core25 min studyLesson 204 in Physics

Thin and thick lens imaging

Use conjugate distances consistently and distinguish effective focal length from vertex-based focal distances in a thick lens.

On this page

What you’ll understand

  • Calculate real and virtual paraxial images using a stated sign convention.
  • Obtain effective and back focal lengths from a reduced-angle matrix.
  • Measure thick-lens conjugate distances from the correct principal planes.

A thin lens assigns all bending to one plane. A thick lens bends rays at separated surfaces, so its focal distance from the last glass surface generally differs from its effective focal length. The distinction matters when placing a detector or combining lenses in a short instrument. We will connect the familiar lens equation to the matrix imaging condition, then interpret a thick element through two principal planes rather than pretending that its physical midpoint is always the correct measuring origin.

Conjugate distances with explicit signs

Continue using light traveling along positive z and reduced ray vector (y,nθ)T(y,n\theta)^{\mathsf T}. For a lens in air, define object distance so=zH1−zos_o=z_{H_1}-z_o and image distance si=zi−zH2s_i=z_i-z_{H_2}. A real object on the incoming side has positive object distance, and a real image on the outgoing side has positive image distance. For a thin lens the two principal planes coincide with its plane; a converging element has positive f.

Multiply translation from object to lens, the lens matrix, and translation to the proposed image. Requiring the total B entry to vanish gives the lens equation. The corresponding A entry is transverse magnification. A negative image distance means the outgoing rays diverge as though they originated from an image on the incoming side; extending them backward locates a virtual image. Such an image can be viewed, but cannot be caught on a screen placed at its apparent position.

1so+1si=1f,m=yiyo=−siso(same external medium: air)\frac{1}{s_o}+\frac{1}{s_i}=\frac{1}{f},\qquad m=\frac{y_i}{y_o}=-\frac{s_i}{s_o}\qquad\text{(same external medium: air)}
Distances and f share length units. For a thick centered system these are principal-plane distances, not distances from the surface vertices. Negative m denotes an inverted transverse image.↔ Scroll wide equations horizontally.

A thick element has separated reference planes

A homogeneous thick lens with surface powers Φ1\Phi_1 and Φ2\Phi_2, axial thickness t, and internal index n has matrix S2T(t,n)S1S_2T(t,n)S_1 between its surface vertices. Its net power is Φ1+Φ2−(t/n)Φ1Φ2\Phi_1+\Phi_2-(t/n)\Phi_1\Phi_2. The separation term shows why simply adding surface powers is only a thin-element approximation. Each surface power uses its own incoming and outgoing indices and the signed radius defined in the preceding lesson.

When the external medium is air on both sides and net power is nonzero, effective focal length is the negative reciprocal of C. A parallel input ray has outgoing height Ay and slope Cy at the last vertex. It reaches the axis after distance −A/C-A/C, which is the back focal length measured from that vertex. The effective focal length is instead measured from the second principal plane. These distances agree only for special placements of that plane.

M=S2T(t,n)S1=(ABCD),P=−C,f=1P,BFL=−AC=AfM=S_2T(t,n)S_1=\begin{pmatrix}A&B\\C&D\end{pmatrix},\qquad P=-C,\qquad f=\frac1P,\qquad \mathrm{BFL}=-\frac AC=Af
P is net optical power in inverse metres for an air-to-air system. BFL is a signed distance from the back surface vertex, whereas f is measured from the second principal plane.↔ Scroll wide equations horizontally.

Principal planes replace an imagined lens center

Principal planes are equivalent reference planes that let a centered first-order system obey a thin-lens-style conjugate equation. They need not coincide with a glass surface, remain inside the glass, or lie at the mechanical center. For an air-to-air matrix, the first plane lies (1−D)/P(1-D)/P to the right of the first vertex, and the second lies (A−1)/P(A-1)/P to the right of the last vertex. These offsets follow by factoring the matrix into two translations and a thin powered element.

