Thin Lens Equation

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.

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.

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.

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.

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.