NBPhO 2026

Overview

A thin converging lens of fixed orientation moves so that its centre $L$ traces a circle in the plane of the figure. The optical axis $\hat{n}$ lies in this plane, and the focal length $f$ is fixed. Two object points $P$ and $Q$ are imaged to closed curves $P'$ and $Q'$.

We will need only two facts about thin-lens imaging:

1. The chief ray is undeviated. A ray from any object point $P$ through the lens centre $L$ continues in a straight line. Hence $P$, $L$, $P'$ are collinear, with $L$ between $P$ and $P'$ whenever the image is real (the case here).

2. Lens equation along the chord $PL$. Let $\alpha$ be the angle between the line $PL$ and the optical axis $\hat{n}$. The axial object and image distances are

$$u = |PL|\cos\alpha,\qquad v = |LP'|\cos\alpha.$$

Substituting these into $1/u + 1/v = 1/f$ gives the form we will use throughout part (ii):

$$\frac{1}{|PL|} + \frac{1}{|LP'|} \;=\; \frac{\cos\alpha}{f}.$$

There are five unknowns hidden in the figure: the centre $O$ and radius $R$ of the lens-centre circle (3 numbers), the direction of the optical axis $\hat{n}$ (1 number), and the focal length $f$ (1 number). The problem only asks for the first four; the fifth ($f$) drops out as a free by-product, as we will see.

The solution proceeds in two essentially independent steps:

• Part (i) uses only fact 1 — the curves $P'$ and $Q'$ live in the angular wedges subtended by the lens circle at $P$ and $Q$.
• Part (ii) uses fact 2 — a harmonic-mean construction collapses the unknown angle $\alpha$ and produces a straight line parallel to the lens plane.

---

Part (i) — The lens-centre circle

Key observation: tangent rays are shared

Because $L$ lies on segment $PP'$, the direction from $P$ to $P'$ is exactly the direction from $P$ to $L$. As $L$ runs around its circle once, the line of sight $PL$ pivots through the angular wedge bounded by the two tangent lines from $P$ to the lens circle, sweeping it twice (once on the way out, once on the way back). The image curve $P'$ is contained in this same wedge, and reaches its boundary at the moment $L$ coincides with one of the tangent points.

The conclusion:

The two tangent lines from $P$ to the image curve $P'$ are exactly the two tangent lines from $P$ to the lens-centre circle.

The same holds for $Q$ and curve $Q'$. So the figure already contains four lines tangent to the lens circle — two from $P$, two from $Q$ — even though the lens circle itself is not drawn.

Construction

                 ℓP+      ℓQ+
    P •──────────●─────────●──────●Q
                  \       /
                   \     /
                    \   /
                     \ /
              bP ─────●───── bQ
                     /O\
                    /   \
                   /     \
                  /       \
    ── ℓP− ──── ●  ────────●─── ℓQ− ──
                  lens
                  circle

1. Draw the four tangent rays. From $P$, sweep a straightedge until it just touches curve $P'$ from each side; mark the two extreme positions $\ell_P^{(+)}$ and $\ell_P^{(-)}$. Repeat from $Q$ on curve $Q'$ to obtain $\ell_Q^{(+)}$ and $\ell_Q^{(-)}$.
2. Bisect the angle at $P$ between $\ell_P^{(+)}$ and $\ell_P^{(-)}$. Call the bisector $b_P$.
3. Bisect the angle at $Q$ between $\ell_Q^{(+)}$ and $\ell_Q^{(-)}$. Call the bisector $b_Q$.
4. The centre is $O = b_P \cap b_Q$.
5. The radius $R$ is the perpendicular distance from $O$ to any one of the four tangent rays.

Why this works

For any point $X$ outside a circle, the two tangent lines drawn from $X$ are reflections of each other across the line $XO$ (where $O$ is the centre). So the angle bisector at $X$ between them passes through $O$. We have two such external points, $P$ and $Q$, hence two bisectors that intersect at $O$. The radius is then the common perpendicular distance from $O$ to all four tangent lines — which provides a four-fold consistency check (all four perpendicular distances must agree).

