Two-lens derivation | Biomedical Optics

1. Geometry and operators

Put lens 1 at z=0 and lens 2 at z=s. Their focal lengths f_1,f_2 are positive. A real object is at z_\mathrm{o}=-d_1 with d_1>0. The final plane has coordinate z=s+d_2. Unlike d_1 and s, d_2 is signed; negative propagation in the final matrix represents a backward extension for locating a virtual image.

The ray vector, propagation operator, and thin-lens operator are

\mathbf r=\begin{pmatrix}x\\\theta\end{pmatrix},\qquad P(\ell)=\begin{pmatrix}1&\ell\\0&1\end{pmatrix},\qquad L(f)=\begin{pmatrix}1&0\\-1/f&1\end{pmatrix}

These are the paraxial operators of Chapter 1.2, with shorter names for this derivation. Their ordered product is

T=P(d_2)L(f_2)P(s)L(f_1)P(d_1)

Each rightmost factor acts first. We work in the same medium on both sides of the lenses, so each factor has determinant one.

2. Multiply the four upstream operators

First propagate from the object to lens 1 and pass through that lens:

L(f_1)P(d_1)=\begin{pmatrix}1&0\\-1/f_1&1\end{pmatrix}\begin{pmatrix}1&d_1\\0&1\end{pmatrix}

=\begin{pmatrix}1&d_1\\-1/f_1&1-d_1/f_1\end{pmatrix}

Next propagate the distance s to lens 2:

P(s)L(f_1)P(d_1)=\begin{pmatrix}1&s\\0&1\end{pmatrix}\begin{pmatrix}1&d_1\\-1/f_1&1-d_1/f_1\end{pmatrix}

=\begin{pmatrix}1-s/f_1&d_1+s(1-d_1/f_1)\\-1/f_1&1-d_1/f_1\end{pmatrix}

For brevity, name the two upper entries a and b:

a=1-\frac{s}{f_1},\qquad b=d_1+s\left(1-\frac{d_1}{f_1}\right)

Now pass through lens 2:

K=L(f_2)P(s)L(f_1)P(d_1)=\begin{pmatrix}1&0\\-1/f_2&1\end{pmatrix}\begin{pmatrix}a&b\\-1/f_1&1-d_1/f_1\end{pmatrix}

=\begin{pmatrix}a&b\\-a/f_2-1/f_1&-b/f_2+1-d_1/f_1\end{pmatrix}=\begin{pmatrix}a&b\\c&d\end{pmatrix}

The lower entries are therefore

c=-\frac{a}{f_2}-\frac{1}{f_1}=\frac{s-f_1-f_2}{f_1f_2},\qquad d=1-\frac{d_1}{f_1}-\frac{b}{f_2}

Here d is a matrix entry, not a distance. In particular it differs from the signed distance d_2. The entries a,d are dimensionless, b has units of length, and c has units of inverse length.

3. Propagate to the image and impose focus

The last propagation gives

T=P(d_2)K=\begin{pmatrix}1&d_2\\0&1\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}a+d_2c&b+d_2d\\c&d\end{pmatrix}

Consequently, the final height is

x_\mathrm{out}=(a+d_2c)x_\mathrm{o}+(b+d_2d)\theta_\mathrm{o}

An image forms when this height does not depend on the initial angle \theta_\mathrm{o}. Set its coefficient to zero:

B=b+d_2d=0\quad\Longrightarrow\quad d_2d=-b\quad\Longrightarrow\quad d_2=-\frac{b}{d}\quad(d\ne0)

At that plane, magnification is the coefficient of object height:

M=A=a+d_2c=a-\frac{bc}{d}=\frac{ad-bc}{d}

We can verify the numerator directly. Substitute the expressions for c and d:

\begin{aligned}ad-bc&=a\left(1-\frac{d_1}{f_1}-\frac{b}{f_2}\right)-b\left(-\frac{a}{f_2}-\frac{1}{f_1}\right)\\&=a\left(1-\frac{d_1}{f_1}\right)-\frac{ab}{f_2}+\frac{ab}{f_2}+\frac{b}{f_1}\\&=\left(1-\frac{s}{f_1}\right)\left(1-\frac{d_1}{f_1}\right)+\frac{d_1+s(1-d_1/f_1)}{f_1}\\&=1-\frac{d_1}{f_1}-\frac{s}{f_1}\left(1-\frac{d_1}{f_1}\right)+\frac{d_1}{f_1}+\frac{s}{f_1}\left(1-\frac{d_1}{f_1}\right)\\&=1.\end{aligned}

