Mirror formula and magnification

A ray diagram tells you roughly where an image forms and roughly how big it is — useful for intuition, useless for a precise answer. The mirror formula and the magnification formula turn that sketch into arithmetic: give them the object distance and focal length, correctly signed, and they hand back the exact image position, its size, and everything about its nature — real or virtual, erect or inverted, enlarged or diminished — without a single line drawn. This pair of formulas is the workhorse of the CBSE Class 10 chapter Light — Reflection and Refraction, and every spherical-mirror numerical you will meet, for concave or convex mirrors alike, reduces to applying them carefully with the New Cartesian Sign Convention.

The core insight

The mirror formula 1/v+1/u=1/f1/v + 1/u = 1/f and the magnification formula m=h′/h=−v/um = h'/h = -v/u convert any spherical-mirror problem into signed algebra that fixes the image's position, size, and nature exactly.

The core idea

The mirror formula. For any spherical mirror, the object distance uu, the image distance vv, and the focal length ff are linked by one compact relation:

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

where: uu is the distance of the object from the pole, vv is the distance of the image from the pole, and ff is the focal length — all measured along the principal axis and substituted with their signed values under the New Cartesian Sign Convention (distances against the incident light are negative, distances with it are positive). This single equation applies without modification to concave mirrors, convex mirrors, real images, and virtual images; the signs of vv, uu, and ff that come out of the algebra automatically tell you which situation you are in. The derivation from similar triangles in a ray diagram is beyond this level — CBSE Class 10 asks you only to state and apply it.

The magnification formula. Linear magnification mm compares the image to the object:

m=h′h=−vum = \frac{h'}{h} = -\frac{v}{u}

where: h′h' is the (signed) image height and hh is the (signed) object height. Because hh is conventionally taken as positive (object upright, above the axis), the sign of mm is fixed entirely by −v/u-v/u, and it reads out the image's story directly:

  • mm positive means the image is erect, and an erect image formed by a mirror is always virtual.
  • mm negative means the image is inverted, and an inverted image is always real.
  • ∣m∣>1|m| > 1 means the image is enlarged; ∣m∣<1|m| < 1 means diminished; ∣m∣=1|m| = 1 means the image is exactly the object's size.

Put the two formulas together and a mirror problem becomes a short pipeline: use the mirror formula to find vv from the known uu and ff, then feed that vv back into the magnification formula to get mm, then read mm's sign and size to describe the image completely.

Figure
Ray diagram of a concave mirror forming a real inverted image beyond the centre of curvature, alongside the sign convention axes with distances against incident light marked negative and distances with incident light marked positive
Sign convention: pole at origin, incident light travels left to right +ve -ve P (pole) F C object real, inverted image
Derivation ASolving a concave mirror numerical from the mirror formula

Given an object placed at a known distance from a concave mirror of known focal length, find the image distance, describe the image fully, and compute its height — a goal that requires substituting signed values into the mirror formula, solving for the unknown, then chaining that result through the magnification formula to extract size and orientation.

Step 1. Fix the signed values using the New Cartesian Sign Convention: the pole is the origin, distances measured against the incident light (to the left, for a mirror facing left) are negative. For a concave mirror the focal length is always negative, since the focus lies in front of the mirror; the object distance is also negative, since real objects sit in front of the mirror.

Step 2. Substitute into the mirror formula and isolate 1/v1/v: 1v=1f−1u\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}. This is the algebraic form you will use every time — solve for whichever of uu, vv, ff is unknown by rearranging this same identity.

Step 3. Combine the fractions on the right using a common denominator (their LCM), simplify to a single fraction 1/v=…1/v = \dots, and invert it to get vv in centimetres. Check the sign of vv: negative means the image is real and forms in front of the mirror; positive means it is virtual and forms behind it.

Step 4. Compute magnification from the same uu and the newly found vv: m=−v/um = -v/u. Read the sign of mm for erect/inverted, and ∣m∣|m| against 11 for enlarged/diminished/same-size. Multiply mm by the object height hh to get the signed image height h′h', and state the complete description: position, real/virtual, erect/inverted, enlarged/diminished/same-size. Every step above is exactly the arithmetic carried out numerically in the worked example below, where u=−20 cmu = -20\ \text{cm} and f=−15 cmf = -15\ \text{cm} give v=−60 cmv = -60\ \text{cm} and m=−3m = -3.