For a real design, label the vertex coordinates and principal-plane coordinates before inserting any object distance. This prevents a physically meaningful but numerically wrong detector placement. Principal-plane equivalence remains paraxial: it does not imply that actual rays bend abruptly at those planes or follow the constructed lines inside the element. If C vanishes the system is afocal, effective focal length is not finite, and this particular finite-power factorization must be replaced by the appropriate matrix description.

Where is the back focus?

A thick lens in air has M=(0.80.04m−9m−10.8)M=\left(\begin{smallmatrix}0.8&0.04\,\mathrm m\\-9\,\mathrm{m^{-1}}&0.8\end{smallmatrix}\right) between its surface vertices. Find effective focal length, back focal length, and the second principal plane’s offset from the back vertex.

  1. First check AD−BC=0.64−(−0.36)=1AD-BC=0.64-(-0.36)=1, consistent with the declared reduced-angle convention. The net power is P=−C=9 m−1P=-C=9\,\mathrm{m^{-1}}.

  2. The effective focal length is 1/9 m=111.1 mm1/9\,\mathrm m=111.1\,\mathrm{mm}. A parallel incoming bundle instead focuses −A/C=0.8/9 m=88.9 mm-A/C=0.8/9\,\mathrm m=88.9\,\mathrm{mm} beyond the last vertex.

  3. The back focus is one effective focal length after the second principal plane. Therefore that plane is 88.9−111.1=−22.2 mm88.9-111.1=-22.2\,\mathrm{mm} relative to the back vertex, on its incoming side.

The result

The detector for a distant object belongs 88.9 mm after the back vertex, although the effective focal length is 111.1 mm. The second principal plane is 22.2 mm before that vertex.

Explore the model

Interactive lab

Interactive lab · matrix

Reference planes affect a linear map

Inspect a generic matrix analogy. Reset, set Diagonal scale a = 1 and Shear b = 0.5, and predict determinant 1. For a separate comparison Reset, set a = 1 and b = 1, and predict the same area scale but a different coordinate map. Playback interpolates geometry; readouts describe the selected final map.

Reference planes affect a linear mapThe selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Current readings: Area scale · det A: 1.4; Singular values · σmax, σmin: 1.597, 0.876; Condition number · κ₂(A): 1.823. All parameters are adjustable with the labeled controls below.Matrix A with first row 1.4, 0.6 and second row zero, one; display interpolates between identity and A while final determinant remains 1.4.−4−2024−2−1012x · equal geometric scaleyA = [ 1.4 0.6 ; 0 1 ]Display blend: 0% · final values below
Interpolated geometryOriginal unit squareSelected final transform
Area scale · det A
1.4
Singular values · σmax, σmin
1.597, 0.876
Condition number · κ₂(A)
1.823
t = 0 s

The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA change in matrix entries need not change determinant. This restricted triangular instrument does not calculate focal length, principal planes, or thick-lens imaging.

Check your understanding

Which distance is the effective focal length of a finite-power thick lens in air?

Try it yourself

A thick system in air has f=0.12 mf=0.12\,\mathrm m, first principal plane at z=0.02 mz=0.02\,\mathrm m, and second at z=0.08 mz=0.08\,\mathrm m. An object is at z=−0.22 mz=-0.22\,\mathrm m. Find image coordinate and magnification.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Find object distance from the first principal plane and add image distance to the second plane’s coordinate.

Compare with the solution

The object distance is 0.02−(−0.22)=0.24 m=2f0.02-(-0.22)=0.24\,\mathrm m=2f. The lens equation gives 1/si=1/0.12−1/0.24=1/0.241/s_i=1/0.12-1/0.24=1/0.24, so si=0.24 ms_i=0.24\,\mathrm m. The image coordinate is zi=0.08+0.24=0.32 mz_i=0.08+0.24=0.32\,\mathrm m, and m=−0.24/0.24=−1m=-0.24/0.24=-1. The image is real, inverted, and equal in transverse size; measuring from either physical vertex would require additional data.

Take it with you

  • Declare the conjugate-distance signs before solving the lens equation.
  • Effective focal length and back focal length use different reference planes.
  • Principal planes encode first-order equivalence and are not literal bending surfaces.

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