Lens and Mirror Formulas — Sign Conventions

Why Sign Conventions Trip Up Even Toppers

Half the marks lost on optics problems aren’t about the physics — they’re about signs. A student writes $1/v + 1/u = 1/f$ for a concave mirror, plugs in numbers, and gets a virtual image when the answer should be real (or vice versa). The mirror formula isn’t wrong; the signs were.

The good news: sign conventions are mechanical once you know the rules. We use the Cartesian sign convention (also called the “New Cartesian convention”) for both mirrors and lenses. Once you internalize it — pole/optical center is the origin, light travels left to right, distances measured against the light are negative — you’ll solve mirror and lens problems in two lines.

This hub covers the formulas, every sign trap, the focal-length convention for concave/convex elements, image-formation logic, magnification rules, and 8+ practice problems with full solutions.

The Cartesian Sign Convention — One Rule

Place the pole (mirror) or optical center (lens) at the origin. Light travels from left to right by default.

Distances measured in the direction of incident light (right) are positive.
Distances measured against the direction of incident light (left) are negative.
Heights measured above the principal axis are positive, below are negative.

That’s it. Every sign in optics flows from this rule.

Mirror Formula

\frac{1}{v} + \frac{1}{u} = \frac{1}{f}

with magnification $m = -v/u = h'/h$ .

For a real object, light comes in from the left, so $u < 0$ always.

For a concave mirror (converges light): $f < 0$ .

For a convex mirror (diverges light): $f > 0$ .

If the image is real (on the same side as the object — left side): $v < 0$ .

If the image is virtual (behind the mirror — right side): $v > 0$ .

Lens Formula

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

with magnification $m = v/u = h'/h$ .

Note the minus sign between $1/v$ and $1/u$ — this is different from the mirror formula. Many students memorize the mirror formula and unconsciously apply it to lenses, getting the wrong answer.

For a real object on the left, $u < 0$ .

For a convex lens (converges): $f > 0$ .

For a concave lens (diverges): $f < 0$ .

For a real image (right side, opposite side from object): $v > 0$ .

For a virtual image (left side, same side as object): $v < 0$ .

Focal Length Memory Trick

Concave mirror → $f$ negative. Convex mirror → $f$ positive. Convex lens → $f$ positive. Concave lens → $f$ negative.

Notice the symmetry: a converging element has different signs for mirror vs lens, because the focal point is on different sides! For mirrors, light reflects back, so the focal point of a converging mirror (concave) is on the incoming-light side → negative. For lenses, light passes through, and the focal point of a converging lens (convex) is on the far side → positive.

Magnification — Two Different Formulas

Mirror: $m = -\dfrac{v}{u}$ — note the minus sign

Lens: $m = \dfrac{v}{u}$ — no minus sign

In both: $m > 0$ means upright image; $m < 0$ means inverted; $|m| > 1$ means enlarged; $|m| < 1$ means reduced.

Worked Example 1 — Concave Mirror

An object is placed $30 \text{ cm}$ in front of a concave mirror of focal length $20 \text{ cm}$ . Find the image position and magnification.

Object is on the incoming-light side (left), so $u = -30 \text{ cm}$ .

Concave mirror, so $f = -20 \text{ cm}$ .

$1/v + 1/(-30) = 1/(-20)$ .

$1/v = -1/20 + 1/30 = -3/60 + 2/60 = -1/60$ .

$v = -60 \text{ cm}$ .

$v < 0$ means image is on the same side as the object — real, in front of the mirror, $60 \text{ cm}$ from the pole.

$m = -v/u = -(-60)/(-30) = -2$ . Image is inverted, twice as large.

Worked Example 2 — Convex Lens (Real Image)

An object $4 \text{ cm}$ tall is placed $24 \text{ cm}$ in front of a convex lens of focal length $8 \text{ cm}$ . Find the image position, size, and nature.

$u = -24$ , $f = +8$ .

$1/v - 1/(-24) = 1/8 \implies 1/v = 1/8 - 1/24 = 3/24 - 1/24 = 2/24 = 1/12$ .

$v = +12 \text{ cm}$ → image on the far side, real.

$m = v/u = 12/(-24) = -0.5$ . Inverted, half-size. Image height = $-0.5 \times 4 = -2 \text{ cm}$ (inverted, $2 \text{ cm}$ tall).

Worked Example 3 — Convex Lens (Virtual Image)

Same lens ( $f = 8 \text{ cm}$ ), but now object placed $5 \text{ cm}$ from the lens.

$u = -5$ , $f = +8$ . $1/v - 1/(-5) = 1/8 \implies 1/v = 1/8 - 1/5 = 5/40 - 8/40 = -3/40$ .

$v = -40/3 \approx -13.3 \text{ cm}$ → image on the same side as the object → virtual.

$m = v/u = -13.3/(-5) = +2.67$ . Upright, magnified $2.67\times$ . This is the magnifying-glass mode.

Lens Maker’s Formula

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

where $R_1$ is the first surface light hits, $R_2$ is the second, $n$ is refractive index of lens material relative to surrounding medium.

Sign rule: $R$ is positive if the center of curvature is on the right (transmission side), negative if on the left.

For a biconvex lens, $R_1 > 0$ , $R_2 < 0$ , so $1/f$ is positive — converging.

For a biconcave lens, $R_1 < 0$ , $R_2 > 0$ , so $1/f$ is negative — diverging.

Combination of Lenses

\frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2}

Power: $P = 1/f$ in dioptres when $f$ is in metres. Powers add: $P_{eq} = P_1 + P_2$ .

If a convex lens of $P_1 = +5$ D is combined with a concave lens of $P_2 = -3$ D, the combination has $P = +2$ D, so $f = 50 \text{ cm}$ (converging).

