Image formation by spherical mirrors

Move an object slowly toward a concave mirror and the image does something surprising: it shrinks, grows, flips upside down, and can vanish into "nowhere" depending on exactly where the object sits. A convex mirror, by contrast, never changes its story at all — the image is always small, upright, and trapped behind the glass, which is precisely why it makes a good rear-view mirror. This topic sits inside the CBSE Class 10 chapter Light — Reflection and Refraction (Unit III), building on spherical-mirror terms (pole, centre of curvature, focus). Beyond memorising the six standard cases, canon-lite adds the tool that makes every one of those cases computable: the mirror formula and magnification, tied together with the New Cartesian Sign Convention so that a "+" or "−" sign alone tells you whether an image is real or virtual.

The core insight

The mirror formula 1/v+1/u=1/f1/v + 1/u = 1/f, read with the New Cartesian Sign Convention, converts every qualitative "real, inverted, enlarged" statement about a spherical mirror into a signed number you can calculate.

The core idea

Sign convention first. All distances are measured from the pole PP along the principal axis. Distances measured in the direction of the incident light (into the mirror, away from the observer) are taken as positive; distances measured against the incident light (back toward the observer) are negative. The object is always placed in front of the mirror, so the object distance uu is always negative. For a concave mirror the focal length ff is negative (focus lies in front, on the same side as the object); for a convex mirror ff is positive (focus lies behind the mirror).

The mirror formula relates object distance uu, image distance vv, and focal length ff:

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

where: uu is the object distance from the pole, vv is the image distance from the pole, and ff is the focal length, with f=R/2f = R/2 for radius of curvature RR. All three are signed according to the convention above; a negative vv places the image in front of the mirror (real, on the object's side), a positive vv places it behind the mirror (virtual).

Magnification tells you the image's size and orientation relative to the object:

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

where: h′h' is the image height, hh is the object height, and mm is the magnification. A negative mm means the image is inverted (real images from a concave mirror are always inverted, so mm is negative); a positive mm means the image is erect (virtual images are always erect, so mm is positive). ∣m∣>1|m| > 1 means enlarged, ∣m∣<1|m| < 1 means diminished, ∣m∣=1|m| = 1 means same size.

Figure
Ray diagram of a concave mirror with an object between the centre of curvature and the focus producing a real inverted enlarged image beyond the centre of curvature
Object between C and F — real inverted enlarged image beyond C P F C object image Two rays from the object tip meet beyond C, forming a real, inverted, larger image there.
Derivation ADeriving the mirror formula from similar triangles

The goal is to show that for a real image formed by a concave mirror, the geometric relationship between the object distance uu, the image distance vv, and the focal length ff collapses to the single algebraic statement 1/v+1/u=1/f1/v + 1/u = 1/f, using only two pairs of similar triangles and the paraxial approximation that rays stay close to the principal axis.

Step 1. Set up the ray diagram for an object beyond CC, so a real, inverted image forms between FF and CC. Let the object ABAB stand on the axis with its foot at BB, and let the image A′B′A'B' form with its foot at B′B'. A ray from AA through the pole PP reflects symmetrically about the axis and passes through A′A', since the two angles it makes with the axis are equal by the law of reflection.

Step 2. Triangle ABPABP and triangle A′B′PA'B'P are similar, because both are right-angled at BB and B′B' and share the equal angle at PP from Step 1. Similar triangles give a ratio of corresponding sides: A′B′AB=PB′PB=vu\dfrac{A'B'}{AB} = \dfrac{PB'}{PB} = \dfrac{v}{u} in magnitude, which is exactly the magnitude of the magnification.

Step 3. Bring in a second pair of similar triangles using a ray from AA parallel to the axis, which reflects through the focus FF and crosses the axis at B′B'. Triangle A′B′FA'B'F and triangle formed by the mirror's pole-height and PFPF are similar (both right-angled, sharing the angle at FF), giving A′B′PN=FB′PF\dfrac{A'B'}{PN} = \dfrac{FB'}{PF}, and since PN≈ABPN \approx AB for rays close to the axis, this becomes A′B′AB=FB′PF=PB′−PFPF=v−ff\dfrac{A'B'}{AB} = \dfrac{FB'}{PF} = \dfrac{PB' - PF}{PF} = \dfrac{v - f}{f}.

Step 4. Both Step 2 and Step 3 equal the same ratio A′B′/ABA'B'/AB, so set them equal: vu=v−ff\dfrac{v}{u} = \dfrac{v - f}{f}. Cross-multiplying gives vf=uv−ufvf = uv - uf, then dividing every term by uvfuvf gives 1u=1f−1v\dfrac{1}{u} = \dfrac{1}{f} - \dfrac{1}{v}, which rearranges to the mirror formula, 1v+1u=1f\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}. Although this derivation used magnitudes, applying the New Cartesian Sign Convention (with uu and ff negative for this case) reproduces exactly the same signed formula used for every mirror and every object position.

