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03 / 05   Thin lenses

Application note

Thin Lens for Imaging

Connect focal length, object distance, and image size with a consistent geometric model.

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The thin-lens model represents refraction through a lens as a direction change at one plane. It estimates image geometry for small ray angles when lens thickness can be neglected, without predicting how sharply a real lens reproduces detail.

Locate the image

For a thin lens surrounded by air:

1f=1s+1s′,s′=fss−f,m=h′h=−s′s\frac{1}{f}=\frac{1}{s}+\frac{1}{s'},\qquad s'=\frac{fs}{s-f},\qquad m=\frac{h'}{h}=-\frac{s'}{s}

Distances use one unit and begin at the ideal lens plane. A real object has s>0s>0. Positive s′s' locates a real image downstream; negative s′s' locates a virtual image upstream. The signed transverse magnification mm compares image and object height; negative values mean inversion.

A converging lens forms a real inverted image when the object is beyond its focal point; rays diverge from a virtual upright image when the object is inside focus.

View diagram at full size

With f=50f=50 mm and s=150s=150 mm, the image lies 75 mm beyond the lens and m=−0.5m=-0.5. A 10 mm feature becomes a 5 mm inverted image. Moving a sensor to that image plane establishes geometric focus for the selected object plane.

Recognize the focal boundary

A converging lens has f>0f>0. Image distance grows as the object approaches the front focal plane from farther away. At s=fs=f, rays from each object point leave parallel, with no finite image plane or image height. Inside focus, the virtual image cannot be recorded by placing a bare sensor at its apparent location.

A diverging lens, f<0f<0, gives a real object a virtual, upright, reduced image. Following optics can still form another image.

Distinguish three magnification quantities

Image size, ray-cone spread, and apparent size describe different optical relationships. Use the quantity that matches the measurement, including its sign and reference conditions.

Transverse image magnification

The transverse magnification mm compares physical image height with object height:

m=h′h=−s′sm=\frac{h'}{h}=-\frac{s'}{s}

Its magnitude gives the size ratio; its sign identifies orientation.

Derivation: transverse image magnification

In the thin-lens model with air on both sides, a ray through the lens center continues without changing direction. Place the object at axial position −s-s with height hh, and the lens center at position zero and height zero. The ray’s slope is:

q=0−h0−(−s)=−hsq=\frac{0-h}{0-(-s)}=-\frac{h}{s}

At the image plane, axial position s′s', the same line has height:

h′=qs′=−s′sh⟹m=h′h=−s′sh'=qs'=-\frac{s'}{s}h \qquad\Longrightarrow\qquad m=\frac{h'}{h}=-\frac{s'}{s}

For a virtual image, extending the ray backward uses negative s′s' and gives an upright image for a real object. The relation describes a finite image; it does not assign a physical image height at infinity.

Signed cone-spread ratio

Consider two rays from the same object point. Let Δθin\Delta\theta_{\mathrm{in}} and Δθout\Delta\theta_{\mathrm{out}} be their signed paraxial angular separations, using one fixed transverse-axis convention. For finite conjugates in air, the ratio is:

γ=ΔθoutΔθin=−ss′=1m\gamma=\frac{\Delta\theta_{\mathrm{out}}}{\Delta\theta_{\mathrm{in}}} =-\frac{s}{s'}=\frac{1}{m}

The cone-width ratio is ∣γ∣|\gamma|. A negative sign means a divergent bundle becomes convergent. At the focal condition, γ=0\gamma=0 because the output is parallel; finite transverse magnification is then undefined.

Derivation: the cone-spread ratio is reciprocal to m

Use paraxial ray slope q=dy/dz≈θq=dy/dz\approx\theta. Two rays from one object point reach the lens with height difference Δy\Delta y. Their incident slope difference is:

Δqin=Δys\Delta q_{\mathrm{in}}=\frac{\Delta y}{s}

A thin lens changes each ray slope by −y/f-y/f. Subtracting the two ray equations and applying the lens equation gives:

Δqout=Δqin−Δyf=Δy(1s−1f)=−Δys′\Delta q_{\mathrm{out}} =\Delta q_{\mathrm{in}}-\frac{\Delta y}{f} =\Delta y\left(\frac{1}{s}-\frac{1}{f}\right) =-\frac{\Delta y}{s'}

Therefore, for a nonzero input spread and a finite image:

γ=ΔqoutΔqin=−ss′=1m\gamma=\frac{\Delta q_{\mathrm{out}}}{\Delta q_{\mathrm{in}}} =-\frac{s}{s'}=\frac{1}{m}

For example, m=−0.5m=-0.5 corresponds to γ=−2\gamma=-2: an inverted image half as tall, with twice the angular cone width and opposite convergence sign. If s=fs=f, directly substituting into the slope difference gives zero output spread. No finite mm is assigned at that boundary.

This ratio concerns rays within one bundle. It does not compare the apparent angular size of an object with and without a viewing instrument.

