17-Phys-A7 Optics · May 2016

Question 3 of 6: Ray trace through a glass cube, full-length mirror, Galilean telescope

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

Question 3: Ray trace through a glass cube, full-length mirror, Galilean telescope (15 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 trace through the cube.

Given. Angle of incidence $\theta_i=45^\circ$ on the centre of the top (horizontal) face of a glass cube, $n=1.414\approx\sqrt2$, surrounding medium air ($n=1$).

Find. The complete ray path through and out of the cube.

Approach. Refract at the top face (Snell's law); track the refracted ray geometrically to see which face it meets next (side or bottom); test that face's angle of incidence against the glass–air critical angle to decide refraction vs. total internal reflection (TIR); repeat until the ray exits.

1. Refraction at the top face. $\sin\theta_r=\dfrac{\sin45^\circ}{n}=\dfrac{\sqrt2/2}{\sqrt2}=\dfrac12\ \Rightarrow\ \theta_r=\boxed{30.0^\circ}$ from the vertical normal, bending toward the normal (entering the denser medium).
2. Critical angle for this glass. $\theta_c=\arcsin(1/n)=\arcsin(1/1.414)=\boxed{45.0^\circ}$ — a clean number because $n=\sqrt2$ makes $1/n=\sin45^\circ$ exactly.
3. Which face is hit next? Entering at the centre of the top face, the ray needs to travel only half the cube's side $L$ horizontally to reach the near side wall, at a depth (from the top) of $\tfrac{L}{2}\cot30^\circ=\tfrac{L}{2}\sqrt3\approx0.866L$ — less than the full depth $L$ to the bottom — so the side wall is reached first.
4. Total internal reflection at the side wall. At that wall the ray's angle of incidence (measured from the wall's own, horizontal normal) is $90^\circ-30^\circ=\boxed{60.0^\circ}$, which exceeds $\theta_c=45.0^\circ$: the ray undergoes total internal reflection and does not exit there.
5. Exit at the bottom face. After TIR the ray continues downward at the same $30^\circ$ from vertical (only its horizontal sense flips), covering the remaining depth $0.134L$ and reaching the bottom face at $x\approx0.923L$ from the entry-side wall — still on the bottom face, not the far side wall. There the angle of incidence in the glass is again $30^\circ<\theta_c$, so the ray refracts out: $n\sin30^\circ=\sin\theta_{\text{out}}\Rightarrow\sin\theta_{\text{out}}=1.414(0.5)=0.707\Rightarrow\theta_{\text{out}}=\boxed{45.0^\circ}$, i.e. the ray leaves the bottom face at the same $45^\circ$ from vertical it entered with at the top (a direct consequence of $n=\sqrt2$ and the pure horizontal-flip nature of TIR, which preserves the vertical angle throughout).

Part (b) — minimum full-length mirror.

Given. Person height $H=6$ ft, eyes at some height $h_e$ below the top of the head (unspecified, but between the feet and the crown).

Find. (i) Minimum mirror height for the person to see their whole reflection; (ii) mirror placement, and whether it depends on viewing distance.

Approach. By the law of reflection, the mirror point that lets the eye see the top of the head is the vertical midpoint between the eye and the crown; the point that lets the eye see the feet is the midpoint between the eye and the feet. The mirror need only span between those two points.

1. Reflection point for the crown. Height above the floor $=\dfrac{h_e+H}{2}$ (the mirror point that images the top of the head to the eye).
2. Reflection point for the feet. Height above the floor $=\dfrac{h_e+0}{2}=\dfrac{h_e}{2}$ (the mirror point that images the feet to the eye).
3. Minimum mirror height. Vertical extent needed: $\dfrac{h_e+H}{2}-\dfrac{h_e}{2}=\dfrac{H}{2}=\boxed{3\ \text{ft}}$ — the eye-height $h_e$ cancels completely, so the result is independent of where the eyes sit between the feet and the crown, and (by the same cancellation) independent of how far the person stands from the mirror: moving away rescales the ray angles but the two reflection-point heights, being simple averages of fixed points (eye, crown, feet), do not move.

Part (c) — Galilean telescope focused for a 30 cm near point.