Educational remark — tangent line, but not tangent point

The tangent line $\ell_P^{(+)}$ touches the lens circle at some point $T_P^{(+)}$, where the lens centre passes at one specific instant. The same line touches the curve $P'$ at the image of $T_P^{(+)}$, which is the point $P + \mu(T_P^{(+)})\,(T_P^{(+)} - P)$. Because $\mu = u/(u-f) > 1$ for real images, the tangent point on the curve $P'$ is further from $P$ than the tangent point on the lens circle, but the two tangent points lie on the same straight tangent line. This is enough for the construction — only the line, not the contact points, matters.

Self-intersection of $Q'$

The figure shows that curve $Q'$ has a self-intersection (it is shaped like a figure-eight), while $P'$ is a simple loop. The construction cares only about the outermost tangent rays from $Q$ — the rays bounding the angular extent of the curve as seen from $Q$. Internal tangent lines that touch interior loops of a self-intersecting curve are not tangent to the lens circle. (A self-intersection of $Q'$ at angle $\theta$ from $Q$ corresponds to two distinct lens positions $L_-, L_+$ on the chord through $Q$ at angle $\theta$ producing the same image — a coincidence in $\mu_-|QL_-| = \mu_+|QL_+|$ that does not affect the angular envelope.)

---

Part (ii) — The lens-plane direction

Key observation: a harmonic-mean point on each chord

For any chord $PL$ at angle $\alpha$ to the optical axis, rearrange the lens equation derived in the overview:

$$\frac{|PL| + |LP'|}{|PL|\cdot|LP'|} \;=\; \frac{\cos\alpha}{f} \quad\Longleftrightarrow\quad \boxed{\;\frac{|PL|\cdot|LP'|}{|PL|+|LP'|} \;=\; \frac{f}{\cos\alpha}.\;}$$

The quantity on the left is computable directly from the figure once we know $L$ (which we do, from part (i)). It is one half of the harmonic mean of the two segment lengths.

Now mark, on segment $PL$, the auxiliary point

$$M \;=\; P \,+\, \frac{|PL|\cdot|LP'|}{|PL|+|LP'|}\;\hat{u}_{PL}, \qquad \hat{u}_{PL} := \frac{L-P}{|PL|}.$$

Two properties of $M$:

• $|PM| = f/\cos\alpha$ by construction.
• The component of $M-P$ along the optical axis is

$$(M - P)\cdot\hat{n} \;=\; |PM|\cos\alpha \;=\; f.$$

The right-hand side is independent of where $L$ sits on its circle. So no matter which lens position we pick, the corresponding $M$ has the same axial coordinate relative to $P$. In other words:

Every $M$ lies on a single straight line — the line perpendicular to $\hat{n}$ at axial distance $f$ from $P$.

That line is parallel to the lens plane, since both are perpendicular to $\hat{n}$. Determining its direction is therefore equivalent to answering part (ii).

Construction

We assume part (i) has produced the lens circle. We do not need to know $\hat{n}$ or $f$ — they emerge.

1. Choose three (or more) well-separated points $P'_1, P'_2, P'_3$ on the curve $P'$.
2. For each $i$, draw the line $PP'_i$. It meets the lens circle in two points; call either one $L_i$. (See the both branches remark below — it does not matter which.)
3. Construct the length

$$|PM_i| \;=\; \frac{|PL_i|\cdot|L_iP'_i|}{|PL_i|+|L_iP'_i|}$$

using the harmonic-mean ruler-and-compass construction below, and mark $M_i$ on segment $PL_i$ at this distance from $P$. 4. The points $M_1, M_2, M_3, \ldots$ are collinear. The line through them is parallel to the lens plane.

The same recipe starting from $Q$ produces a parallel line (also perpendicular to $\hat{n}$, but at axial distance $f$ from $Q$ rather than from $P$). Doing both halves and confirming they come out parallel is a strong consistency check on the entire construction.

