17-Phys-A7 Optics · December 2016

Question 3 of 6: Apparent depth, ray-transfer matrix arrays, and a two-lens system

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

Question 3: Apparent depth, ray-transfer matrix arrays, and a two-lens system (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) — refractive index from the apparent-depth-doubling raise.

Given. Plate thickness $t=1.50$ mm (top and bottom surfaces parallel). Microscope objective stays fixed, focused on the top-surface scratch at its original height $H_0$. The plate (with the scratch still on its top surface) is then raised by $R=2.00$ mm, and the SAME fixed microscope refocuses on the scratch again.

Find. Refractive index $n$ of the glass.

Approach. Once raised, the scratch's direct image (straight up through air) is now $R$ above the microscope's fixed focal plane, so it is out of focus. What refocuses instead is the faint image formed by a ray that leaves the scratch, travels down through the glass (thickness $t$), partially reflects off the bottom (glass–air) surface, and travels back up through the glass to exit at the top surface — a round trip of physical path $2t$ inside index $n$. "Unfolding" the reflection (replacing it with a straight-line path to the mirror image of the scratch, a standard trick for a source viewed via a plane reflector through a refracting layer) turns this into an ordinary apparent-depth problem: an object at real depth $2t$ inside a medium of index $n$, viewed from air, appears at depth $2t/n$. For the microscope's fixed focal plane to catch this virtual image, the raised top surface's apparent-image depth $2t/n$ must exactly cancel the raise $R$.

1. Set up the refocus condition. New top-surface height $=H_0+R$. The round-trip virtual image appears a depth $2t/n$ below that surface, i.e. at height $(H_0+R)-2t/n$. Refocus at the original fixed plane $H_0$ requires $(H_0+R)-\dfrac{2t}{n}=H_0$, i.e. $R=\dfrac{2t}{n}$.
2. Solve for $n$. $n=\dfrac{2t}{R}=\dfrac{2(1.50\ \text{mm})}{2.00\ \text{mm}}=\boxed{1.50}.$

Part (b) — physical meaning of the ray-transfer array elements. In paraxial (Gaussian) optics, a ray crossing a reference plane is fully specified, to first order, by two numbers packed into the column vector $[y,\alpha]^T$: $y$ is the ray's transverse height above the optical axis at that plane, and $\alpha$ is the ray's angle (paraxially, its slope $dy/dz$) measured from the axis. The system matrix $[M]$ is the $2\times2$ linear map that propagates the input pair $[y_{in},\alpha_{in}]^T$, measured at the input reference plane, to the output pair $[y_{out},\alpha_{out}]^T$, measured at the output reference plane, through every intervening translation and refraction.

Part (c) — matrix analysis of the two-lens system.

Given. $f_1=+10$ cm, $f_2=-10$ cm, separation $d=5$ cm (both lenses thin, air on both sides).

Find. Equivalent focal length $f_{eq}$; linear magnification; angular magnification; locations of the front and back focal planes (and principal planes).

Approach. Build the system matrix as the product of a thin-lens matrix $L=\begin{bmatrix}1&0\\-1/f&1\end{bmatrix}$, a translation matrix $T(d)=\begin{bmatrix}1&d\\0&1\end{bmatrix}$, and a second thin-lens matrix, applied in the order light actually travels: $M=L_2\,T(d)\,L_1=\begin{bmatrix}A&B\\C&D\end{bmatrix}$. Once $M$ is known, every cardinal quantity follows from standard matrix-optics formulas: $f_{eq}=-1/C$; back focal distance (from lens 2) $=-A/C$; front focal distance (from lens 1, object side) $=-D/C$; and the two special-case ray inputs $\theta_{in}=0$ (a collimated beam, giving $y_{out}=A\,y_{in}$) and $y_{in}=0$ (a ray through the centre of lens 1 from an off-axis object at infinity, giving $\theta_{out}=D\,\theta_{in}$) read off $A$ and $D$ directly as the two magnifications.

1. Multiply the matrices. $L_1=\begin{bmatrix}1&0\\-0.100&1\end{bmatrix}$, $T(5)=\begin{bmatrix}1&5\\0&1\end{bmatrix}$, $L_2=\begin{bmatrix}1&0\\0.100&1\end{bmatrix}$ (all lengths in cm, $-1/f_2=+0.100\ \text{cm}^{-1}$ since $f_2$ is negative). Multiplying in propagation order, $M=L_2\,T(5)\,L_1=\begin{bmatrix}0.500&5.00\\-0.0500&1.500\end{bmatrix}$ (checked: $AD-BC=0.500(1.500)-5.00(-0.0500)=1.000$, as required for a lossless system).
2. Equivalent focal length. $f_{eq}=-\dfrac{1}{C}=-\dfrac{1}{-0.0500\ \text{cm}^{-1}}=\boxed{20.0\ \text{cm}}$ (converging overall — a compact "telephoto" pairing of a strong positive front lens with a weaker-net negative rear element).
3. Magnifications from the two special rays. Collimated input ($\theta_{in}=0$): $y_{out}=A\,y_{in}$, so the linear (beam-width) magnification is $m_{lin}=A=\boxed{0.500}$ (the beam is compressed to half its input width). Object at infinity, chief ray through lens 1's centre ($y_{in}=0$): $\theta_{out}=D\,\theta_{in}$, so the angular magnification is $m_{ang}=D=\boxed{1.500}$.
4. Focal-plane locations. Back focal distance from lens 2: $\mathrm{BFD}=-A/C=-0.500/(-0.0500)=10.0\ \text{cm}$, so the back focal point $F'$ sits $10.0$ cm behind lens 2, i.e. $15.0$ cm from lens 1. Front focal distance from lens 1: $\mathrm{FFD}=-D/C=-1.500/(-0.0500)=30.0\ \text{cm}$, so the front focal point $F$ sits $\boxed{30.0\ \text{cm in front of lens 1}}$, and $F'$ sits $\boxed{10.0\ \text{cm behind lens 2}}$.
5. Principal planes (for reference). $z_H=(D-1)/C=(0.500)/(-0.0500)=-10.0$ cm and $z_{H'}=d+(1-A)/C=5.00+(0.500)/(-0.0500)=-5.00$ cm (both measured from lens 1, negative = in front of it) — both principal planes lie in front of lens 1 itself, with $H'$ in front of $H$ (a "crossed" arrangement), the signature of a telephoto-type pairing that packs a focal length longer than the physical lens spacing into a short barrel.