Thin Lens Formula and Magnification

For a thin lens of focal length $f$ in air, the object and image distances satisfy the lens equation.

Concept

Thin lens formula:

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

Linear magnification:

$m = \frac{h'}{h} = \frac{v}{u}$

(Both $u$ and $v$ are measured with the Cartesian sign convention; signs of $h$ and $h'$ follow.)

• A converging lens has $f > 0$.
• A diverging lens has $f < 0$.
• For a real object ($u < 0$) in a converging lens, the image is real and inverted when $|u| > f$; virtual and erect when $|u| < f$.

Derivation

The lens formula follows directly from the lens-maker's derivation: adding the two single-surface relations gave

$\frac{1}{v} - \frac{1}{u} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \frac{1}{f}$

For the magnification, two rays construct the image: one parallel to the axis bending through $F$, and one through the optical centre going undeviated. Similar triangles (object and image heights with $u$ and $v$) give

$\frac{h'}{h} = \frac{v}{u}$

with the appropriate signs absorbed.

Two lenses in contact (focal lengths $f_1, f_2$):

$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$

equivalently $P = P_1 + P_2$.

Worked Example

Object 30 cm in front of converging lens of $f = 20$ cm. Find image position and magnification.

$u = -30$ cm, $f = +20$ cm.

$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{20} + \frac{1}{-30} = \frac{3 - 2}{60} = \frac{1}{60}$

$v = +60$ cm (real image, on the opposite side of the lens).

$m = \frac{v}{u} = \frac{60}{-30} = -2$

Image is real, inverted, twice the object size.

Common Confusions

• For mirrors $m = -v/u$, for lenses $m = +v/u$. The signs come from the geometry of how rays go through versus reflect.
• The lens formula here uses the convention $1/v - 1/u = 1/f$ (with signed quantities); some textbooks write the unsigned version $1/v + 1/u = 1/f$ for a real object — both are equivalent if used consistently.
• The optical centre is the point where a ray passes undeviated. For a thin symmetric lens it coincides with the geometric centre.

Key Takeaways

• $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ for a thin lens.
• Magnification $m = v/u$ (signed).
• Converging lens: $f > 0$; diverging lens: $f < 0$.
• Lenses in contact add powers: $P = P_1 + P_2$.