Physics · Optics

Thin Lens Equation

Q: How do you calculate with the lens equation 1/f = 1/g + 1/b?

Solve for the required reciprocal and only take the reciprocal at the end. Example: converging lens with f = 10 cm, object at g = 30 cm. Then 1/b = 1/f − 1/g = 1/10 − 1/30 = 3/30 − 1/30 = 2/30 = 1/15, so b = 15 cm, where a real, inverted image forms. It is safest to work with fractions over a common denominator instead of rounded decimals. All three distances must be in the same unit (all cm or all m). The classic slip: computing 1/b = 0.0667 and writing it down as b, forgetting the reciprocal.

Q: What sign convention applies to the lens equation?

In the school convention for thin lenses: the object distance g is positive when the object stands really in front of the lens (the normal case). The image distance b is positive for real images behind the lens, which can be caught on a screen. A negative b means a virtual image on the object side: it appears when looking through the lens, as with a magnifying glass. The focal length f is positive for converging lenses and negative for diverging lenses, which fundamentally produce only virtual, reduced images. If you carry the signs consistently, the image type comes for free: the numerical result itself says whether the image is real or virtual; you only have to interpret it correctly.

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

Then the converging lens works as a magnifying glass. Mathematically the image distance becomes negative: at g = 5 cm and f = 10 cm, 1/b = 1/10 − 1/5 = −1/10, so b = −10 cm. The negative sign means there is no real image that could be caught on a screen. Instead a virtual, upright and magnified image forms on the object side; the rays leaving the lens diverge, and the eye extends them backwards to an apparent image. The magnification |b|/g = 10/5 = 2 shows twice the size. Exactly the three cases g > 2f (reduced), 2f > g > f (magnified, real) and g < f (magnifier) are the standard material of exam problems.

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

The magnification follows from the same similar triangles as the lens equation: B/G = b/g; image size to object size behaves like image distance to object distance. Example: a 2 cm object at g = 15 cm in front of a lens with f = 10 cm gives b = 30 cm from the lens equation; the magnification is b/g = 2, so the image is 4 cm tall, real and inverted. Special cases help with orientation: at g = 2f, b = 2f and the image is exactly the same size, which is also a quick experimental way to find the focal length. For g → ∞ (distant objects) the image moves into the focal plane, the principle of every camera: the sensor sits roughly at b ≈ f.

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

Yes, with a negative focal length. A diverging lens with f = −10 cm and an object at g = 20 cm gives 1/b = −1/10 − 1/20 = −3/20, so b ≈ −6.7 cm: always a virtual, upright, reduced image, wherever the object stands. Opticians calculate with the refractive power D = 1/f in dioptres (f in metres) instead of focal lengths: glasses of −2 dpt have f = −0.5 m and correct short-sightedness by pushing the prematurely focused image back onto the retina; long-sighted people wear converging lenses with positive dioptres. The dioptre measure is practical: for lenses placed closely one behind the other the powers simply add, D_total = D₁ + D₂.

The lens equation links focal length, object distance and image distance of a thin lens, the fundamental equation of optical imaging.

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Formula

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

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

f, g, b in metres [m] or centimetres [cm], all in the same unit

Construction at a converging lens: the parallel ray and the central ray meet at the real, inverted image. 1/f = 1/g + 1/b.

Variables & units – Thin Lens Equation

Symbol	Meaning	Unit
f	Focal length of the lens	m or cm
g	Object distance (object to lens)	m or cm
b	Image distance (lens to image)	m or cm

Derivation & background – Thin Lens Equation

The equation follows geometrically from similar triangles for the parallel, central and focal rays. Sign convention (thin lens): g > 0 for real objects, b > 0 for real images behind the lens, b < 0 for virtual images (magnifying glass: g < f), f > 0 for converging lenses, f < 0 for diverging lenses. The magnification is B/G = b/g.

Exam blueprint

Validity range

Holds for thin lenses and paraxial rays. Sign convention: g > 0 for real objects, b > 0 for real images, b < 0 for virtual images, f < 0 for diverging lenses.

Derivation steps

From similar triangles for the parallel and focal rays at the lens centre.

1Similar triangles give B/G = b/g and B/G = (b−f)/f.
2Equating and rearranging leads to 1/f = 1/g + 1/b.

Rearrangements

Image distance from focal length and object distance

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

For g < f, b becomes negative; the image is virtual (magnifying glass).

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

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

Image size to object size behaves like image distance to object distance.

Task variant

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

1/b = 1/10 − 1/15 = 1/30 → b = 30 cm, real, inverted, magnified (b/g = 2).

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

1/b = 1/10 − 1/5 = −1/10 → b = −10 cm: a virtual, upright, magnified image, the magnifying glass.

Common mistakes

Forgetting the reciprocal at the end and reporting 1/b as b.

First compute 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.

