Physics · Optics

Thin-lens equation

Q: How do you calculate with the lens equation $\frac{1}{f} =\frac{ 1}{g} +\frac{ 1}{b}$?

First solve for the unknown reciprocal, then take the reciprocal at the end. All distances must use the same length unit. For a converging lens with $f=10 cm$ and a real object at $g=30 cm$, reciprocal image distance is the difference between reciprocals of 10 cm and 30 cm; this gives $b=15 cm$. Positive b indicates a real image on the emerging side. Signed magnification $m=-\frac bg$ is−0.5, so the image is inverted and half-sized. Check by substitution and the expected position between f and 2 f. Only the sum $g+b$ gives total separation between object and screen.

Q: What sign convention applies to the lens equation?

In this convention g is positive for a real object on the incident side. b is positive for a real image on the emerging side and negative for a virtual image on the incident side. f is positive for converging lenses and negative for diverging lenses. Already converging input can create a virtual object, giving negative g. Image heights also use signs: $m=-\frac bg$ is negative for an inverted image. The claim that every diverging lens produces only virtual images assumes real objects. Draw propagation direction and lens plane before translating positive geometric segment lengths into signed quantities.

Q: What happens when the object is inside the focal length?

A real object inside a converging lens’s positive focal distance gives a virtual, upright enlarged image. Reciprocal image distance becomes negative. With $f=10 cm$ and $g=5 cm$, $b=-10 cm$ and $m=+2$. The emerging rays actually diverge; their backward extensions meet at the virtual image location. A screen placed there cannot capture that image. Transverse magnification 2 is also not automatically the magnifier’s angular magnification, which depends on eye position and a chosen reference viewing condition. At $g=f$, ideal image distance tends to infinity rather than giving a finite virtual image. These statements assume the thin paraxial model.

Q: How is the magnification related to the lens equation?

The central ray and similar triangles give signed magnification $m=\frac BG=-\frac bg$. The minus sign carries orientation: real objects with real images are inverted. Use absolute values only when discussing unsigned sizes. A 2 cm object at $g=15 cm$ with $f=10 cm$ gives $b=30 cm$ and $B=-4 cm$, a 4 cm inverted image. At $g=2 f$, $m=-1$ and size is unchanged. For very distant objects, b approaches focal length. These relations predict geometric size and location, not diffraction resolution or the number of visible details. A longer focal length can enlarge an image without changing the aperture’s angular resolving criterion.

Q: Does the lens equation also hold for diverging lenses and glasses?

Yes, provided signs and the optical model are appropriate. For $f=-10 cm$ and a real object at $g=20 cm$, $b\approx{}-6.667 cm$ and $m\approx{}+0.3333$: virtual, upright and reduced. Optical power is reciprocal focal length in meters, for example−2 diopters when $f=-0.5 m$. Closely contacting thin lenses in the same medium have approximately additive powers; finite spacing needs an extended system calculation. Real corrective optics also involve the eye and vertex distance. The school equation explains basic imaging but does not replace individual fitting. A diverging lens with a virtual object can even form a real image, so object type must be stated explicitly.

The thin-lens relation locates real or virtual paraxial images. Signed magnification $m=-\frac bg$ describes their orientation.

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Formula

Calculate

\frac{1}{f} = \frac{1}{g} + \frac{1}{b}

Use one length unit for f,g,b; optical power $\Phi=\frac1f$ requires f in meters, giving m⁻¹ or diopters.

Variables & units

Variables & units
Symbol	Meaning	Unit
$f$	Signed paraxial focal length: positive converging, negative diverging	m or cm
$g$	Signed object distance from thin-lens plane; positive for a real object	m or cm
$b$	Signed image distance from thin-lens plane; positive for a real image	m or cm

LaTeX

\frac{1}{f} = \frac{1}{g} + \frac{1}{b}

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Ray diagram: an upright object on the left at distance g, a converging lens, focal points F; the parallel ray and the central ray form an inverted real image on the right at distance b. — Construction at a converging lens: the parallel ray and the central ray meet at the real, inverted image. 1/f = 1/g + 1/b.

Thin lens equation calculator

Find the image distance, object distance or focal length of a thin lens; negative values stand for virtual images or diverging lenses.

Find

Focal length f (signed)

Focal length, positive for a converging lens, negative for a diverging lens, e.g. 10 cm.

Object distance g (signed)

Distance from the object to the lens, positive for a real object in front of the lens, e.g. 30 cm.

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for b, substitute, evaluate

1. Rearrange for b
$b=\frac{fg}{g-f}$
2. Substitute values (in SI units)
- f = 10 cm = 0.1 m
- g = 30 cm = 0.3 m
$b=\frac{0.1\cdot0.3}{0.3-0.1}=0.15\,\mathrm m$
3. Result
b = 15 cm

Positive distance: real image or real object under the stated convention.

Result: Image distance b (signed)

b = 15 cm

With f = 10 cm and g = 30 cm, b = 15 cm.

Quick tips

Converging lens: at g = 2f the image is at b = 2f and the same size. The further away the object, the closer b moves to f; for g < f the image becomes virtual.
Rearrange with reciprocals: 1/b = 1/f − 1/g. Do not forget the final reciprocal: 1/10 − 1/30 = 1/15, so b = 15 cm.

