Lens Formula, Magnification and Sign Convention

Sign Convention for Lenses

Lenses use the same New Cartesian Sign Convention as mirrors, but all distances are measured from the optical centre O:

• The object is on the left, so $u$ is negative.
• Distances measured in the direction of incident light (to the right) are positive; against it (left) are negative.
• Heights above the axis are positive; below are negative.

Crucially:

• Convex lens: focal length $f$ is positive.
• Concave lens: focal length $f$ is negative.

The Lens Formula

The object distance ($u$), image distance ($v$) and focal length ($f$) of a lens are related by the lens formula: $$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$

Note the minus sign — this is the key difference from the mirror formula (which has a plus). The lens formula works for all lenses and all object positions, with the correct signs.

[NEET Important] Compare: mirror $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$; lens $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$. Mixing them up is the most common exam mistake.

Magnification for Lenses

For a lens, magnification is again the ratio of image height to object height, and is related to $v$ and $u$ by: $$m = \frac{h'}{h} = \frac{v}{u}$$

(Notice: for a lens it is $+\dfrac{v}{u}$, with no minus sign, unlike the mirror's $-\dfrac{v}{u}$.)

Reading the sign:

• $m$ positive → image virtual and erect (convex lens with object inside F, or any concave lens).
• $m$ negative → image real and inverted.
• $|m| > 1$ enlarged; $|m| < 1$ diminished.

Key Point: Lens magnification $m = \dfrac{v}{u}$ (no minus). Mirror magnification $m = -\dfrac{v}{u}$.

Memory Capsule — Section 7

Quick revision: lens formula and magnification.

1. Sign convention: measure from O; $u$ negative; convex $f$ positive, concave $f$ negative. 2. Lens formula: $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ (note the minus — opposite of mirror). 3. Magnification: $m = \frac{h'}{h} = \frac{v}{u}$ (no minus, unlike mirrors). 4. $m>0$ virtual/erect; $m<0$ real/inverted; $|m|>1$ enlarged. 5. Remember: mirror has $+$ and $-v/u$; lens has $-$ and $+v/u$.

Solved Examples

Example 1: NCERT — Concave Lens (Example 9.3)

A concave lens has focal length 15 cm. At what distance should the object be placed so the image forms 10 cm from the lens? Find the magnification.

Solution: Given: $v = -10$ cm (virtual, same side), $f = -15$ cm (concave).

$\dfrac{1}{u} = \dfrac{1}{v} - \dfrac{1}{f} = \dfrac{1}{-10} - \dfrac{1}{-15} = -\dfrac{1}{10} + \dfrac{1}{15} = \dfrac{-3 + 2}{30} = \dfrac{-1}{30}$

$$u = -30\ \text{cm}$$

The object is 30 cm from the lens. Magnification: $m = \dfrac{v}{u} = \dfrac{-10}{-30} = +0.33$.

The positive $m$ shows the image is virtual and erect, one-third the object's size.

Takeaway: A concave lens gives a positive m (virtual, erect) and a diminished image.

Example 2: NCERT — Convex Lens (Example 9.4)

A 2.0 cm tall object is placed 15 cm from a convex lens of focal length 10 cm. Find the nature, position, size and magnification of the image.

Solution: Given: $h = +2.0$ cm, $u = -15$ cm, $f = +10$ cm (convex).

$\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} = \dfrac{1}{10} + \dfrac{1}{-15} = \dfrac{1}{10} - \dfrac{1}{15} = \dfrac{3 - 2}{30} = \dfrac{1}{30}$

$$v = +30\ \text{cm}$$

Positive $v$ ⇒ image on the other side, real and inverted.

Magnification: $m = \dfrac{v}{u} = \dfrac{+30}{-15} = -2$. Height $h' = m \cdot h = -2 \times 2.0 = -4.0$ cm.

So the image is real, inverted, enlarged (2×), 4.0 cm tall, 30 cm from the lens.

Takeaway: Negative m and h' confirm a real, inverted image; |m| = 2 means twice enlarged.

Example 3: NCERT Exercise — Converging Lens, Object at 25 cm

A 5 cm long object is held 25 cm from a converging lens of focal length 10 cm. Find the position, size and nature of the image.

Solution: Given: $h = +5$ cm, $u = -25$ cm, $f = +10$ cm.

$\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} = \dfrac{1}{10} - \dfrac{1}{25} = \dfrac{5 - 2}{50} = \dfrac{3}{50}$

$$v = +16.67\ \text{cm} \approx +16.7\ \text{cm}$$

$m = \dfrac{v}{u} = \dfrac{16.7}{-25} = -0.67$; height $h' = -0.67 \times 5 = -3.3$ cm.

The image is real, inverted and diminished, about 3.3 cm tall, 16.7 cm on the other side of the lens.

Takeaway: Object beyond 2F (25 > 20) gives a diminished, real, inverted image.

Example 4: NCERT Exercise — Concave Lens, Image at 10 cm

A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object from the lens?

Solution: Given: $v = -10$ cm, $f = -15$ cm.

$\dfrac{1}{u} = \dfrac{1}{v} - \dfrac{1}{f} = \dfrac{1}{-10} - \dfrac{1}{-15} = -\dfrac{1}{10} + \dfrac{1}{15} = \dfrac{-1}{30}$

$$u = -30\ \text{cm}$$

The object is placed 30 cm from the lens.

Takeaway: Same setup as Example 9.3 — a concave lens forms a virtual image between the object and the lens.

Example 5: Magnification and Image Height

A convex lens forms an image with magnification $m = -3$ of an object 2 cm tall. What is the image height and nature?

Solution: Image height: $h' = m \cdot h = -3 \times 2 = -6$ cm.

Since $m$ is negative, the image is real and inverted; $|m| = 3 > 1$, so it is enlarged.

The image is 6 cm tall, real, inverted and 3× enlarged.

Takeaway: $h' = m \times h$; the sign of m tells you real/virtual, its size tells you enlarged/diminished.