Harmonic-mean construction (similar triangles)

Given collinear $P, L, P'$ with $L$ between, construct $M$ on segment $PL$ with $|PM| = |PL||LP'|/(|PL|+|LP'|)$:

        A •
          |\
          | \
          |  \
          |   \
        B •    \
          |\    \
          | \    \
          |  \    \
          P • M ── L ───── P'

• Through $P$, draw any auxiliary line not collinear with $PP'$.
• On it, mark $A$ with $|PA| = |PP'| = |PL|+|LP'|$.
• On it, mark $B$ with $|PB| = |LP'|$, between $P$ and $A$.
• Draw segment $AL$.
• Through $B$, draw the line parallel to $AL$. It meets segment $PL$ at $M$.

By similar triangles $PBM \sim PAL$:

$$\frac{|PM|}{|PL|} \;=\; \frac{|PB|}{|PA|} \;=\; \frac{|LP'|}{|PL|+|LP'|},$$

so $|PM| = |PL|\cdot|LP'|/(|PL|+|LP'|)$ as required.

Why both branches give the same $M$

A line through $P$ in a direction inside the wedge crosses the lens circle in two points $L_-$ and $L_+$ on the same chord — hence at the same angle $\alpha$ from $\hat{n}$. The formula $|PM| = f/\cos\alpha$ depends only on $\alpha$, so both $L_\pm$ produce the same $M$ on the same ray from $P$.

Concretely: the two branches of curve $P'$ along the chord through $P$ at angle $\theta$ — the two intersection points of the chord-line with $P'$ — yield identical auxiliary points $M$. This is consistency check #2: pick a chord, build $M$ from each branch, and confirm they coincide.

Educational remark — why the harmonic mean appears

In a 1D problem along the axis, the lens equation reads $1/u + 1/v = 1/f$, i.e. $f$ is half the harmonic mean of $u$ and $v$. The angle $\alpha$ enters because the chord $PL$ is not along the axis: $|PL|=u/\cos\alpha$ replaces $u$, and similarly for $v$. The factor $\cos\alpha$ appears as a uniform stretch on both sides of the equation, producing $f/\cos\alpha = |PM|$. Marking that distance along the chord is geometrically equivalent to projecting onto the axis a fixed offset of $f$ from $P$ — and projection onto a fixed direction is exactly what produces a straight line locus perpendicular to that direction.

What this also gives us — the focal length

The construction does not require $f$, but once the line of $M$‘s is drawn, $f$ can be read off in two ways:

• $f = |PM|\cos\alpha$ for any one chord, where $\alpha$ is now measurable as the angle between $PL$ and the constructed axial direction $\hat{n}$.
• $f$ is the perpendicular distance from $P$ to the line of $M$‘s.

This was not asked, but it confirms that the curves contain enough information to fully determine the optical setup (lens circle, axis direction, focal length).

Sanity check — limiting case $f \to \infty$

In the limit $f\to\infty$ (no lens, “imaging” is just a clear sightline), $u \to f$ from above gives $v \to \infty$, and the image disappears. So $f$ finite was essential. Less trivially: as the lens circle shrinks, the two tangent lines from $P$ converge to a single line of sight to $O$, and the image curve $P'$ shrinks to a single point — the construction degenerates but remains consistent.

---

Summary of constructions

Part	Object	Construction
(i)	The lens-centre circle	(a) Draw the two tangent rays from $P$ to curve $P'$ and the two tangent rays from $Q$ to curve $Q'$. (b) Bisect the angle between the two rays at $P$, and again at $Q$. (c) The centre $O$ of the lens circle is the intersection of the two bisectors; the radius $R$ is the perpendicular distance from $O$ to any of the four tangent rays.
(ii)	A line parallel to the lens plane	For three or more points $P'_i$ on curve $P'$: locate $L_i$ on the line $PP'_i$ (using the lens circle from part (i)), and mark $M_i$ on segment $PL_i$ at distance $\dfrac{

The reasoning behind each step:

• Part (i): $P', L$ lie on the same ray from $P$, so the angular extent of curve $P'$ at $P$ equals that of the lens circle, and their tangent rays from $P$ coincide. Two angle bisectors from external points pin the centre.
• Part (ii): The lens-equation rearrangement $\frac{|PL||LP'|}{|PL|+|LP'|} = \frac{f}{\cos\alpha}$ means that the harmonic-mean point on each chord lies at axial distance $f$ from $P$, irrespective of $L$. The locus is therefore a straight line perpendicular to the optical axis, which is parallel to the lens plane.