Visual angular magnification

Visual magnification MM compares the angle an object subtends through the optics with its unaided angle at a stated reference distance DD. For an upright virtual image, s′<0s'<0, with the eye a distance e≥0e\geq0 behind the lens:

M=αopticsαreference≈mDe−s′M=\frac{\alpha_{\mathrm{optics}}}{\alpha_{\mathrm{reference}}} \approx\frac{mD}{e-s'}

Unlike the cone-spread ratio, this result depends on the viewing geometry and comparison distance. A positive magnifier gives M=D/fM=D/f for a relaxed view at infinity; with the eye at the lens and the virtual image at distance DD, it gives M=1+D/fM=1+D/f.

Derivation: visual magnification and its viewing conditions

For object height hh and upright virtual-image height h′=mhh'=mh, the image lies a distance e−s′e-s' from the eye. In the small-angle approximation:

αoptics≈mhe−s′,αreference≈hD\alpha_{\mathrm{optics}}\approx\frac{mh}{e-s'}, \qquad \alpha_{\mathrm{reference}}\approx\frac{h}{D}

Taking their ratio yields M≈mD/(e−s′)M\approx mD/(e-s'). All lengths use the same units, and DD is measured from the unaided eye to the comparison object. The chosen rays must reach the eye without being clipped.

For relaxed viewing, place the object at the front focus of a positive lens. Its outgoing angle has magnitude approximately h/fh/f, so:

M∞=h/fh/D=DfM_{\infty}=\frac{h/f}{h/D}=\frac{D}{f}

For a virtual image at the reference distance with the eye at the lens, e=0e=0 and s′=−Ds'=-D. Then:

1s=1f+1D,M=m=Ds=1+Df\frac{1}{s}=\frac{1}{f}+\frac{1}{D}, \qquad M=m=\frac{D}{s}=1+\frac{D}{f}

The common choice D=250D=250 mm is a comparison convention. It does not establish an individual’s near point or accommodation range. Moving the eye or changing the virtual-image location changes the general result.

Three distinct ratios compare physical image height, signed angular spread within one ray bundle, and apparent angular size at a specified eye position.

View diagram at full size

Relate shape to power

For a thin lens of refractive index nn in air:

1f=(n−1)(1R1−1R2)\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)

Signed surface curvatures determine the optical power of a thin lens in air; a plane surface has zero curvature.

View diagram at full size

Light travels left to right; a radius is positive when its center of curvature lies to the right of that surface. A plane contributes zero curvature. For n=1.5n=1.5, R1=+50R_1=+50 mm and R2=−50R_2=-50 mm, the focal length is 50 mm. Use the material index at the operating wavelength.

Real assemblies introduce thickness, separated principal planes, aberrations, and aperture limits. Confirm the required field, working distance, and image quality with the lens specification before turning a first estimate into a measurement setup.

Explore the model

Thin-lens image

Explore real, virtual, and collimated image conditions.

Change the example values to update the results and diagram. Use the stated units and sign convention.

Positive for converging; negative for diverging. Zero is invalid.

Positive; measured upstream from the surface or thin lens.

Signed image distance, s′
75mm
Transverse magnification
-0.5×
Signed cone-spread ratio, γ
-2×
Cone angular-width ratio, |γ|
2×
Image classification
real, inverted

Distances follow the sign convention described below.

Thin-lens image: updated diagramObject is upstream. Signed image distance 75 mm and transverse magnification -0.5. Real downstream image.ObjectLensReal images′ = 75 mm · m = -0.5
Input-responsive schematic; not to scale. Scroll horizontally on narrow screens. The numerical results define the distances and angles.
Formula & assumptions

1/f = 1/s + 1/s′. Transverse magnification m = −s′/s. Signed cone-spread ratio γ = −s/s′ = 1/m; positive angular-width ratio = |γ|.

Thin lens in air, paraxial rays, and a real object. γ compares outgoing and incoming angular spreads between rays from the same object point; it is not visual angular magnification. Positive s′ is real; negative s′ is virtual. At s = f the outgoing spread is zero (γ = 0), while finite transverse magnification is undefined.

Explore the model

Thin lensmaker estimate

Relate refractive index and signed surface curvature to focal length in air.

Change the example values to update the results and diagram. Use the stated units and sign convention.

Positive center downstream. Enter 0 for a plane; radius +50 mm gives +0.02/mm.

Same sign convention: radius −50 mm gives −0.02/mm.

Signed focal length
50mm
Optical power
20diopters
Lens behavior
converging
Thin lensmaker estimate: updated diagramFirst curvature 0.02 per mm; second curvature -0.02 per mm. Focal length 50 mm.C₁ 0.02/mmC₂ -0.02/mmn = 1.5f = 50 mm
Input-responsive schematic; not to scale. Scroll horizontally on narrow screens. The numerical results define the distances and angles.
Formula & assumptions

1/f = (n − 1)(1/R₁ − 1/R₂). Power [diopters] = 1000/f [mm].

Thin homogeneous lens surrounded by air with index approximated as 1. Curvature, rather than radius, makes a planar surface explicit. Thickness, dispersion across wavelengths, aberrations, and manufacturing limits are omitted. Zero power has no finite focal length in this model.

From estimate to application

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