17-Phys-A7 Optics · May 2015

Question 2 of 6: Thin-lens ray tracing and a Galilean telescope

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

Question 2: Thin-lens ray tracing and a Galilean telescope (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.

Part (a) — real object at $u=-2f$. The two-ray construction uses (i) the ray leaving the object parallel to the axis, which after the lens is bent to pass through (or, for a diverging lens, to appear to diverge from) the back focal point, and (ii) the ray through the lens centre, which is undeviated. Their intersection locates the image; the Cartesian thin-lens equation $1/v-1/u=1/f$ confirms each geometric result.

With $f>0$ and $u=-2f$: $v=+2f$, $m=v/u=-1$. The image is real, inverted, and the same size as the object, formed $2f$ beyond the lens — the classic symmetric "2f–2f" point.

With $f<0$ and $u=-2f$ (i.e. $u=-2|f|$): $v=-\tfrac{2}{3}|f|$, $m=+\tfrac13$. The image is virtual, upright, and reduced to $1/3$ size, formed between the lens and the object — as a diverging lens always produces for a real object.

Part (b) — virtual object at $u=+2f$. A virtual object means the incident rays are already converging toward a point $2f$ behind the lens when they arrive at it (as they would if a prior converging system, e.g. another lens, were sending light there). The same two construction rays still apply: the ray that (extended straight through the lens plane) is heading toward the virtual-object height parallel to the axis refracts through the focal point exactly as before; the ray aimed at the lens centre is undeviated.

With $f>0$ and $u=+2f$: $v=+\tfrac23 f$, $m=+\tfrac13$. A real image forms $\tfrac23 f$ beyond the lens, reduced to $1/3$ size and erect relative to the converging incident bundle — a converging lens can turn a virtual object into a real image closer than $f$.

With $f<0$ and $u=+2|f|$: $v=-2|f|$, $m=-1$. The image is virtual, inverted, the same size as the (virtual) object, and forms $2|f|$ in front of the lens.

Part (c) — telescope. The objective ($f_o=+12$ cm) and eyepiece ($f_e=-4$ cm) form a Galilean telescope (converging objective, diverging eyepiece). Aligning it "for a near point of 30 cm" means the eyepiece is set so the final virtual image forms $30$ cm from the eyepiece — not at infinity, the usual relaxed-eye case.

Given. $f_o=+12$ cm, $f_e=-4$ cm, final virtual image at $v=-30$ cm from the eyepiece.

Find. (i) lens separation $d$; (ii) angular magnification $M$; (iii) meaning of the two entries of the system-matrix ray vector.

Approach. A distant object is imaged by the objective at its focal point $F'_o$; that image is a virtual object for the eyepiece. Apply the thin-lens equation to the eyepiece with the required final image distance to fix the eyepiece's object distance, hence $d$; then trace the chief ray (through the objective centre) with the paraxial ray-transfer matrix to get the emergent ray angle, and hence $M$.

1. Locate the intermediate image. A distant object is brought to a real image by the objective at its back focal point, $x=f_o=+12$ cm (measured from the objective). This image is the object for the eyepiece.
2. Solve the eyepiece equation for its object distance. With $1/v-1/u=1/f_e$ and $v=-30$ cm, $f_e=-4$ cm: $$\frac{1}{u}=\frac{1}{v}-\frac{1}{f_e}=\frac{1}{-30}-\frac{1}{-4}=\frac{13}{60}\ \Rightarrow\ u=\frac{60}{13}=4.62\ \text{cm}.$$ This positive $u$ confirms the intermediate image sits $4.62$ cm beyond the eyepiece — a virtual object for it, as required.
3. Separation of the lenses. The intermediate image is at $x=f_o=12$ cm from the objective and at $x=u=4.62$ cm beyond the eyepiece, so the eyepiece sits at $d=f_o-u$: $$d = 12-\frac{60}{13}=\frac{96}{13}=\boxed{7.38\ \text{cm}}.$$
4. Angular magnification — chief-ray construction. Trace the ray from a distant off-axis point that passes through the objective's centre: it is undeviated there (height $0$, angle $\alpha$ unchanged), then travels a distance $d$ to reach the eyepiece at height $y=d\alpha$, still at angle $\alpha$. The eyepiece then bends it: $\theta_{\text{out}}=\alpha-y/f_e=\alpha(1-d/f_e)$. With the eye placed at the eyepiece, this emergent angle IS the angle at which the final (virtual) image is seen, so $$M=\frac{\theta_{\text{out}}}{\alpha}=1-\frac{d}{f_e}=1-\frac{96/13}{-4}=1+\frac{24}{13}=\frac{37}{13}=\boxed{2.85}.$$ Check: for the ordinary relaxed-eye telescope, $d=f_o+f_e=8$ cm, and the same formula gives $M=1-8/(-4)=3=-f_o/f_e$ — the familiar Galilean-telescope result, recovered as the special case $v\to-\infty$.

5. System-matrix column vector. In paraxial (Gaussian) matrix optics a ray at any plane is represented by the 2-vector $\binom{y}{\theta}$: the first element $y$ is the ray's height above the optical axis, and the second element $\theta$ is the ray's angle (slope) with respect to the axis, in the small-angle approximation. It is exactly this vector, propagated through a translation matrix $\begin{pmatrix}1&d\\0&1\end{pmatrix}$ and a thin-lens matrix $\begin{pmatrix}1&0\\-1/f&1\end{pmatrix}$, that generated the bend used step-by-step above.