17-Phys-A7 Optics · December 2018

Question 8 of 10: Astronomical telescope — aperture stop, system matrix, resolving power

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 8: Astronomical telescope — aperture stop, system matrix, resolving power (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 trajectory and the aperture stop.

Given. Objective $f_1=+12$ cm, $\phi_1=14$ cm (radius $7$ cm); eyepiece $f_2=+6$ cm, $\phi_2=10$ cm (radius $5$ cm); relaxed-eye (afocal) arrangement, so lens separation $d=f_1+f_2=18$ cm; object (star) at infinity.

Find. Which element limits the beam (the aperture stop).

Approach. A ray entering parallel to the axis at height $h$ (from a star on-axis at infinity) is bent by the objective toward its focal point $F_1'$, which coincides with the eyepiece's own focal point $F_2$ in the afocal arrangement; using the thin-lens matrix from Question 7(b) over the translation $d$, its height at the eyepiece is $h'=h\left(1-\dfrac{d}{f_1}\right)$. Compare the ray height each element's own edge would allow: whichever is reached first (smaller required object-side height) is the aperture stop.

Height-transfer factor to the eyepiece. $1-\dfrac{d}{f_1}=1-\dfrac{18}{12}=-0.5$, i.e. $|h'|=0.5\,h$ — the ray height at the eyepiece is always half its height at the objective.
Height needed to just clear each element's own edge. Objective edge: $h=7$ cm. Eyepiece edge: $|h'|=5$ cm $\Rightarrow h=5/0.5=10$ cm at the objective.
Compare. A ray starting at the objective is clipped by the objective's own $7$ cm radius before it could reach the $10$ cm height that would be needed to be clipped by the eyepiece instead — so the objective lens is the aperture stop (and, since the object is at infinity, it is also the entrance pupil).

Afocal telescope, star at infinity: the marginal ray at the objective's own edge (height 7 cm) maps to only 5 cm at the eyepiece — well inside its 5 cm-radius aperture — confirming the objective limits the beam.

Part (b) — symbolic telescope system matrix.

Given. $f_1$ (objective), $f_2$ (eyepiece), separation $d$, unevaluated.

Find. System matrix from just left of the objective to just right of the eyepiece.

Order of operations (right-to-left for column vectors). Refract at the objective, translate the separation $d$, refract at the eyepiece: $$M=L_{f_2}\,T_d\,L_{f_1}=\begin{bmatrix}1&0\\-\dfrac1{f_2}&1\end{bmatrix}\begin{bmatrix}1&d\\0&1\end{bmatrix}\begin{bmatrix}1&0\\-\dfrac1{f_1}&1\end{bmatrix}.$$
Multiply the last two first. $$T_dL_{f_1}=\begin{bmatrix}1-\dfrac{d}{f_1}&d\\-\dfrac1{f_1}&1\end{bmatrix}.$$
Apply $L_{f_2}$ on the left. $$M=\begin{bmatrix}1-\dfrac{d}{f_1}&d\\[4pt]-\dfrac1{f_2}\!\left(1-\dfrac{d}{f_1}\right)-\dfrac1{f_1}&1-\dfrac{d}{f_2}\end{bmatrix}.$$ For the afocal case $d=f_1+f_2$ this reduces (substituting only as a check, not as the requested symbolic answer) to a pure angular-magnification matrix with zero lower-left element — a bundle of parallel input rays exits as a bundle of parallel output rays, the defining property of an afocal telescope.

Part (c) — angular resolving power.

Given. $\lambda=0.5\ \mu\text{m}$; limiting aperture $=$ objective diameter $D=\phi_1=14$ cm $=0.14$ m (part a).

Find. Minimum resolvable angular separation $\theta_{\min}$.

Approach. Two close stars are resolved by the same Rayleigh criterion as Question 3's circular-aperture pattern, applied to the objective — the element identified in part (a) as the stop, and hence the aperture that sets the diffraction limit: $\theta_{\min}=1.22\lambda/D$. A larger objective diameter gives a smaller (better) $\theta_{\min}$; this is why the resolving power of an astronomical telescope is fundamentally set by objective diameter, not by focal length or magnification.

Angular resolving power. $\theta_{\min}=\dfrac{1.22\lambda}{D}=\dfrac{1.22(0.5\times10^{-6})}{0.14}=4.357\times10^{-6}\ \text{rad}$. Converting to the astronomically conventional unit: $\theta_{\min}=4.357\times10^{-6}\times206265\ \text{arcsec/rad}=\boxed{0.899\ \text{arcsec}\approx0.9''}.$

Quantity	Value
Aperture stop	Objective lens ($D=14$ cm)
Angular resolving power $\theta_{\min}$	$4.36\times10^{-6}$ rad $=0.90''$

Concept check

Topic. Aperture stop identification in a two-lens system; the telescope system matrix; diffraction-limited resolving power of a telescope.

Key relations. Height-transfer $h'=h(1-d/f_1)$; $M=L_{f_2}T_dL_{f_1}$; $\theta_{\min}=1.22\lambda/D$.

Why this works. The aperture stop is whichever physical aperture, once its clipping effect is propagated to a common reference (here, the eyepiece), most restricts the ray bundle; because it is also the entrance pupil for an object at infinity, it is exactly the aperture that sets the telescope's diffraction-limited resolving power via the Rayleigh criterion of Question 3(b).

Common pitfall. Assuming the smaller-diameter element (here the eyepiece) is automatically the stop — what matters is each element's clipping effect after propagating ray heights through the intervening optics, not the raw diameters compared directly.

Source. Hecht, Optics, Ch. 5 (stops and pupils), Ch. 6 (matrix methods), Ch. 10 (Rayleigh criterion, telescope resolving power).

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