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Module 01 · Part II · JEE Advanced

Spherical mirrors — concave & convex

Light striking a smooth surface obeys two simple laws — yet from them fall out every image a plane, concave or convex mirror can make. We derive the mirror equation from scratch, then let you drag the object through every case on a live bench.

The two laws of reflection

1 · Equal angles

The angle of incidence equals the angle of reflection, both measured from the normal at the point of incidence: ∠i = ∠r.

2 · Coplanar

The incident ray, the reflected ray and the normal all lie in one plane. On a curved mirror the normal is simply the line drawn to the centre of curvature C.

For a plane mirror these two laws already fix the image: rays from a point source, on reflection, appear to diverge from a single point exactly as far behind the mirror as the source is in front. That image is virtual, erect and the same size — switch the bench below to plane to see the construction.

Anatomy of a spherical mirror

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The coordinate sign convention

Every formula on this page is written in one convention. Get these four rules right and the algebra takes care of real vs. virtual by itself.

◦ The pole P is the origin; the principal axis is the x-axis.
◦ The +x direction is the direction the incident light travels.
◦ Distances measured against the incident light are negative (a real object gives u < 0).
◦ Heights above the axis are positive, below are negative. So f is for concave, + for convex.

Explore the bench

Drag to orbit in 3D, or flip to the 2D ray diagram. Slide the object, switch mirrors, and hit a preset to jump between cases. Arrows nudge · P presenter · C clean-view · R reset (while hovering the bench).

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u = {{ uTxt }}
v = {{ vTxt }}
f = {{ fTxt }}
m = {{ mTxt }}
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v = −u{{ vDen }} = −( {{ uDen }} )
1v+ 1u= 1f 1{{ vDen }}+ 1{{ uDen }}= 1{{ fDen }}
object → {{ farLabel }}

Deriving the mirror equation

Take a paraxial ray from an on-axis object O striking a concave mirror at A, and a second ray straight down the axis. Let α, β, γ be the angles AOP, ACP, AIP. The exterior-angle theorem plus the small-angle approximation does the rest.

1v+ 1u= 2R = 1f
Step 1. CA is the normal at A, so ∠OAC = ∠CAI. Using the exterior angle of a triangle: from △OAC,  β = α + θ   and from △OAI,  γ = α + 2θ.
Step 2. Eliminate θ between them:  2β = α + γ.
Step 3. For paraxial rays A is close to P, so the angles are small: α ≈ AP/PO, β ≈ AP/PC, γ ≈ AP/PI. Substituting and cancelling AP:
2PC= 1PO+ 1PI
Step 4. Apply the sign convention. O, I, C sit on the −x side, so PO = −u, PI = −v, PC = −R. The minus signs cancel throughout, leaving the mirror equation — and it now holds in every case, real or virtual.

Focal length: f = R/2

Send the object to infinity: the incident rays are parallel and the image lands at the focus, so u → ∞ and v = f. The mirror equation gives 1/f = 2/R, i.e. the focus sits exactly midway between pole and centre of curvature.

Magnification: m = −v/u

The chief ray to the pole reflects with ∠BPA = ∠APB′, making triangles ABP and A′B′P similar. With signs, h₂/h₁ = −v/u.

Similar triangles give A′B′/AB = PA′/PA. Substituting A′B′ = h₂, AB = h₁, PA′ = −v, PA = −u yields m = h₂/h₁ = −v/u. A negative m means an inverted image; |m| > 1 means enlarged.

Concave mirror — the six positions

Tap any row to send the bench above to that case.

Object position
Image position
Nature
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{{ r.img }}
{{ r.nature }}
try it →

A convex mirror is simpler: for a real object it always makes a virtual, erect, diminished image between P and F. A plane mirror is the R → ∞ limit — virtual, erect, same size, equidistant behind.

Going deeper — beyond the six cases

The mirror equation holds more than static pictures. Differentiate it and it tells you how fast images move; square the magnification and it tells you how a solid object is stretched along the axis. These are the results JEE Advanced leans on.

Velocity of the image

Differentiate 1/v + 1/u = 1/f with f fixed: −dv/v² − du/u² = 0. Along the axis this gives

vimage = − m² · vobject

Motion across the axis scales only by m: v⊥,image = m · v⊥,object. Near the focus m is huge, so a slow object throws a blazing-fast image — the optical-lever trick.

Longitudinal magnification

Because dv/du = −m², a short rod lying along the axis images to a rod of length

Limage = m² · Lobject

Transverse scale m, axial scale m² — so a small sphere near a concave mirror images to a squashed egg, reversed front-to-back.

Uses — and the flaw

Concave: shaving & makeup mirrors (object inside F → magnified, erect), headlamps and torches (source at F → parallel beam), reflecting telescopes. Convex: vehicle and shop mirrors (wide, erect field of view).

The flaw: a real sphere shows spherical aberration — rim rays focus nearer the pole than paraxial rays. A parabolic mirror removes it, which is why telescope and torch mirrors are parabolas.

Worked examples

Example 1 — concave

A concave mirror has focal length 15 cm. An object stands 25 cm in front of it. Find the image distance, magnification and nature.

Signs: u = −25 cm, f = −15 cm. From 1/v = 1/f − 1/u = −1/15 + 1/25 = −2/75, so v = −37.5 cm (in front → real).

m = −v/u = −(−37.5)/(−25) = −1.5.

The image is real, inverted and 1.5× enlarged, 37.5 cm in front of the mirror. (This is the C→F case — try it on the bench.)

Example 2 — convex

A convex mirror of radius of curvature 30 cm has an object 10 cm in front. Locate the image and give its nature.

Convex: f = +R/2 = +15 cm, u = −10 cm. 1/v = 1/f − 1/u = 1/15 + 1/10 = 1/6, so v = +6 cm (behind → virtual).

m = −v/u = −6/(−10) = +0.6.

Image is virtual, erect and diminished, 6 cm behind the mirror — as always for a convex mirror.

Practice arena

A warm-up on the basics, then a JEE Advanced bank that pushes into image velocity, aberration and the tricky sign cases. Keep the sign convention in hand.

Warm-up
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Exam bank 10 single-correct · JEE Advanced grade · {{ examSolved }}/10 correct
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Recap card

1v+ 1u= 1f
m=vu
f=R2
concave f < 0 · convex f > 0
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