Assumptions and limits

Thin paraxial lens with the same medium on both sides. f positive for converging, negative for diverging lens; g positive for real, negative for virtual object; b positive for real, negative for virtual image. No image-quality prediction or corrective-lens prescription.

Derivation & background

Similar-triangle geometry of a central ray and a parallel ray passing through the focus gives $\frac1f=\frac1g+\frac1b$ . g is positive for real and negative for virtual objects; b is positive for real and negative for virtual images; f is positive for converging and negative for diverging lenses. Signed heights give $\frac BG=-\frac bg$ ; unsigned sizes use absolute values. Diverging lenses form virtual reduced images for real objects but can form real images from sufficiently converging input.

Exam blueprint

Validity range

Thin-lens paraxial imaging with a consistent sign convention. Thick lenses use principal-plane distances; marginal rays, aberrations and diffraction require additional models. A diverging lens’s virtual reduced image statement assumes a real object. Both sides have the same surrounding medium.

Derivation steps

Intersect central and parallel rays using signed similar triangles.

1Use $g>0$ for real objects, $b>0$ for real images and $f>0$ for converging lenses. Virtual positions and diverging focal lengths carry corresponding negative signs.
2The thin paraxial central ray gives $\frac BG=-\frac bg$ for signed transverse heights.
3The incoming parallel ray passes through the image-side focus; at the image position $B=G\left(1-\frac bf\right)$ .
4Both rays belong to the same image point: $-\frac bg=1-\frac bf$ .
5Rearrange and divide by b to obtain $\frac1f=\frac1g+\frac1b$ . Use one length unit throughout.
6Take the reciprocal to find b and interpret its sign. Magnification $m=-\frac bg$ gives orientation and height; $g=f$ places the ideal image at infinity.

Rearrangements

Image distance from focal length and object distance

b = \frac{f \cdot g}{g - f}

For a converging lens with $f>0$ and a real object inside the focal length, $0<g<f$ , b is negative and the image is virtual. Other sign combinations require the full calculation.

Focal length from g and b

f = \frac{g \cdot b}{g + b}

This is how the focal length is determined experimentally (Bessel method as an alternative).

Magnification

m=\frac{B}{G}=-\frac{b}{g}

Magnification uses signed heights. The minus sign records orientation; unsigned sizes satisfy $\frac{|B|}{|G|}=\left|\frac bg\right|$ .

Task variant

Converging lens $f = 10 cm$ , object at $g = 15 cm$ . Where does the image form?

Approach: $\frac1b=\frac1f-\frac1g$ .
Using the given centimetre values, $\frac1b=\frac1{10\,\mathrm{cm}}-\frac1{15\,\mathrm{cm}}=\frac1{30\,\mathrm{cm}}$ .
The image is at $b = 30 cm$ : real, inverted and twice the size, since $m=-\frac bg=-2$ .

Object at $g = 5 cm$ in front of a lens with $f = 10 cm$ . What results?

Approach: $\frac1b=\frac1f-\frac1g$ .
Thus $\frac1b=\frac1{10\,\mathrm{cm}}-\frac1{5\,\mathrm{cm}}=-\frac1{10\,\mathrm{cm}}$ .
Hence $b = -10 cm$ and $m=-\frac{-10}{5}=+2$ : virtual, upright and twice the size.
A screen at the computed location does not capture this virtual image.

Common mistakes

Forgetting the reciprocal at the end and reporting $\frac{1}{b}$ as b.

First compute $\frac{1}{b}$ , then invert.

Ignoring the sign convention.

$b < 0$ means a virtual image, $f < 0 a$ diverging lens; the signs carry the physics.

Substituting g and b in different units.

Use the same unit (cm or m) for all distances.

Expecting an image at $g = f$ .

For $g = f$ the image lies at infinity; the rays emerge parallel.

Exam context

Typical: ray construction plus calculation, case analysis $g > 2f$ , $f < g < 2f$ , $g < f$ , and application to the eye and camera.

Formula cluster

Geometrical optics

Refraction explains the lens, the lens equation its imaging.

n₁·sin(α₁) = n₂·sin(α₂)

Snell's Law of Refraction

c = λ·f

Wave Equation (c = λ·f)

λ = h/p

De Broglie Wavelength

Worked example

Converging $f=10 cm$ and $g=30 cm$ gives $b=15 cm$ , $m=-0.5$ , real and inverted. The new example $f=80 mm$ , $g=240 mm$ , $G=30 mm$ gives $b=120 mm$ and $B=-15 mm$ ; at $g=50 mm$ the magnifier gives $b=-133.333 mm$ and $m=+2.66667$ .

Applications

Plan cameras, projectors, magnifiers, microscopes and telescopes geometrically; describe eyes and corrective lenses within an appropriate optical system model.

How do you calculate with Thin Lens Equation?

