17-Phys-A7 Optics · Undated paper

Question 7 of 10: Expanded-beam fiber coupler — system matrix (choice with Q8)

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 7: Expanded-beam fiber coupler — system matrix (choice with Q8) (10 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. Two identical thin lenses, focal length $f$ (stated as 4 mm; diameter 6.0 mm, not needed here), lens separation $D$; input fiber at the front focal point of lens 1, output fiber at the back focal point of lens 2. Translation matrix $T_D=\begin{pmatrix}1&D\\0&1\end{pmatrix}$, thin-lens matrix $L_f=\begin{pmatrix}1&0\\-1/f&1\end{pmatrix}$ (as given on the exam’s own formula sheet).

a) System matrix, input fiber to output fiber.

Find. $M_{\text{sys}}$ relating the ray vector at fiber 1’s end-face to the ray vector at fiber 2’s end-face, in symbolic $f,D$.

Approach. Chain the five elements the ray passes through, left-multiplying in propagation order (rightmost matrix acts first): translate distance $f$ from fiber 1 to lens 1, refract through lens 1, translate the lens separation $D$, refract through lens 2, translate distance $f$ to fiber 2.

Write the matrix chain. $M_{\text{sys}}=T_f\,L_f\,T_D\,L_f\,T_f\;=\;\begin{pmatrix}1&f\\0&1\end{pmatrix}\begin{pmatrix}1&0\\-1/f&1\end{pmatrix}\begin{pmatrix}1&D\\0&1\end{pmatrix}\begin{pmatrix}1&0\\-1/f&1\end{pmatrix}\begin{pmatrix}1&f\\0&1\end{pmatrix}.$
Multiply out. $M_{\text{sys}}=\boxed{\begin{pmatrix}-1&0\\[2pt]\dfrac{D-2f}{f^2}&-1\end{pmatrix}}.$

Two features of this result are worth noting. The $B$ element is exactly zero for any $D$ — the input and output fiber end-faces are always conjugate (imaging) planes of one another, which is exactly why the design places the fibers at the two focal points in the first place. The $A=D=-1$ diagonal means unit-magnitude, inverted coupling (no size change between the two fiber end-faces) regardless of the lens spacing. The lens separation only shows up in the $C$ element (the system’s optical power); at $D=2f$ (lenses separated by twice the focal length, the classic confocal/afocal "4f" relay, matching the collimated ray path drawn between the two lenses in the source figure) $C=0$ too, and the system becomes perfectly afocal.

b) Focal length of the thin lenses.

Find. $f$.

Approach. The value is given directly in the problem statement (4 mm); cross-check it independently by reading the fiber-to-lens spacing off the source figure’s own 0.5 mm grid, since the input fiber sits exactly at that lens’s front focal point by construction.

Grid measurement. The input-fiber tip to lens-1 spacing spans 8 grid squares $\times\,0.5\ \text{mm/square}=4.0\ \text{mm}$.
Result. $f=\boxed{4.0\ \text{mm}}$, matching the value stated in the problem text exactly.

Quantity	Value
System matrix $M_{\text{sys}}$	$\begin{pmatrix}-1&0\\(D-2f)/f^2&-1\end{pmatrix}$
Focal length $f$	$4.0$ mm

Concept check

Topic. ABCD (ray-transfer) matrix optics; conjugate imaging condition; the confocal/4f relay.

Key relations. $T_D=\begin{pmatrix}1&D\\0&1\end{pmatrix}$, $L_f=\begin{pmatrix}1&0\\-1/f&1\end{pmatrix}$; imaging condition $B=0$; system power $=-C$.

Why this works. Placing an object at a lens’s front focal point always makes the corresponding image plane sit at infinity (collimated output) for the first lens, and a second lens used in reverse re-images that collimated beam back to a point at its own back focal point — the algebra above is just the ABCD bookkeeping confirming that geometric argument for arbitrary lens spacing.

Common pitfall. Substituting numerical values before finishing the symbolic matrix multiplication, which the question explicitly asks to avoid; and mis-ordering the matrix product (ray-transfer matrices apply right-to-left in the direction of propagation).

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

← Question 6 Next: Question 8 →