17-Phys-A7 Optics · December 2013

Question 1 of 6: Definitions and short concept questions

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

Notes on this paper

Paper format. National Examinations, December 2013, 98-Phys-A7 Optics. Three hours, closed book, one approved Casio or Sharp calculator. Question 1 (20 short parts) is mandatory; the rubric asks for any four of Questions 2–6 — all six are worked here, because the set is a study resource.

Reference texts. E. Hecht, Optics, 5th ed.; F. L. Pedrotti, L. M. Pedrotti and L. S. Pedrotti, Introduction to Optics, 3rd ed.

Question 1: Definitions and short concept questions (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.

Twenty independent one-mark parts; each is answered in turn.

(a) Light. Electromagnetic radiation over the band of wavelengths (roughly 380–750 nm) to which the human eye responds; in the broader sense used throughout physical optics, any electromagnetic radiation treated with the same wave/photon formalism (UV through IR).

(b) Optics. The branch of physics concerned with the generation, propagation and detection of light and its interaction with matter.

(c) Geometrical optics. The short-wavelength ($\lambda\to 0$) approximation in which light is treated as straight rays that change direction only at interfaces (reflection/refraction); valid whenever every aperture and obstacle is much larger than $\lambda$.

(d) Physical optics. The treatment of light as an electromagnetic wave, retaining the finite wavelength so that interference, diffraction and polarization can be described.

(e) Why physical optics was needed. Geometrical optics cannot account for diffraction (bending of light around edges), interference fringes from multiple coherent paths, the diffraction-limited resolution of instruments, or polarization — all of which appear once an aperture or feature size becomes comparable to $\lambda$.

(f) Plane of incidence. The plane containing the incident ray, the reflected/refracted ray, and the normal to the interface at the point of incidence.

(g) Law of reflection. $\theta_r=\theta_i$ (measured from the normal), with the reflected ray in the plane of incidence.

(h) Law of refraction (Snell’s law). $n_1\sin\theta_1 = n_2\sin\theta_2$, incident ray, refracted ray and normal coplanar.

(i) Wavelength/frequency ranges (all four rows, using $f=c/\lambda$; two suffice for the mark):

Wavelength/frequency bands

Quantity	Symbol	Value
UV light	—	100–400 nm  ↔  7.5×1014–3.0×1015 Hz
Blue light	—	450–495 nm  ↔  6.06×1014–6.66×1014 Hz
Red light	—	620–750 nm  ↔  4.00×1014–4.84×1014 Hz
IR light	—	750 nm–1 mm  ↔  3.0×1011–4.00×1014 Hz

(j) Transverse plane wave. A wave whose field vectors oscillate perpendicular to the direction of propagation, with surfaces of constant phase (wavefronts) that are infinite planes normal to the propagation vector $k$.

(k) Is monochromatic plane-wave analysis restrictive? No. Any physically realizable field, of any shape or spectrum, can be written as a Fourier superposition of monochromatic plane waves, so plane-wave analysis is fully general once superposition is invoked.

(l) Ray ↔ plane wave. A ray is the trajectory along the local wavevector $k$ — the normal to the physical-optics wavefront at each point — in the short-wavelength (eikonal) limit.

(m)–(p) The wave $f(x,t)=A\exp(ax)\sin(2\pi(at+bx))$. Matching the sinusoidal part to the standard travelling-wave form $\sin(2\pi(\nu t + x/\lambda))$ (a wave moving in the $-x$ direction) identifies $a=\nu$ (temporal frequency) and $b=1/\lambda$ (reciprocal wavelength).

1. Part (m) — speed of propagation. Constant phase requires $2\pi(at+bx)=\text{const}\Rightarrow a\,dt+b\,dx=0$, so $$\boxed{v=\left|\frac{dx}{dt}\right|=\frac{a}{b}}\ \text{, travelling in the } -x \text{ direction.}$$
2. Part (n) — interpretation of $a$. At fixed $x$, $f\propto\sin(2\pi a t+\text{const})$ oscillates $a$ times per unit time, so $\boxed{a=\nu}$, the temporal frequency (Hz).
3. Part (o) — interpretation of $b$. At fixed $t$, $f\propto\sin(2\pi b x+\text{const})$ repeats every $\Delta x=1/b$, so $\boxed{b=1/\lambda}$, the reciprocal wavelength.
4. Part (p) — refractive index. The frequency is unchanged on entering the medium, so $\lambda_o=c/\nu=c/a$; the medium wavelength is $\lambda=1/b$. Hence $$\boxed{n=\frac{\lambda_o}{\lambda}=\frac{cb}{a}}$$ — identical to $n=c/v$ with $v=a/b$ from (m).

(q) Interference vs. diffraction. Both are superposition effects; by convention, “interference” describes the pattern from a small number of discrete coherent sources or paths (e.g. two slits), while “diffraction” describes the pattern from a continuum of secondary (Huygens) wavelets across an aperture or obstacle.

(r) Poynting vector. $$\vec{S}=\vec{E}\times\vec{H}$$ SI units: W/m² (power per unit area).

(s) Fraunhofer diffraction. The far-field limit, in which source and/or observation point are effectively at infinity (or brought there by a lens), so the diffraction integral reduces to the Fourier transform of the aperture function.

(t) Fresnel diffraction. The near-field regime, in which finite source/observation distances leave a residual quadratic (wavefront-curvature) phase term in the diffraction integral, so it does not reduce to a simple Fourier transform.

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