17-Phys-A7 Optics · December 2013

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

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

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

Given. Thin-lens (Gaussian) formula $\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}$, distances measured from the lens with the real object taken as negative; object height 1 unit; object distance $|u|=1.5f$ in all four cases; telescope $f_o=20$ cm, $f_e=5$ cm, wall at $d=30$ m.

Find. Image position, nature (real/virtual) and magnification for the four lens/object combinations, and the graticule’s angular scale and system matrix.

Approach. Apply the thin-lens equation with each lens’s signed focal length and each object’s signed distance, and confirm every result with the standard two-principal-ray construction (parallel ray through the far focal point; central ray undeviated).

1. Part (a)(i) — convex lens, real object. $u=-1.5f$: $\dfrac1v=\dfrac1f-\dfrac1{1.5f}=\dfrac1{3f}\Rightarrow \boxed{v=+3f}$, $m=v/u=-2$: real, inverted, magnified $\times2$, beyond the far focal point — the classic case of an object between $F$ and $2F$.
2. Part (a)(ii) — concave lens, real object. $f=-f_c$, $u=-1.5f_c$: $\dfrac1v=-\dfrac1{f_c}-\dfrac1{1.5f_c}=-\dfrac{5}{3f_c}\Rightarrow \boxed{v=-0.6f_c}$, $m=+0.4$: virtual, upright, reduced — a diverging lens always gives this result, regardless of object position.
3. Part (b)(i) — convex lens, virtual object. Converging light aimed at a point $1.5f$ behind the lens is a virtual object, $u=+1.5f$: $\dfrac1v=\dfrac1f+\dfrac1{1.5f}=\dfrac{5}{3f}\Rightarrow \boxed{v=+0.6f}$, $m=+0.4$: real, upright, reduced.
4. Part (b)(ii) — concave lens, virtual object. $f=-f_c$, $u=+1.5f_c$: $\dfrac1v=-\dfrac1{f_c}+\dfrac1{1.5f_c}=-\dfrac1{3f_c}\Rightarrow \boxed{v=-3f_c}$, $m=-2$: virtual and inverted — a diverging lens overpowers the incoming convergence and throws the apparent source back onto the object side, still flipped.

Part (c) — the graticule. The chief ray through the centre of the objective is undeviated, so its height at any axial position is exactly linear in that position for every object distance, not only at focus.

1. Part (c)(i) — wall interval per mm. A ray through the objective centre subtends the same angle $\theta=y_{wall}/d = y_{grat}/f_o$ at the graticule (fixed at the common focal plane) as at the wall. With $y_{grat}=1$ mm, $d=30\,000$ mm, $f_o=200$ mm: $$y_{wall}=y_{grat}\frac{d}{f_o}=1\times\frac{30\,000}{200}=\boxed{150\text{ mm}}$$ of wall per mm mark (15 cm/mm).
2. Part (c)(ii) — system matrix. Using the paper’s own matrix forms, with $T(L)=\begin{bmatrix}1&L\\0&1\end{bmatrix}$ (translation) and $\text{Lens}(f)=\begin{bmatrix}1&0\\-1/f&1\end{bmatrix}$, the wall-to-eye system is $$M = T(\ell_{eye})\cdot \text{Lens}(f_e)\cdot T(f_o+f_e)\cdot \text{Lens}(f_o)\cdot T(d)$$ left as the indicated product (the objective and eyepiece are separated by $f_o+f_e=25$ cm since they share one focal plane).
3. Part (c)(iii) — column-vector elements. The two-element ray vector acted on by $M$ is $\begin{bmatrix}y\\ \theta\end{bmatrix}$: $y$ is the ray’s height above the optical axis and $\theta$ is its angle (slope) with the axis, both evaluated at the corresponding plane.