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Formula cluster

Geometrical optics

Refraction explains the lens, the lens equation its imaging.

n_{1} \cdot sin (α_{1}) = n_{2} \cdot sin (α_{2})

Snell's Law of Refraction

c = λ \cdot f

Wave Equation (c = λ·f)

λ = \frac{h}{p} = \frac{h}{m v}

De Broglie Wavelength

Worked example

Converging lens f = 10 cm, object at g = 30 cm: 1/b = 1/10 − 1/30 = 2/30, so b = 15 cm, a real, inverted image.

Applications

Camera and lens design, fitting spectacles, microscope and telescope, projectors, the eye (accommodation)

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How do you rearrange 1/f = 1/g + 1/b for Image distance from focal length and object distance?

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Which common mistake happens with Thin Lens Equation?

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Scientific sources

[1]Hecht, E. (2017). Optics, 5th Ed. Pearson.

Common notations & search queries

1/f=1/g+1/bLinsengleichung FormelAbbildungsgleichung LinseBrennweite berechnenBildweite berechnendünne Linse Formelthin lens equationSammellinse Formel

Related formulas

n_{1} \cdot sin (α_{1}) = n_{2} \cdot sin (α_{2})

Snell's Law of Refraction

c = λ \cdot f

Wave Equation (c = λ·f)

λ = \frac{h}{p} = \frac{h}{m v}

De Broglie Wavelength

More Physics formulas

F = m \cdot a

Newton's Second Law

F_{G} = G \cdot \frac{m _{1} \cdot m _{2}}{r ^{2}}

Newton's Law of Universal Gravitation

E_{kin} = \frac{1}{2} m v^{2}

Kinetic Energy

E_{p o t} = m \cdot g \cdot h

Potential Energy (Gravitational)

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Frequently asked questions about Thin Lens Equation

How do you calculate with the lens equation 1/f = 1/g + 1/b?+

Solve for the required reciprocal and only take the reciprocal at the end. Example: converging lens with f = 10 cm, object at g = 30 cm. Then 1/b = 1/f − 1/g = 1/10 − 1/30 = 3/30 − 1/30 = 2/30 = 1/15, so b = 15 cm, where a real, inverted image forms. It is safest to work with fractions over a common denominator instead of rounded decimals. All three distances must be in the same unit (all cm or all m). The classic slip: computing 1/b = 0.0667 and writing it down as b, forgetting the reciprocal.

What sign convention applies to the lens equation?+

In the school convention for thin lenses: the object distance g is positive when the object stands really in front of the lens (the normal case). The image distance b is positive for real images behind the lens, which can be caught on a screen. A negative b means a virtual image on the object side: it appears when looking through the lens, as with a magnifying glass. The focal length f is positive for converging lenses and negative for diverging lenses, which fundamentally produce only virtual, reduced images. If you carry the signs consistently, the image type comes for free: the numerical result itself says whether the image is real or virtual; you only have to interpret it correctly.

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

Then the converging lens works as a magnifying glass. Mathematically the image distance becomes negative: at g = 5 cm and f = 10 cm, 1/b = 1/10 − 1/5 = −1/10, so b = −10 cm. The negative sign means there is no real image that could be caught on a screen. Instead a virtual, upright and magnified image forms on the object side; the rays leaving the lens diverge, and the eye extends them backwards to an apparent image. The magnification |b|/g = 10/5 = 2 shows twice the size. Exactly the three cases g > 2f (reduced), 2f > g > f (magnified, real) and g < f (magnifier) are the standard material of exam problems.

How is the magnification related to the lens equation?+

The magnification follows from the same similar triangles as the lens equation: B/G = b/g; image size to object size behaves like image distance to object distance. Example: a 2 cm object at g = 15 cm in front of a lens with f = 10 cm gives b = 30 cm from the lens equation; the magnification is b/g = 2, so the image is 4 cm tall, real and inverted. Special cases help with orientation: at g = 2f, b = 2f and the image is exactly the same size, which is also a quick experimental way to find the focal length. For g → ∞ (distant objects) the image moves into the focal plane, the principle of every camera: the sensor sits roughly at b ≈ f.

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

Yes, with a negative focal length. A diverging lens with f = −10 cm and an object at g = 20 cm gives 1/b = −1/10 − 1/20 = −3/20, so b ≈ −6.7 cm: always a virtual, upright, reduced image, wherever the object stands. Opticians calculate with the refractive power D = 1/f in dioptres (f in metres) instead of focal lengths: glasses of −2 dpt have f = −0.5 m and correct short-sightedness by pushing the prematurely focused image back onto the retina; long-sighted people wear converging lenses with positive dioptres. The dioptre measure is practical: for lenses placed closely one behind the other the powers simply add, D_total = D₁ + D₂.

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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/b = 1/10 − 1/15 = 1/30 → b = 30 cm, real, inverted, magnified (b/g = 2).
2
Task
Object at g = 5 cm in front of a lens with f = 10 cm. What results?
Solution path
1/b = 1/10 − 1/5 = −1/10 → b = −10 cm: a virtual, upright, magnified image, the magnifying glass.