This also follows from the product of the unit determinants of the operators. Thus

d_2=-\frac{b}{d},\qquad M=\frac{1}{d}\quad(d\ne0)

4. Normalize around the 4-f baseline

Define the baseline magnification and relative deviations:

M_0=-\frac{f_2}{f_1},\qquad \alpha=\frac{d_1}{f_1}-1,\qquad\beta=\frac{s}{f_1+f_2}-1

Solving these definitions for the physical distances gives

d_1=f_1(1+\alpha),\qquad s=(f_1+f_2)(1+\beta),\qquad 1-\frac{d_1}{f_1}=-\alpha

Substitute into b and expand each term:

\begin{aligned}b&=d_1+s\left(1-\frac{d_1}{f_1}\right)\\&=f_1(1+\alpha)-(f_1+f_2)(1+\beta)\alpha\\&=f_1+f_1\alpha-(f_1+f_2)\alpha-(f_1+f_2)\beta\alpha\\&=f_1-f_2\alpha-(f_1+f_2)\beta\alpha.\end{aligned}

Now substitute this result into d:

\begin{aligned}d&=1-\frac{d_1}{f_1}-\frac{b}{f_2}\\&=-\alpha-\frac{f_1-f_2\alpha-(f_1+f_2)\beta\alpha}{f_2}\\&=-\alpha-\frac{f_1}{f_2}+\alpha+\frac{f_1+f_2}{f_2}\beta\alpha\\&=-\frac{f_1}{f_2}\left[1-\frac{f_1+f_2}{f_1}\beta\alpha\right].\end{aligned}

Because M_0-1=-f_2/f_1-1=-(f_1+f_2)/f_1, define

Q=1+(M_0-1)\beta\alpha=1-\frac{f_1+f_2}{f_1}\beta\alpha

The matrix entry becomes d=-(f_1/f_2)Q=Q/M_0. Hence

M=\frac{1}{d}=\frac{M_0}{Q},\qquad \frac{M}{M_0}=\frac{1}{Q}

For the image distance, use d_2=-b/d:

d_2=\frac{f_2b}{f_1Q}\quad\Longrightarrow\quad\frac{d_2}{f_2}=\frac{b/f_1}{Q}

Divide the expanded b by f_1 and identify its terms:

\begin{aligned}\frac{b}{f_1}&=1-\frac{f_2}{f_1}\alpha-\frac{f_1+f_2}{f_1}\beta\alpha\\&=1+M_0\alpha+(M_0-1)\beta\alpha\\&=Q+M_0\alpha.\end{aligned}

Finally,

\frac{d_2}{f_2}=\frac{Q+M_0\alpha}{Q}=1+\frac{M_0\alpha}{Q}

\boxed{\frac{M}{M_0}=\frac{1}{1+(M_0-1)\beta\alpha}},\qquad\boxed{\frac{d_2}{f_2}-1=\frac{M_0\alpha}{1+(M_0-1)\beta\alpha}}

Both formulas require Q\ne0. No expansion in small \alpha or \beta was used. Their range of physical accuracy is set by the paraxial thin-lens assumptions and whether the chosen rays pass through the actual optical components.

5. Check the baseline limits

At \alpha=\beta=0, Q=1. The full baseline matrix is

T_0=\begin{pmatrix}-f_2/f_1&0\\0&-f_1/f_2\end{pmatrix}=\begin{pmatrix}M_0&0\\0&1/M_0\end{pmatrix}

At \beta=0, the formulas reduce to M=M_0 and d_2=f_2(1+M_0\alpha). At \alpha=0, they give M=M_0 and d_2=f_2 even when lens separation changes.

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