Here is how to work through a typical Thin Lens Equation (1/f = 1/g + 1/b) task step by step:

1
Task
Converging lens $f = 10 cm$ , object at $g = 15 cm$ . Where does the image form?
Solution path
1. Approach: $\frac1b=\frac1f-\frac1g$ .
2. Using the given centimetre values, $\frac1b=\frac1{10\,\mathrm{cm}}-\frac1{15\,\mathrm{cm}}=\frac1{30\,\mathrm{cm}}$ .
3. The image is at $b = 30 cm$ : real, inverted and twice the size, since $m=-\frac bg=-2$ .
2
Task
Object at $g = 5 cm$ in front of a lens with $f = 10 cm$ . What results?
Solution path
1. Approach: $\frac1b=\frac1f-\frac1g$ .
2. Thus $\frac1b=\frac1{10\,\mathrm{cm}}-\frac1{5\,\mathrm{cm}}=-\frac1{10\,\mathrm{cm}}$ .
3. Hence $b = -10 cm$ and $m=-\frac{-10}{5}=+2$ : virtual, upright and twice the size.
4. A screen at the computed location does not capture this virtual image.

Understand and apply the equation

A converging lens can project onto a screen or show a nearby object as a magnifier. The same lens therefore need not always form the same kind of image. The thin-lens equation connects object distance, image distance and focal length. First identify both image sides and the signs, then calculate a screen image, a virtual image and a designed magnification.

Before calculating

A real image forms where emerging rays meet and can be captured on a screen. For a virtual image, rays diverge but appear to originate at one point. Extend them backward to locate it. A virtual image can be seen, but cannot be directly projected onto a screen at that location.[1]

You need reciprocals, similar triangles and one sign convention. The thin lens has the same medium on both sides and rays close to the optical axis. Measure distances from the idealized lens plane. Focal length characterizes the lens, not its currently adjusted screen distance.[1][2]

What is the thin-lens equation with this sign convention?

For a thin lens in the paraxial model with the same medium on both sides, $\frac1f=\frac1g+\frac1b$ . Use the signed distances defined below rather than their magnitudes. The equation gives image location; signed magnification also gives orientation and size.[1]

Use $g>0$ for a real object before the lens and $b>0$ for a real image on the emerging side. A virtual image on the incident side has $b<0$ . Converging lenses have $f>0$ , diverging lenses $f<0$ . A virtual object, an incoming bundle already converging toward a point beyond the lens, has $g<0$ . Compound systems can create that situation; the main examples below use real objects. Both sides of the ideal lens use the same surrounding medium here.[1]

\begin{aligned}&\frac1f=\frac1g+\frac1b\\&m=\frac BG=-\frac bg\end{aligned}

B and G are signed transverse heights; m<0 means inverted, m>0 upright. For unsigned sizes use $\frac{|B|}{|G|}=\left|\frac bg\right|$ .[1]

Distances start at the ideal thin-lens position. A thick real lens instead uses its corresponding principal planes for object and image distances. A large front lens housing is not automatically a valid reference plane. For positive g,b, object-to-screen separation is $g+b$ , not b alone.[2]

How do you derive the thin-lens equation and signed magnification?

For a real object of height G and a converging paraxial lens, a central ray approximately passes straight through the lens center. Similar triangles give $\frac BG=-\frac bg$ . A parallel incoming ray reaches the lens at height G and then passes through the image-side focus. At the image plane its height is $B=G\left(1-\frac bf\right)$ . Both rays must meet the same image point.[1]

\begin{aligned}-\frac bg&=1-\frac bf\\\frac bf-\frac bg&=1\\\frac1f-\frac1g&=\frac1b\\\frac1f&=\frac1g+\frac1b\end{aligned}

Both principal rays belong to one image point. The third line divides by finite image distance b.[1]

This is paraxial imaging, using first-order small-angle and small-height approximations. The relation extends with the same signs to virtual images and diverging lenses. Diverging rays must be extended backward to locate a virtual intersection; it is not a position where actual beam energy focuses onto a screen. Snell’s law of refraction first describes local ray bending at an interface; the thin-lens model combines the two refracting surfaces into an imaging model.[2]

How do you calculate image distance, object distance, focal length and dioptres?

\begin{aligned}&b=\frac{fg}{g-f}\\&g=\frac{fb}{b-f}\\&f=\frac{gb}{g+b}\end{aligned}

Use one length unit and check denominators. For $g=f$ , the ideal image lies at infinity.[1]

For $f>0$ and g>2 f, the image is real, inverted and reduced. At g=2 f it lies at b=2 f with m=−1. For f<g<2 f it is real and enlarged; for $0<g<f$ it is virtual, upright and enlarged. A diverging lens with a real object gives $b<0$ and $0<m<1$ . Without the real-object qualification, saying every diverging lens always forms a virtual image is too broad.[1]

Optical power $\Phi=\frac1f$ has units m⁻¹ when f is in meters, conventionally called diopters. It is different from camera f-number. Thin lenses in close contact in the same surrounding medium have additive powers; finite spacing introduces another term and requires a system model.[2]

How do you calculate image distance, height and orientation for a screen image?

An inverted image of a 30-mm object

Given: f = 80.0 mm; g=240.0 mm; G=30.0 mm; thin converging lens.
Find: Find image position, image height and total separation.

Approach: Use the thin-lens equation for image position; signed magnification gives orientation and height.
$\begin{aligned}\frac1b&=\frac1f-\frac1g\\m&=-\frac bg\\B&=mG\end{aligned}$
First obtain reciprocal image distance.
$\frac1b=\frac1{80\,\mathrm{mm}}-\frac1{240\,\mathrm{mm}}=\frac1{120\,\mathrm{mm}}$
Form signed magnification and apply it to height.
$\begin{aligned}b&=120\,\mathrm{mm}\\m&=-\frac{120\,\mathrm{mm}}{240\,\mathrm{mm}}=-0.5\\B&=-15.0\,\mathrm{mm}\end{aligned}$
Object and screen lie on opposite sides.
$L=g+b=360\,\mathrm{mm}$

Result: The real image lies 120 mm after the lens,15 mm tall and inverted. Object and screen are 360 mm apart.