Worked example

An object 4 cm tall is placed 20 cm in front of a concave mirror of focal length 15 cm. Find the position, nature, and size of the image.

Step 1. Write the signed values: u=−20 cmu = -20\ \text{cm}, f=−15 cmf = -15\ \text{cm} (concave mirror), h=4 cmh = 4\ \text{cm}.

Step 2. Apply the mirror formula:

1v=1f−1u=1−15−1−20=−115+120\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-15} - \frac{1}{-20} = -\frac{1}{15} + \frac{1}{20}

Taking the LCM of 15 and 20, which is 60:

1v=−460+360=−160⇒v=−60 cm\frac{1}{v} = -\frac{4}{60} + \frac{3}{60} = -\frac{1}{60} \quad\Rightarrow\quad v = -60\ \text{cm}

Since vv is negative, the image forms 60 cm in front of the mirror — a real image.

Step 3. Compute magnification:

m=−vu=−−60−20=−3m = -\frac{v}{u} = -\frac{-60}{-20} = -3

Since mm is negative, the image is inverted; since ∣m∣=3>1|m| = 3 > 1, the image is enlarged — three times the object's size.

Step 4. Find image height: h′=m×h=−3×4 cm=−12 cmh' = m \times h = -3 \times 4\ \text{cm} = -12\ \text{cm}. The negative sign confirms inversion. The image is 12 cm tall, real, inverted, enlarged three times, formed 60 cm in front of the mirror.

Competitive stretch

A convex mirror used for shop security has focal length 25 cm, and an object stands 60 cm from it. Here u=−60 cmu = -60\ \text{cm}, f=+25 cmf = +25\ \text{cm} (convex mirror, focus behind). The mirror formula gives 1/v=1/25−1/(−60)=1/25+1/601/v = 1/25 - 1/(-60) = 1/25 + 1/60; with LCM 300, 1/v=12/300+5/300=17/3001/v = 12/300 + 5/300 = 17/300, so v=300/17≈17.6 cmv = 300/17 \approx 17.6\ \text{cm}, positive — a virtual image 17.6 cm behind the mirror. Magnification m=−v/u=−17.6/(−60)≈0.29m = -v/u = -17.6/(-60) \approx 0.29, positive and less than 1, so the image is erect and diminished — exactly the standard rule that a convex mirror always forms a virtual, erect, diminished image, whatever the object distance.

Common traps

  • Using unsigned magnitudes. Both formulas demand signed values throughout — plugging in a plain positive ff for a concave mirror gives a meaningless answer.
  • Dropping the negative sign in m=−v/um = -v/u. Writing m=v/um = v/u (the lens-formula convention) flips the predicted orientation for a mirror problem.
  • Reading the sign of mm as size. The sign of mm fixes erect vs. inverted; only the magnitude ∣m∣|m| compared with 1 tells you enlarged, diminished, or same size.
  • Sign slips in the LCM arithmetic. Since uu and ff are usually both negative for concave mirrors, mishandled double negatives are the most common source of a wrong final vv — verify each substitution before combining fractions.
Why is f always negative for a concave mirror but positive for a convex mirror?

The New Cartesian Sign Convention measures every distance from the pole, with distances against the incident light taken as negative. A concave mirror's focus lies in front of the mirror, on the same side as the object, so it gets a negative distance. A convex mirror's focus lies behind the mirror, so it gets a positive distance. This single rule is what lets one formula, 1/v+1/u=1/f1/v + 1/u = 1/f, work correctly for both mirror types without needing separate versions.

Do I need to memorise the derivation of the mirror formula for the CBSE exam?

No. The CBSE Class 10 syllabus explicitly states and applies the mirror formula and the magnification formula without requiring their derivation from similar triangles in a ray diagram — that derivation belongs to a later stage. What you must be fluent in is substituting correctly signed values, solving the resulting algebra, and interpreting the sign and magnitude of the answer.

My computed v came out positive for a concave mirror problem — did I make a mistake?

Not necessarily a sign-convention mistake, but check your object distance: for a concave mirror, if the object is placed between the pole and the focus, the image is virtual, forms behind the mirror, and vv is genuinely positive. If instead the object lies beyond the focus (as in the worked example above), a real image forms in front of the mirror and vv must be negative — so a positive vv there would signal an arithmetic slip. Always sanity-check the sign against where the object sits relative to FF.

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