17-Phys-A7 Optics · Undated paper

Question 8 of 10: Ball-lens fiber coupler — refractive index (choice with Q7)

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

Notes on this paper

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).

Note on question choice

Q7/Q8 and Q9/Q10 are each an either/or pair on the printed paper (only one of each counts toward the mark total, and the printed 67-mark total matches the mandatory questions plus the Q8+Q10 pairing). As a complete study resource, all four (7, 8, 9, 10) are solved in full below.

Note on the numbers

The numbers used here are internally consistent (mark totals sum correctly, angle/critical-angle values cross-check, see the Concept boxes below).

Question 8: Ball-lens fiber coupler — refractive index (choice with Q7) (8 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.

Given. Full glass sphere (“ball lens”), diameter 1 mm (radius $R=0.5$ mm) — both spherical surfaces share this same radius; surrounding medium air, $n=1.000$; parallel (collimated) input beam; output fiber placed directly against the back surface of the ball, i.e. the beam must focus exactly at the rear vertex. Refraction matrix $R_T=\begin{pmatrix}1&0\\-(n_{out}-n_{in})/R&1\end{pmatrix}$, translation $T_D$, as given.

Find. $n_{\text{glass}}$, the refractive index of the ball lens.

Approach. Chain the two refractions (air→glass at the front surface, radius $+R$; glass→air at the rear surface, radius $-R$) with a translation of the full diameter $2R$ through the glass in between, using the given matrices exactly as stated. The physical design requirement — the fiber sits directly on the back surface, so parallel rays entering at any height $x_0$ must all converge to $x=0$ right at the exit vertex — means the system matrix’s $A$ element (which maps input height to output height for an input ray parallel to the axis) must vanish.

Front-surface refraction (air $\to$ glass, $R_1=+R$). $R_{T1}=\begin{pmatrix}1&0\\-(n-1)/R&1\end{pmatrix}.$
Translate the full diameter through the glass. $T_{2R}=\begin{pmatrix}1&2R\\0&1\end{pmatrix}.$
Rear-surface refraction (glass $\to$ air, $R_2=-R$). $R_{T2}=\begin{pmatrix}1&0\\-(n-1)/R&1\end{pmatrix}$ (same form as step 1, since the two sign flips — $n_{out}-n_{in}\to1-n$ and $R\to-R$ — cancel).
Multiply the chain. $M_{\text{ball}}=R_{T2}\,T_{2R}\,R_{T1}=\begin{pmatrix}3-2n & 2R\\[2pt]\dfrac{2(n^2-3n+2)}{R} & 3-2n\end{pmatrix}.$
Impose the focusing condition $A=0$. $3-2n=0\ \Rightarrow\ n=\boxed{1.5}.$
Cross-check via the stated design fact ("focal length is twice the radius"). The system’s effective focal length is $f=-1/C$. At $n=1.5$: $C=\dfrac{2(1.5^2-3(1.5)+2)}{R}=\dfrac{2(-0.25)}{R}=-\dfrac{0.5}{R}$, so $f=-1/C=2R$ — exactly the stated design condition, confirming $n=1.5$ from an independent equation.

The ball diameter (1 mm) cancels completely out of both the $A=0$ condition and the $f=2R$ check — the refractive index that puts the focus exactly on the back surface of a full sphere is a fixed number independent of the sphere’s size, so the "take the diameter to be 1 mm" instruction is not actually needed for this sub-part; a real design would use it downstream to translate the answer into a numerical aperture or spot size.

Quantity	Value
Ball-lens refractive index $n_{\text{glass}}$	$1.5$

Concept check

Topic. Matrix optics through a full-sphere (ball) lens; the classical $n=2$-focuses-on-surface result and its variant here.

Key relations. $R_T=\begin{pmatrix}1&0\\-(n_{out}-n_{in})/R&1\end{pmatrix}$; focusing condition $A=0$; system focal length $f=-1/C$.

Why this works. A ball lens has exactly two refracting surfaces of the same radius magnitude but opposite sign convention (convex toward the beam on entry, convex away from it on exit), so its system matrix reduces to a single free parameter $n$ once $R$ is fixed — both the "focus at the back surface" condition and the "$f=2R$" statement collapse to the same equation, which is a useful internal check on the algebra.

Common pitfall. Confusing this result with the well-known "$n=2$ focuses a sphere exactly on its own back surface" textbook fact — that classical result assumes the rigorous reduced-angle ray vector (translation matrix scaled by $1/n$); the exam’s own formula sheet defines $T_D$ without that scaling, and using the matrices exactly as given (as this solution does) is what makes $n=1.5$, not $n=2$, the answer consistent with the exam’s own stated design fact.

Source. Hecht, Optics, Ch. 6 (matrix methods, thick lenses/spheres); Pedrotti, Introduction to Optics, Ch. on Gaussian optics.

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