Check: Reciprocal check: $\frac1{240\,\mathrm{mm}}+\frac1{120\,\mathrm{mm}}=\frac1{80\,\mathrm{mm}}$ . For g=3f the reduced real image lies between f and 2f. Both 120 mm and magnification −0.5 match that expectation.

When does a magnifier produce an upright virtual image?

Object inside the focal length

Given: f = 80.0 mm; g=50.0 mm; G=12.0 mm.
Find: Find virtual image position and signed transverse size.

Approach: The same thin-lens equation applies inside the focal length; negative b locates the virtual image on the incident side.
$\begin{aligned}\frac1b&=\frac1f-\frac1g\\m&=-\frac bg\\B&=mG\end{aligned}$
Compute the negative reciprocal image distance.
$\frac1b=\frac1{80}-\frac1{50}=-\frac3{400}\;\mathrm{mm}^{-1}$
Take the reciprocal and calculate magnification.
$\begin{aligned}&b=-133.333\,\mathrm{mm}\\&m=-\frac{b}{g}=2.66667\\&B=32.0\,\mathrm{mm}\end{aligned}$

Result: The virtual image lies 133.3 mm before the lens on the object side, upright and 2.667 times as large.

Check: The check $\frac1{50\,\mathrm{mm}}-\frac1{133.333\,\mathrm{mm}}\approx\frac1{80\,\mathrm{mm}}$ uses negative image distance. Positive magnification means upright. Transverse size magnification is not yet a magnifier’s angular magnification; that needs eye position and a reference viewing condition.

Explore further: designing a projection

A fourfold screen image over 500 mm

Given: Real object and image; object-to-screen distance L=500 mm; target signed magnification m=−4. Thin lens, same medium.
Find: Find object distance, image distance and required focal length.

Approach: a real image places object and screen on opposite sides. Combine separation and magnification before finding focal length.
$\begin{aligned}L&=g+b\\m&=-\frac bg=-4\\b&=4g\\g&=\frac L5\\f&=\frac{gb}{g+b}\end{aligned}$
Substitute the two distances and then focal length.
$\begin{aligned}g&=\frac{500\,\mathrm{mm}}5=100\,\mathrm{mm}\\b&=400\,\mathrm{mm}\\f&=\frac{(100\,\mathrm{mm})(400\,\mathrm{mm})}{500\,\mathrm{mm}}=80\,\mathrm{mm}\end{aligned}$

Result: The model lens has f = 80 mm and sits 100 mm after the object, leaving 400 mm to the screen. The image is four times as large and inverted.

Check: $\frac1{100}+\frac1{400}=\frac1{80}$ in inverse millimetres and $m=-\frac{400}{100}=-4$ check both constraints. The lens lies nearer the object, as expected for an enlarged real image.

Why is image distance so sensitive near the focal plane?

\frac{db}{dg}=-\frac{f^2}{(g-f)^2}

For fixed f, small changes of g shift b oppositely. The linear approximation requires sufficiently small changes.[1]

For f = 80 mm and g = 90 mm, b = 720 mm and $\frac{db}{dg}=-64$ . A 1 mm change is already ten percent of g−f; exact values b(89 mm) = 791.11 mm and b(91 mm) = 661.82 mm show asymmetric shifts. A linear error estimate is therefore limited. Many decimal places must not hide measurement uncertainty near the focal plane.[1]

A useful measurement check uses several(g,b) pairs and tests for consistent $f=\frac{gb}{g+b}$ . If image position and size behave unexpectedly, first inspect lens reference position, screen location, signs and the focus criterion. Blur can also arise from aberration or diffraction; changing the reciprocal arithmetic cannot explain those effects.[1]

View diagram: image distance near the focal length

Image distance on both sides of the focal plane

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The gap marks the asymptote, not an image disappearing at b=0. Lines join only the displayed measurement ranges. All distances follow the article’s sign convention.[1]

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An imaging equation does not determine all image quality

The simple relation requires a thin lens or appropriate principal-plane treatment, with paraxial rays. Large apertures, strong off-axis angles and compound objectives need more detailed models. Chromatic and spherical aberrations shift or spread imaging; a real broadband system does not automatically focus every color identically.[2]

The lens equation predicts geometric image position and magnification. It does not determine whether two close points are resolved: diffraction, aperture, noise and sampling matter too. A simple optical-power model helps describe corrective lenses, but actual fitting also depends on the eye, vertex distance and additional optical parameters. A calculation alone is not a prescription. The Rayleigh criterion for a circular aperture describes a particular diffraction scale for two point sources; it determines neither geometric image position nor, by itself, the quality of a real camera system.[1][3]

Five exercises from a camera to a virtual object

Calculate · Foundation

Converging lens f=50 mm, object g=200 mm: find b and m.

Show hint

Use $b=\frac{fg}{g-f}$ .

Learning goal: Find image distance and signed magnification of a converging lens.

