{\partial t}\qquad\text{(Amp\`ere's law, Maxwell's correction)}$$Equivalently, in integral form:
$$\oint_C\mathbf E\cdot d\boldsymbol\ell=-\dfrac{d}{dt}\int_S\mathbf B\cdot d\mathbf a \qquad\qquad \oint_C\mathbf H\cdot d\boldsymbol\ell=I_{f,\text{enc}}+\dfrac{d}{dt}\int_S\mathbf D\cdot d\mathbf a$$Faraday’s law says a time-varying magnetic flux induces a circulating electric field; Ampère’s law (with Maxwell’s displacement-current term $\partial\mathbf D/\partial t$) says a time-varying electric flux, or a real current, induces a circulating magnetic field. Together they are the mechanism that lets an EM wave regenerate itself and propagate through vacuum.
e) Rainbow formation. Sunlight enters a spherical raindrop and refracts at the front (air→water) surface; different wavelengths refract by slightly different amounts because water is dispersive ($n$ decreases with $\lambda$, so violet bends more than red). The refracted ray then undergoes one total/near-total internal reflection off the back inner surface of the drop, and refracts a second time on exiting the front surface. This double-refraction-plus-reflection deviates each colour by a slightly different total angle (minimum deviation ≈138° for red, ≈140° for violet, measured from the antisolar point), so an observer with the sun behind them sees a coloured arc — each raindrop at a given angular position sends only one colour to the eye, and different drops at different angles supply the rest of the spectrum, red on the outside.
f) Detector irradiance. The instantaneous irradiance (intensity) is simply the power divided by the illuminated area.
Given. $P=5\ \text{W}$; $A=2\ \text{mm}^2=2\times10^{-6}\ \text{m}^2$.
Find. Irradiance $I=P/A$.
g) Why the sky is blue, sunsets red. Air molecules are far smaller than visible wavelengths, so sunlight scatters off them by Rayleigh scattering, whose cross-section scales as $\lambda^{-4}$ — blue light ($\lambda\approx450$ nm) scatters roughly $(700/450)^4\approx 5.8\times$ more strongly than red ($\lambda\approx700$ nm). Looking at any part of the sky away from the sun, what reaches the eye is mostly this scattered short-wavelength light, so the sky appears blue. At sunset the direct solar ray travels a much longer path through the atmosphere; the strongly-scattered blue (and green) components are progressively removed from the direct beam along that long path, leaving the transmitted direct light enriched in the weakly-scattered long wavelengths — orange and red.
h) Spherical aberration (one of the five third-order/Seidel aberrations — the others being coma, astigmatism, field curvature, and distortion). For a single spherical refracting/reflecting surface, only paraxial rays (close to the axis) obey the ideal thin-lens focusing relation; rays striking the lens farther from the axis are refracted more strongly and cross the axis closer to the lens than the paraxial rays do. There is therefore no single sharp focal point on-axis — each annular zone of the lens has its own focus, and a point object images as a blurred disc (the circle of least confusion) rather than a point.