Exam-Specific Tips

JEE Main: 1-2 questions per year on optics. Lens combinations and silvered lenses are common JEE Main 2024 favourites. Memorize $P_{eq} = P_1 + P_2$ for contact and the silvered-lens trick: $P_{equivalent} = 2P_{lens} + P_{mirror}$ .

NEET: Heavy on lens formula numericals — usually $f$ given, $u$ given, find $v$ and magnification. 2-3 marks. Watch for “bi-convex with $R_1 = R_2 = R$ ” which simplifies to $1/f = 2(n-1)/R$ .

CBSE Class 12: Derivation of lens formula from refraction at curved surfaces is a 5-mark question. Lens maker’s formula derivation is 3 marks.

When an MCQ has all four options as numbers, plug in $u, f$ with proper signs and chase the answer. Don’t waste time on “is it real or virtual?” up front — let the sign of $v$ tell you at the end.

Common Mistakes to Avoid

Mistake 1: Using positive $u$ for a real object. Real objects are on the incoming-light side → $u$ is always negative for a real object in standard problems.

Mistake 2: Mixing mirror and lens formulas. Mirror has $1/v + 1/u$ ; lens has $1/v - 1/u$ . Different.

Mistake 3: Wrong sign of $f$ for converging vs diverging. Concave mirror: $f < 0$ . Convex lens: $f > 0$ . They’re different conventions for the same word “converging”.

Mistake 4: Forgetting that magnification of a lens has no minus sign, but magnification of a mirror does. $m_{lens} = v/u$ , $m_{mirror} = -v/u$ .

Mistake 5: Plugging the magnitude of $u$ instead of the signed value. If $u = -30 \text{ cm}$ , write $-30$ , not $30$ , in the formula.

Practice Questions

Q1. Object at $20 \text{ cm}$ from concave mirror, $f = 15 \text{ cm}$ . Find $v$ .

$u = -20$ , $f = -15$ . $1/v = -1/15 + 1/20 = (-4 + 3)/60 = -1/60$ . $v = -60 \text{ cm}$ — real, $60 \text{ cm}$ in front.

Q2. Convex mirror, $f = 30 \text{ cm}$ , object at $20 \text{ cm}$ . $v$ and $m$ .

$u = -20$ , $f = +30$ . $1/v = 1/30 + 1/20 = 5/60 = 1/12$ . $v = +12 \text{ cm}$ (virtual, behind mirror). $m = -v/u = -12/(-20) = +0.6$ — upright, reduced.

Q3. Concave lens $f = 20 \text{ cm}$ , object at $30 \text{ cm}$ . Image position?

$u = -30$ , $f = -20$ . $1/v = 1/(-20) + 1/(-30) = -3/60 - 2/60 = -5/60 = -1/12$ . $v = -12 \text{ cm}$ — virtual, same side as object.

Q4. Two thin lenses, $+10 \text{ cm}$ and $-15 \text{ cm}$ , in contact. Equivalent focal length?

$1/f_{eq} = 1/10 - 1/15 = 3/30 - 2/30 = 1/30$ . $f_{eq} = +30 \text{ cm}$ (converging).

Q5. Biconvex lens, $R_1 = R_2 = 20 \text{ cm}$ , $n = 1.5$ . Find $f$ .

Sign convention: $R_1 = +20$ , $R_2 = -20$ . $1/f = (1.5 - 1)(1/20 - 1/(-20)) = 0.5 \times 2/20 = 1/20$ . $f = +20 \text{ cm}$ .

Q6. A real image is twice the object size, formed by a convex lens of $f = 20 \text{ cm}$ . Find $u$ .

$m = -2$ (real image inverted), so $v/u = -2 \implies v = -2u$ . From lens formula: $1/(-2u) - 1/u = 1/20$ . $-1/(2u) - 2/(2u) = 1/20 \implies -3/(2u) = 1/20$ . $u = -30 \text{ cm}$ .

Q7. Power of a lens is $-2 \text{ D}$ . What is its $f$ ?

$P = 1/f$ (in metres). $f = 1/(-2) = -0.5 \text{ m} = -50 \text{ cm}$ . Concave lens.

Q8. A convex mirror has $f = 25 \text{ cm}$ . Object placed $50 \text{ cm}$ away. Find $v$ .

$u = -50$ , $f = +25$ . $1/v = 1/25 + 1/50 = 2/50 + 1/50 = 3/50$ . $v = +50/3 \approx 16.7 \text{ cm}$ (virtual, behind mirror).

FAQs

Q: Why is the focal length of a concave mirror negative if it’s the “useful” one?

It’s a convention, not a value judgment. The pole is at the origin and the focal point is on the same side as the incoming light, which by the Cartesian convention is the negative side. The negative sign just records that geometric fact.

Q: Can magnification be greater than 1 for a convex mirror?

No — convex mirrors always give virtual, upright, diminished images. $|m| < 1$ always.

Q: When is the image of a convex lens virtual?

When the object is placed within the focal length: $|u| < f$ . The lens can’t converge the rays enough to form a real image.

Q: Does the lens formula assume thin lens?

Yes. For thick lenses, we treat each surface separately or use the lens-maker’s formula at each interface.

Q: How does the sign of $h$ (image height) relate to magnification?

$m = h'/h$ . If $m > 0$ , $h'$ has the same sign as $h$ (same orientation, upright). If $m < 0$ , opposite sign (inverted).

Q: What if the object is on the right side of a lens (light going left to right anyway)?

Then by the convention, $u > 0$ (object beyond the optical center in the transmission direction). Such cases arise for “virtual objects” formed by another lens — handle them carefully.