Show solution and calculation

Approach: use millimetres consistently and keep the positive focal length.
The lens equation gives b≈66.6667 mm.
Magnification $m=-\frac bg=-\frac13$ indicates a reduced inverted real image.

Check: $\frac1{200\,\mathrm{mm}}+\frac1{66.6667\,\mathrm{mm}}\approx\frac1{50\,\mathrm{mm}}$ . Since g=4f, b lies between f and 2f as expected.

b=\frac{50\cdot200}{200-50}\,\mathrm{mm}

Calculate · Application

Diverging lens f=−100 mm, real object g=250 mm: determine b and m.

Show hint

Keep negative f in the equation.

Learning goal: Use negative focal length to find a virtual image and its magnification.

Show solution and calculation

Approach: retain the negative focal length of the diverging lens.
The result is b≈−71.4286 mm and $m=\frac27\approx0.285714$ : virtual, upright and reduced.
Optical power is −10 diopters.

Check: inverse image distance plus inverse object distance equals negative inverse focal length; positive b would not fit this real object.

b=\frac{(-100)250}{250+100}\,\mathrm{mm}

Transfer · Challenge

An already converging bundle would meet 50 mm beyond a diverging lens: g=−50 mm, f=−100 mm. Where is the image?

Show hint

A virtual object has $g<0$ .

Learning goal: Handle a virtual object consistently for a diverging lens.

Show solution and calculation

Approach: already converging input means a virtual object, so g=−50 mm.
With f=−100 mm, inverse image distance is positive and b=+100 mm.
A real image forms despite the diverging lens.

Check: $\frac1{-50\,\mathrm{mm}}+\frac1{100\,\mathrm{mm}}=\frac1{-100\,\mathrm{mm}}$ . The input converges strongly enough; the lens moves its intersection farther downstream.

\begin{aligned}\frac1b&=\frac1{-100\,\mathrm{mm}}-\frac1{-50\,\mathrm{mm}}\\&=\frac1{100\,\mathrm{mm}}\\b&=+100\,\mathrm{mm}\end{aligned}

Interpret · Application

A real object before a thin converging lens is moved from g=150 mm to g=75 mm. Focal length is +100 mm and both sides have the same medium. Find image distance and magnification for each position using $m=-\frac bg$ . Which position permits projection onto a screen, and where is the other image?

Show hint

Solve for b at each object distance. Its sign identifies image side in this model; the sign of m identifies orientation.

Learning goal: Understand the change from real to virtual image as object distance crosses focal length.

Show solution and calculation

Approach: solve the thin-lens equation separately for both positions.
At 150 mm, b=+300 mm and m=−2: a real inverted image on the outgoing side.
At 75 mm, b=−300 mm and m=+4: a virtual upright image on the incident side.
Only the first can be directly projected onto a screen.

Check: image distance diverges as g passes through f rather than passing continuously through zero.

\begin{aligned}b&=\frac{fg}{g-f}\\{}m&=-\frac bg\\{}b_1&=\frac{100\cdot150}{150-100}\,\mathrm{mm}=+300\,\mathrm{mm}\\{}m_1&=-2\\{}b_2&=\frac{100\cdot75}{75-100}\,\mathrm{mm}=-300\,\mathrm{mm}\\{}m_2&=+4\end{aligned}

[1]

Error analysis · Application

A thin diverging lens has f=−80 mm and a real object at g=+120 mm. A solution replaces f by its magnitude and finds b=+240 mm. Correct image distance and magnification. Substitute into the original thin-lens equation to show why the positive image distance does not describe this lens.

Show hint

The focal-length sign carries physical information about lens type. It cannot be made positive without changing the model.

Learning goal: Use a diverging lens’s negative focal length consistently.

Show solution and calculation

Approach: retain the given negative focal length.
This gives b=−48 mm and m=+0.40: a reduced upright virtual image on the incident side.

Check: inverse object and image distances sum to $-\frac1{80}\,\mathrm{mm^{-1}}$ . Using +240 mm would instead yield $+\frac1{80}\,\mathrm{mm^{-1}}$ , describing a converging lens rather than the specified diverging one.

\begin{aligned}b&=\frac{(-80)(120)}{120-(-80)}\,\mathrm{mm}=-48\,\mathrm{mm}\\m&=-\frac{-48}{120}=+0.40\\\frac1{120\,\mathrm{mm}}+\frac1{-48\,\mathrm{mm}}&=-\frac1{80\,\mathrm{mm}}\\\frac1{120\,\mathrm{mm}}+\frac1{240\,\mathrm{mm}}&=+\frac1{80\,\mathrm{mm}}\end{aligned}

[1][2]

What to take away

Set the signs and solve for image distance first. Then magnification’s sign and magnitude determine orientation and size. Negative b means a virtual image, not a negative distance magnitude. At $g=f$ the ideal image is at infinity, and nearby small changes strongly move it. Geometric image position and actual detail resolution are separate questions.[1][2]

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 1, 2026Clarified the sign convention and thin paraxial model directly; exposed existing image cases, rearrangements and sensitivity, with relevant refraction and diffraction links.
October 1, 2026Introduced sign convention before calculation and explained the derivation; added inverse projection design, image-distance curves and a summary; clarified exercise checks.
October 1, 2026Added prerequisites and model assumptions; clarified derivation steps and arithmetic checks; included an additional original example, diagram and summary.

