17-Phys-A7 Optics · December 2018

Question 7 of 10: Thin-lens imaging — ray diagram and matrix optics

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

Notes on this paper

17-Phys-A7, Optics — National Exams, December 2018. 3 hours; closed book (approved Sharp/Casio calculator only). Total 78 marks. 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. 9, EM waves in matter, for Maxwell's equations).

Note on question choice

Q7/Q8 and Q9/Q10 are each an either/or pair on the printed paper (only one of each counted toward the 78-mark total). As a complete study resource, all four (7, 8, 9, 10) are solved in full below.

Question 7: Thin-lens imaging — ray diagram and matrix optics (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.

Part (a) — ray-diagram image position and magnification.

Given. Thin (converging) lens, $f=7$ cm, diameter $14$ cm; object $14$ cm to the left of the lens, height $4.5$ cm (base on axis).

Find. Image position and transverse magnification $m$.

Approach. Trace two principal rays from the tip of the object: (1) parallel to the axis, refracting through the far focal point $F'$; (2) straight through the lens centre (undeviated). Their intersection locates the image tip.

Recognize the object distance. The object sits at $u=14$ cm $=2f$ (since $f=7$ cm) — the classic "object at $2f$" case, so the ray diagram is expected to place the image symmetrically at $2f$ on the far side.
Trace the rays and read the image. Ray 1 leaves the object tip parallel to the axis, meets the lens at height $4.5$ cm, and bends through $F'$ (at $7$ cm right of the lens). Ray 2 leaves the object tip and passes straight through the lens centre. The two rays cross $14$ cm to the right of the lens, at a height of $4.5$ cm below the axis (inverted) — confirmed algebraically by the thin-lens equation, $\dfrac1v=\dfrac1f+\dfrac1u$ with the Cartesian sign convention $u=-14$, $f=+7$: $\dfrac1v=\dfrac17-\dfrac1{14}=\dfrac1{14}\Rightarrow v=\boxed{+14\ \text{cm}}$ (real image, right of the lens).
Transverse magnification. $m=\dfrac{v}{u}=\dfrac{14}{-14}=\boxed{-1}$ — image height $=m\times4.5=-4.5$ cm: same size as the object, inverted.

Ray-diagram construction: object at $2f$ (14 cm) produces a real, inverted, same-size image at $2f$ on the far side — the parallel ray (through $F'$) and the centre ray cross at the image tip.

Quantity	Value
Image distance $v$	$+14$ cm (real, right of lens)
Transverse magnification $m$	$-1$
Image height	$-4.5$ cm (inverted, same size)

Part (b) — symbolic system matrix.

Given. Object-to-lens distance $s_o$, lens-to-image distance $s_i$ (both unevaluated symbols), focal length $f$, air throughout ($n=1$).

Find. The $2\times2$ system matrix $M$ relating the ray vector at the object plane to the ray vector at the image plane, as a matrix product, no values substituted.

Approach. The ray travels: translate distance $s_o$ to the lens, refract at the thin lens, translate distance $s_i$ to the image plane. In matrix optics the operations are applied in the order the ray experiences them, but because each matrix acts on a column vector from the left, the matrix product is written in reverse physical order: $M=T_{s_i}\,L_f\,T_{s_o}$.

Write the product symbolically. $$M=T_{s_i}L_fT_{s_o}=\begin{bmatrix}1&s_i\\0&1\end{bmatrix}\begin{bmatrix}1&0\\-\dfrac1f&1\end{bmatrix}\begin{bmatrix}1&s_o\\0&1\end{bmatrix}.$$
Multiply the last two matrices first. $$L_fT_{s_o}=\begin{bmatrix}1&s_o\\-\dfrac1f&1-\dfrac{s_o}{f}\end{bmatrix}.$$
Apply $T_{s_i}$ on the left. $$M=\begin{bmatrix}1-\dfrac{s_i}{f}&s_o+s_i-\dfrac{s_os_i}{f}\\[4pt]-\dfrac1f&1-\dfrac{s_o}{f}\end{bmatrix}.$$ The lower-left element $-1/f$ is the lens power (unchanged by the translations, as expected); the upper-left element $1-s_i/f$ vanishes exactly when $s_i$ satisfies the thin-lens equation for the given $s_o$, which is the imaging condition (a ray leaving the object at any angle arrives at the same image height, independent of angle) — the matrix method reproduces the same object/image relation as the ray diagram in part (a), symbolically.

Concept check

Topic. Ray-diagram imaging and $2\times2$ ray-transfer (ABCD) matrix optics for a single thin lens.

Key relations. $1/v=1/f+1/u$; $m=v/u$; $M=T_{s_i}L_fT_{s_o}$, with the imaging condition equivalent to the matrix's upper-left element vanishing.

Why this works. A thin lens is, to paraxial order, a linear map on $(x,n\alpha)$; composing a translation, a refraction, and another translation as ordered matrix multiplication reproduces exactly the same thin-lens/magnification results derived geometrically, but stays valid for compound systems where geometric ray-tracing becomes unwieldy (Question 8).

Common pitfall. Writing the matrix product in the physical left-to-right order of the elements instead of the required right-to-left (first-operation-rightmost) order for column-vector matrix multiplication.

Source. Hecht, Optics, Ch. 6 (matrix methods in paraxial optics).

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