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

Find. (i) Lens separation $d$; (ii) angular magnification $M$ from ray tracing; (iii) the system (ray-transfer) matrix as a product; (iv) how $M$ and the image distance are read off that matrix.

Approach. The distant object is imaged by the objective at its own back focal point; that real image is a virtual object for the eyepiece. Solve the eyepiece's thin-lens equation for its required object distance, hence $d$. Then trace the chief ray (through the objective's centre, undeviated there) with the paraxial 2×2 ray-transfer matrices to get both the emergent angle (for $M$) and the full system matrix.

1. Intermediate image and eyepiece object distance. The objective forms the distant object's image at $x=f_o=12$ cm. Applying $1/v-1/u=1/f_e$ to the eyepiece with $v=-30$ cm, $f_e=-4$ cm: $\dfrac1u=\dfrac1v-\dfrac1{f_e}=\dfrac1{-30}-\dfrac1{-4}=\dfrac{13}{60}\Rightarrow u=\dfrac{60}{13}=4.615$ cm (a virtual object for the eyepiece, as required for a Galilean design).
2. Lens separation. $d=f_o-u=12-\dfrac{60}{13}=\dfrac{96}{13}=\boxed{7.38\ \text{cm}}.$
3. Angular magnification — chief-ray construction. A ray from a distant off-axis point passes through the objective's centre undeviated (height 0, angle $\alpha$), travels a distance $d$ to reach the eyepiece at height $y=d\alpha$ (still angle $\alpha$), and is bent by the eyepiece to $\theta_{\text{out}}=\alpha-y/f_e=\alpha(1-d/f_e)$. With the eye at the eyepiece, this IS the angle the final image subtends, so $M=\theta_{\text{out}}/\alpha=1-d/f_e=1-\dfrac{96/13}{-4}=1+\dfrac{24}{13}=\dfrac{37}{13}=\boxed{2.85}$ (upright, since $M>0$). Check: the ordinary relaxed-eye telescope has $d=f_o+f_e=8$ cm, giving $M=1-8/(-4)=3=-f_o/f_e$, recovered as the $v\to-\infty$ limit of the same formula.
4. System matrix. With the ray vector $\binom{y}{\theta}$ (height, angle) and thin-lens/translation matrices $L(f)=\begin{pmatrix}1&0\\-1/f&1\end{pmatrix}$, $T(d)=\begin{pmatrix}1&d\\0&1\end{pmatrix}$, light crosses the objective, then travels $d$, then crosses the eyepiece, so $$M_{\text{sys}}=L(f_e)\,T(d)\,L(f_o)=\begin{pmatrix}1&0\\ \tfrac14&1\end{pmatrix}\begin{pmatrix}1&\tfrac{96}{13}\\0&1\end{pmatrix}\begin{pmatrix}1&0\\ -\tfrac1{12}&1\end{pmatrix}=\begin{pmatrix}0.3846&7.385\\0.01282&2.846\end{pmatrix}.$$ Its determinant is exactly $1$ (as required of any lossless paraxial system built from unit-determinant elementary matrices).
5. Reading $M$ and the image distance off the matrix. Write $M_{\text{sys}}=\begin{pmatrix}A&B\\C&D\end{pmatrix}$ for a ray entering the objective parallel to the axis at height $y_0$ (angle $0$): the emergent ray is $\binom{Ay_0}{Cy_0}$, at angle $Cy_0$; for a ray that entered at angle $\alpha$ through the front node (the construction used above), the emergent angle is $D\alpha$ — so the $D$ element of the system matrix is the angular magnification, $M=D=2.846$, matching Step 3 exactly. The image location is found by asking how far past the eyepiece ($z$) a further translation $T(z)$ must be inserted so that an axial input ray with $y_0\ne0,\ \theta_0=0$ emerges at height $0$ (crosses the axis, i.e. reaches focus): the combined matrix's top-left entry must vanish, $A+Cz=0\Rightarrow z=-A/C$ — i.e. the image distance is read from $-A/C$ of the (possibly further-propagated) system matrix, the paraxial generalization of solving $1/v-1/u=1/f$ lens by lens.