Cite this reference page

Quanta. Thin Lens Equation. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/linsengleichung

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Frequently asked questions about Thin Lens Equation

How do you calculate with the lens equation

\frac{1}{f} =\frac{ 1}{g} +\frac{ 1}{b}

First solve for the unknown reciprocal, then take the reciprocal at the end. All distances must use the same length unit. For a converging lens with $f=10 cm$ and a real object at $g=30 cm$ , reciprocal image distance is the difference between reciprocals of 10 cm and 30 cm; this gives $b=15 cm$ . Positive b indicates a real image on the emerging side. Signed magnification $m=-\frac bg$ is−0.5, so the image is inverted and half-sized. Check by substitution and the expected position between f and 2 f. Only the sum $g+b$ gives total separation between object and screen.

What sign convention applies to the lens equation?+

In this convention g is positive for a real object on the incident side. b is positive for a real image on the emerging side and negative for a virtual image on the incident side. f is positive for converging lenses and negative for diverging lenses. Already converging input can create a virtual object, giving negative g. Image heights also use signs: $m=-\frac bg$ is negative for an inverted image. The claim that every diverging lens produces only virtual images assumes real objects. Draw propagation direction and lens plane before translating positive geometric segment lengths into signed quantities.

What happens when the object is inside the focal length?+

A real object inside a converging lens’s positive focal distance gives a virtual, upright enlarged image. Reciprocal image distance becomes negative. With $f=10 cm$ and $g=5 cm$ , $b=-10 cm$ and $m=+2$ . The emerging rays actually diverge; their backward extensions meet at the virtual image location. A screen placed there cannot capture that image. Transverse magnification 2 is also not automatically the magnifier’s angular magnification, which depends on eye position and a chosen reference viewing condition. At $g=f$ , ideal image distance tends to infinity rather than giving a finite virtual image. These statements assume the thin paraxial model.

How is the magnification related to the lens equation?+

The central ray and similar triangles give signed magnification $m=\frac BG=-\frac bg$ . The minus sign carries orientation: real objects with real images are inverted. Use absolute values only when discussing unsigned sizes. A 2 cm object at $g=15 cm$ with $f=10 cm$ gives $b=30 cm$ and $B=-4 cm$ , a 4 cm inverted image. At $g=2 f$ , $m=-1$ and size is unchanged. For very distant objects, b approaches focal length. These relations predict geometric size and location, not diffraction resolution or the number of visible details. A longer focal length can enlarge an image without changing the aperture’s angular resolving criterion.

Does the lens equation also hold for diverging lenses and glasses?+

Yes, provided signs and the optical model are appropriate. For $f=-10 cm$ and a real object at $g=20 cm$ , $b\approx{}-6.667 cm$ and $m\approx{}+0.3333$ : virtual, upright and reduced. Optical power is reciprocal focal length in meters, for example−2 diopters when $f=-0.5 m$ . Closely contacting thin lenses in the same medium have approximately additive powers; finite spacing needs an extended system calculation. Real corrective optics also involve the eye and vertex distance. The school equation explains basic imaging but does not replace individual fitting. A diverging lens with a virtual object can even form a real image, so object type must be stated explicitly.

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Thin-lens equation

The thin-lens relation locates real or virtual paraxial images. Signed magnification

m=-\frac bg

describes their orientation.

BasicExam-relevantLower secondaryUpper secondary

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Symbol

Meaning

Unit

f

Signed paraxial focal length: positive converging, negative diverging

m or cm

g

Signed object distance from thin-lens plane; positive for a real object

m or cm

b

Signed image distance from thin-lens plane; positive for a real image

m or cm

Derivation & background

Exam blueprint

Validity range

Derivation steps

Intersect central and parallel rays using signed similar triangles.

1Use $g>0$ for real objects, $b>0$ for real images and $f>0$ for converging lenses. Virtual positions and diverging focal lengths carry corresponding negative signs.
2The thin paraxial central ray gives $\frac BG=-\frac bg$ for signed transverse heights.
3The incoming parallel ray passes through the image-side focus; at the image position $B=G\left(1-\frac bf\right)$ .
4Both rays belong to the same image point: $-\frac bg=1-\frac bf$ .
5Rearrange and divide by b to obtain $\frac1f=\frac1g+\frac1b$ . Use one length unit throughout.
6Take the reciprocal to find b and interpret its sign. Magnification $m=-\frac bg$ gives orientation and height; $g=f$ places the ideal image at infinity.

How do you calculate with Thin Lens Equation?

Here is how to work through a typical Thin Lens Equation (1/f = 1/g + 1/b) task step by step:

Task

Converging lens $f = 10 cm$ , object at $g = 15 cm$ . Where does the image form?

Solution path

Approach: $\frac1b=\frac1f-\frac1g$ .
Using the given centimetre values, $\frac1b=\frac1{10\,\mathrm{cm}}-\frac1{15\,\mathrm{cm}}=\frac1{30\,\mathrm{cm}}$ .
The image is at $b = 30 cm$ : real, inverted and twice the size, since $m=-\frac bg=-2$ .

Task

Object at $g = 5 cm$ in front of a lens with $f = 10 cm$ . What results?