Worked example

An object 2 cm tall is placed 20 cm in front of a concave mirror of focal length 15 cm — a position between CC (at 30 cm) and FF (at 15 cm), matching the qualitative case from the core idea. Find the image position, size, and nature.

Step 1. Assign signs. Since the object is in front of the mirror, u=−20 cmu = -20\ \text{cm}. Since the mirror is concave, f=−15 cmf = -15\ \text{cm}.

Step 2. Apply the mirror formula. 1v=1f−1u=1−15−1−20=−115+120=−4+360=−160\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = \dfrac{1}{-15} - \dfrac{1}{-20} = -\dfrac{1}{15} + \dfrac{1}{20} = \dfrac{-4 + 3}{60} = -\dfrac{1}{60}, so v=−60 cmv = -60\ \text{cm}.

Step 3. Interpret vv. Since vv is negative, the image forms 60 cm in front of the mirror — on the same side as the object, so it is real, and since 60 cm>30 cm=∣PC∣60\ \text{cm} > 30\ \text{cm} = |PC|, it lies beyond CC, exactly matching the standard table.

Step 4. Find the magnification. m=−vu=−−60−20=−3m = -\dfrac{v}{u} = -\dfrac{-60}{-20} = -3. The negative sign confirms the image is inverted, and ∣m∣=3|m| = 3 means it is three times enlarged, so the image height is h′=mh=−3×2 cm=−6 cmh' = mh = -3 \times 2\ \text{cm} = -6\ \text{cm} (6 cm tall, inverted).

Answer: the image forms 60 cm in front of the mirror, real, inverted, and 6 cm tall — three times the object's size.

Competitive stretch

For a convex mirror of the same focal length, f=+15 cmf = +15\ \text{cm}, and the same object at u=−20 cmu = -20\ \text{cm} gives 1v=115−1−20=115+120=760\dfrac{1}{v} = \dfrac{1}{15} - \dfrac{1}{-20} = \dfrac{1}{15}+\dfrac{1}{20} = \dfrac{7}{60}, so v≈+8.6 cmv \approx +8.6\ \text{cm} — positive, confirming the image is virtual, behind the mirror, and m=−v/u≈+0.43m = -v/u \approx +0.43, positive and less than 1, confirming erect and diminished, exactly as the "always virtual, erect, diminished" rule for convex mirrors demands, for every value of uu.

Common traps

  • Forgetting the sign convention entirely. Plugging in uu or ff as positive numbers gives a wrong or nonsensical answer; the object distance is always negative, and the focal length is negative only for a concave mirror.
  • Misreading the sign of vv. A negative vv means real (in front of the mirror); a positive vv means virtual (behind the mirror) — students frequently reverse this.
  • Assuming a concave mirror always gives a real image. Only true when the object is beyond FF; between PP and FF, vv comes out positive, meaning virtual and erect.
  • Confusing "inverted" with "diminished/enlarged." The sign of mm tells you orientation (inverted vs erect); the magnitude of mm tells you size (enlarged, same, or diminished) — these are two separate readings of one number.

Ask the tutor: If the mirror formula's sign convention keeps tripping you up, ask the AI tutor to walk through two or three quick u,f→v,mu, f \to v, m calculations with you, one for each mirror type.

Why is the focal length negative for a concave mirror but positive for a convex mirror?

The sign convention measures every distance from the pole, taking the direction of the incident light as positive. A concave mirror's focus lies in front of the mirror, on the same side the light is coming from, so it is measured against the incident light and gets a negative sign. A convex mirror's focus lies behind the mirror, in the same direction the light is travelling, so it gets a positive sign. This single rule is what lets one formula, 1/v+1/u=1/f1/v + 1/u = 1/f, work for both mirror types without switching equations.

Do I need to memorise the six-case table if I know the mirror formula?

The formula is more powerful because it handles any object distance, not just the six standard positions, but the table is still worth knowing by heart for speed in objective questions and to sanity-check a calculated answer instantly. In an exam, use the table to predict the answer's character (real or virtual, bigger or smaller) in seconds, then use the formula when an exact numerical position or magnification is asked for.

What happens to v in the mirror formula when the object is placed exactly at the focus?

At u=fu = f, the formula gives 1/v=1/f−1/f=01/v = 1/f - 1/f = 0, so vv is infinite — the image forms at infinity. Physically, rays from an object at the focus reflect off a concave mirror and emerge exactly parallel to the principal axis, never converging to a finite point, which is why a torch or headlight bulb is placed at the focus of its reflector to produce a parallel beam.

Before this

Where this leads