Solution path

Approach: $\frac1b=\frac1f-\frac1g$ .
Thus $\frac1b=\frac1{10\,\mathrm{cm}}-\frac1{5\,\mathrm{cm}}=-\frac1{10\,\mathrm{cm}}$ .
Hence $b = -10 cm$ and $m=-\frac{-10}{5}=+2$ : virtual, upright and twice the size.
A screen at the computed location does not capture this virtual image.

Understand and apply the equation

Before calculating

What is the thin-lens equation with this sign convention?

\begin{aligned}&\frac1f=\frac1g+\frac1b\\&m=\frac BG=-\frac bg\end{aligned}

B and G are signed transverse heights; m<0 means inverted, m>0 upright. For unsigned sizes use $\frac{|B|}{|G|}=\left|\frac bg\right|$ .[1]

How do you derive the thin-lens equation and signed magnification?

\begin{aligned}-\frac bg&=1-\frac bf\\\frac bf-\frac bg&=1\\\frac1f-\frac1g&=\frac1b\\\frac1f&=\frac1g+\frac1b\end{aligned}

Both principal rays belong to one image point. The third line divides by finite image distance b.[1]

How do you calculate image distance, object distance, focal length and dioptres?

\begin{aligned}&b=\frac{fg}{g-f}\\&g=\frac{fb}{b-f}\\&f=\frac{gb}{g+b}\end{aligned}

Use one length unit and check denominators. For $g=f$ , the ideal image lies at infinity.[1]

How do you calculate image distance, height and orientation for a screen image?

An inverted image of a 30-mm object

Given: f = 80.0 mm; g=240.0 mm; G=30.0 mm; thin converging lens.
Find: Find image position, image height and total separation.

Approach: Use the thin-lens equation for image position; signed magnification gives orientation and height.
$\begin{aligned}\frac1b&=\frac1f-\frac1g\\m&=-\frac bg\\B&=mG\end{aligned}$
First obtain reciprocal image distance.
$\frac1b=\frac1{80\,\mathrm{mm}}-\frac1{240\,\mathrm{mm}}=\frac1{120\,\mathrm{mm}}$
Form signed magnification and apply it to height.
$\begin{aligned}b&=120\,\mathrm{mm}\\m&=-\frac{120\,\mathrm{mm}}{240\,\mathrm{mm}}=-0.5\\B&=-15.0\,\mathrm{mm}\end{aligned}$
Object and screen lie on opposite sides.
$L=g+b=360\,\mathrm{mm}$

Result: The real image lies 120 mm after the lens,15 mm tall and inverted. Object and screen are 360 mm apart.

When does a magnifier produce an upright virtual image?

Object inside the focal length

Given: f = 80.0 mm; g=50.0 mm; G=12.0 mm.
Find: Find virtual image position and signed transverse size.

Approach: The same thin-lens equation applies inside the focal length; negative b locates the virtual image on the incident side.
$\begin{aligned}\frac1b&=\frac1f-\frac1g\\m&=-\frac bg\\B&=mG\end{aligned}$
Compute the negative reciprocal image distance.
$\frac1b=\frac1{80}-\frac1{50}=-\frac3{400}\;\mathrm{mm}^{-1}$
Take the reciprocal and calculate magnification.
$\begin{aligned}&b=-133.333\,\mathrm{mm}\\&m=-\frac{b}{g}=2.66667\\&B=32.0\,\mathrm{mm}\end{aligned}$

Result: The virtual image lies 133.3 mm before the lens on the object side, upright and 2.667 times as large.

Explore further: designing a projection

A fourfold screen image over 500 mm

Given: Real object and image; object-to-screen distance L=500 mm; target signed magnification m=−4. Thin lens, same medium.
Find: Find object distance, image distance and required focal length.

Approach: a real image places object and screen on opposite sides. Combine separation and magnification before finding focal length.
$\begin{aligned}L&=g+b\\m&=-\frac bg=-4\\b&=4g\\g&=\frac L5\\f&=\frac{gb}{g+b}\end{aligned}$
Substitute the two distances and then focal length.
$\begin{aligned}g&=\frac{500\,\mathrm{mm}}5=100\,\mathrm{mm}\\b&=400\,\mathrm{mm}\\f&=\frac{(100\,\mathrm{mm})(400\,\mathrm{mm})}{500\,\mathrm{mm}}=80\,\mathrm{mm}\end{aligned}$

Result: The model lens has f = 80 mm and sits 100 mm after the object, leaving 400 mm to the screen. The image is four times as large and inverted.

Check: $\frac1{100}+\frac1{400}=\frac1{80}$ in inverse millimetres and $m=-\frac{400}{100}=-4$ check both constraints. The lens lies nearer the object, as expected for an enlarged real image.

Why is image distance so sensitive near the focal plane?

\frac{db}{dg}=-\frac{f^2}{(g-f)^2}

For fixed f, small changes of g shift b oppositely. The linear approximation requires sufficiently small changes.[1]

View diagram: image distance near the focal length

Image distance on both sides of the focal plane

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The gap marks the asymptote, not an image disappearing at b=0. Lines join only the displayed measurement ranges. All distances follow the article’s sign convention.[1]

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An imaging equation does not determine all image quality

Five exercises from a camera to a virtual object

Calculate · Foundation

Converging lens f=50 mm, object g=200 mm: find b and m.

Show hint

Use $b=\frac{fg}{g-f}$ .

Learning goal: Find image distance and signed magnification of a converging lens.

Show solution and calculation

Approach: use millimetres consistently and keep the positive focal length.
The lens equation gives b≈66.6667 mm.
Magnification $m=-\frac bg=-\frac13$ indicates a reduced inverted real image.

Check: $\frac1{200\,\mathrm{mm}}+\frac1{66.6667\,\mathrm{mm}}\approx\frac1{50\,\mathrm{mm}}$ . Since g=4f, b lies between f and 2f as expected.

b=\frac{50\cdot200}{200-50}\,\mathrm{mm}

Calculate · Application

Diverging lens f=−100 mm, real object g=250 mm: determine b and m.

Show hint

Keep negative f in the equation.

Learning goal: Use negative focal length to find a virtual image and its magnification.

Show solution and calculation

Approach: retain the negative focal length of the diverging lens.
The result is b≈−71.4286 mm and $m=\frac27\approx0.285714$ : virtual, upright and reduced.
Optical power is −10 diopters.

Check: inverse image distance plus inverse object distance equals negative inverse focal length; positive b would not fit this real object.

b=\frac{(-100)250}{250+100}\,\mathrm{mm}

Transfer · Challenge

An already converging bundle would meet 50 mm beyond a diverging lens: g=−50 mm, f=−100 mm. Where is the image?

Show hint

A virtual object has $g<0$ .

Learning goal: Handle a virtual object consistently for a diverging lens.

Show solution and calculation

Approach: already converging input means a virtual object, so g=−50 mm.
With f=−100 mm, inverse image distance is positive and b=+100 mm.
A real image forms despite the diverging lens.

Check: $\frac1{-50\,\mathrm{mm}}+\frac1{100\,\mathrm{mm}}=\frac1{-100\,\mathrm{mm}}$ . The input converges strongly enough; the lens moves its intersection farther downstream.

\begin{aligned}\frac1b&=\frac1{-100\,\mathrm{mm}}-\frac1{-50\,\mathrm{mm}}\\&=\frac1{100\,\mathrm{mm}}\\b&=+100\,\mathrm{mm}\end{aligned}

Interpret · Application

A real object before a thin converging lens is moved from g=150 mm to g=75 mm. Focal length is +100 mm and both sides have the same medium. Find image distance and magnification for each position using $m=-\frac bg$ . Which position permits projection onto a screen, and where is the other image?

Show hint

Solve for b at each object distance. Its sign identifies image side in this model; the sign of m identifies orientation.

Learning goal: Understand the change from real to virtual image as object distance crosses focal length.

Show solution and calculation

Approach: solve the thin-lens equation separately for both positions.
At 150 mm, b=+300 mm and m=−2: a real inverted image on the outgoing side.
At 75 mm, b=−300 mm and m=+4: a virtual upright image on the incident side.
Only the first can be directly projected onto a screen.

Check: image distance diverges as g passes through f rather than passing continuously through zero.

\begin{aligned}b&=\frac{fg}{g-f}\\{}m&=-\frac bg\\{}b_1&=\frac{100\cdot150}{150-100}\,\mathrm{mm}=+300\,\mathrm{mm}\\{}m_1&=-2\\{}b_2&=\frac{100\cdot75}{75-100}\,\mathrm{mm}=-300\,\mathrm{mm}\\{}m_2&=+4\end{aligned}

[1]

Error analysis · Application

A thin diverging lens has f=−80 mm and a real object at g=+120 mm. A solution replaces f by its magnitude and finds b=+240 mm. Correct image distance and magnification. Substitute into the original thin-lens equation to show why the positive image distance does not describe this lens.

Show hint

The focal-length sign carries physical information about lens type. It cannot be made positive without changing the model.

Learning goal: Use a diverging lens’s negative focal length consistently.

Show solution and calculation

Approach: retain the given negative focal length.
This gives b=−48 mm and m=+0.40: a reduced upright virtual image on the incident side.

\begin{aligned}b&=\frac{(-80)(120)}{120-(-80)}\,\mathrm{mm}=-48\,\mathrm{mm}\\m&=-\frac{-48}{120}=+0.40\\\frac1{120\,\mathrm{mm}}+\frac1{-48\,\mathrm{mm}}&=-\frac1{80\,\mathrm{mm}}\\\frac1{120\,\mathrm{mm}}+\frac1{240\,\mathrm{mm}}&=+\frac1{80\,\mathrm{mm}}\end{aligned}

[1][2]

What to take away

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 1, 2026Clarified the sign convention and thin paraxial model directly; exposed existing image cases, rearrangements and sensitivity, with relevant refraction and diffraction links.
October 1, 2026Introduced sign convention before calculation and explained the derivation; added inverse projection design, image-distance curves and a summary; clarified exercise checks.
October 1, 2026Added prerequisites and model assumptions; clarified derivation steps and arithmetic checks; included an additional original example, diagram and summary.

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Quanta. Thin Lens Equation. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/linsengleichung

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Frequently asked questions about Thin Lens Equation

How do you calculate with the lens equation

\frac{1}{f} =\frac{ 1}{g} +\frac{ 1}{b}

What sign convention applies to the lens equation?+

What happens when the object is inside the focal length?+

How is the magnification related to the lens equation?+

Does the lens equation also hold for diverging lenses and glasses?+