Plane Mirrors & Image Formation

The plane mirror is the most-used optical component on Earth, precisely because its behaviour is so exactly predictable. Every periscope, from a toy to a submarine's, is two plane mirrors at forty-five degrees relaying your line of sight over an obstacle without magnification. A dressing mirror is the half-height rule made retail. Dental and inspection mirrors are plane (or gently curved) mirrors reaching sight around corners. In a single-lens-reflex camera a plane mirror flips up to send the image alternately to your eye and to the sensor, and the pentaprism above it is a folded set of reflections that un-inverts the view. A kaleidoscope is equation (2) turned into a toy, and the mirrored walls of a laser cavity are plane mirrors bouncing light thousands of times to build up a beam. Even the humble rear-view mirror marked "objects are closer than they appear" is a reminder that its passenger-side twin is not plane but convex — the subject two articles from here — which is exactly why it warns you. Flat mirrors are simple, and their simplicity is why they are everywhere.

Light does not go behind a mirror — the silvering stops it dead. So how can there be an image back there? The answer is the whole idea of a virtual image, and it is worth slowing down for. Take a point on the object, say the tip of a candle flame, and let it spray rays in all directions. A cone of those rays strikes the mirror and reflects, each ray obeying $\theta_i = \theta_r$. The reflected rays are now diverging — spreading apart — exactly as if they had started from a single point behind the mirror. Your eye, catching that diverging cone, cannot tell the difference between light that truly came from a point and light that merely appears to; it traces the rays straight back and places the flame-tip at the phantom point. That phantom point is the image. To locate it, you need only two rays. Any two reflected rays from a single object point, extended backwards through the mirror, cross at exactly one place — the image of that point. Do it for the top of the object and the bottom, join them, and you have the whole image. The construction is the entire method, and it never changes for the rest of the chapter; only the mirror shape does.

Carry out the ray trace for any object and the same five properties come out every time. It is worth stating them as a block, because boards award a mark for each and students routinely give three.

where: using the sign convention that measures distances from the mirror with "in front" negative, the object distance $u$ and image distance $v$ are equal in size and opposite in sign — the image is as far behind as the object is in front — and the magnification $m$ (image height divided by object height) is exactly $+1$. In words, the five facts are: the image is (i) virtual — formed by the apparent, not actual, meeting of rays, so it cannot be projected onto a screen; (ii) erect — the right way up; (iii) the same size as the object; (iv) laterally inverted; and (v) located as far behind the mirror as the object is in front. A candle 15 cm in front of the glass has its image 15 cm behind it, so object and image are 30 cm apart, and the image flame is exactly as tall as the real one. The contrast that earns marks is virtual versus real. A real image is one where reflected (or refracted) rays actually cross and pile up, so you could hold a screen there and catch a bright picture — a concave mirror or a lens can make one, and the next articles are full of them. A plane mirror never can: its reflected rays always diverge, so its image is always virtual, always behind the glass, always uncatchable. Kepler was the first to draw this line cleanly, and it is the single most examined distinction in the whole optics chapter.

Hold up your right hand to a mirror and the image raises what looks like its left. Print holds a "b" up and it reads "d". This is lateral inversion, and almost every textbook explains it wrongly, saying "a mirror swaps left and right but not up and down". That cannot be the real rule — a mirror has no way of knowing which way is up, so why would it treat the horizontal and vertical directions differently? It does not. What the mirror actually reverses is the direction along the line of sight — front to back. Your nose, closest to the glass, has its image closest to the glass; the back of your head, furthest away, images furthest away — so the image is a version of you turned inside-out along the viewing axis, like a rubber mask pushed through itself. Your "left hand looks like the image's right" only because you mentally rotate yourself to face the way the image faces; the mirror never touched left or right at all. The clean test: raise the hand nearest a window, and your image raises the hand nearest the same window — no left-right swap has occurred until you turn around to compare. State it as "front-to-back reversal along the line of sight" and you will never be caught by the exam question that asks why text reads backwards.

Stand an object between two mirrors hinged at an angle and each mirror makes an image, then makes an image of the other mirror's image, and so on around the wedge — the effect a kaleidoscope turns into a flower of reflections. When the mirrors close to an angle $\theta$ such that $360/\theta$ is a whole number, the images arrange themselves evenly around a circle and their count is exactly:

where: $\theta$ is the angle between the mirrors and $N$ the number of images. Read it as: the object plus its $N$ images make $360/\theta$ equally spaced points around the hinge, and you subtract one because one of those points is the object itself. So $\theta = 90^\circ$ gives $360/90 - 1 = 3$ images; $\theta = 60^\circ$ gives $5$; $\theta = 45^\circ$ gives $7$; and $\theta = 30^\circ$ gives $11$. Push the mirrors parallel ($\theta \to 0^\circ$) and $360/\theta \to \infty$: two facing mirrors make an endless corridor of images, each dimmer than the last as a little light is lost at every bounce.

There is a caveat the sharpest exam questions turn on. Equation (2) is exact only when $360/\theta$ is an even whole number (like $4, 6, 8$ for $\theta = 90^\circ, 60^\circ, 45^\circ$). When $360/\theta$ is an odd whole number, the count depends on where the object sits: on the bisector of the angle you get $360/\theta - 1$ images, but off the bisector you get one more, $360/\theta$. And when $360/\theta$ is not a whole number at all, the working rule is to take the whole-number part of $360/\theta$. A JEE multiple-choice item that quotes an angle like $50^\circ$ or an off-centre object is testing exactly this; the safe move is to state which case you are in before you write the number.

Here is the plane-mirror result that most surprises people: to see yourself from the top of your head to the soles of your feet, the mirror on the wall need only be half your height, and it makes not the slightest difference how close or far you stand.

where: $H$ is your full height and $L_{\min}$ the minimum vertical length of a plane mirror that lets you see your whole self. For a person of height $H = 1.72\ \mathrm{m}$, that is $L_{\min} = 0.86\ \mathrm{m}$ — a mirror well under a metre tall, hung at the right height, is enough for anyone that size. The reason distance drops out is that stepping back shrinks the angles your head and feet subtend at the mirror in exactly the proportion needed to keep the required mirror length fixed; the derivation below makes that precise. This is why the tall wall mirrors in clothing shops are pure showmanship — half the glass would do the same job.

Board level — image position and counting images (CBSE and ICSE pattern)

(a) An object is placed 15 cm in front of a plane mirror. State the position, nature and size of the image, and the distance between object and image. (b) The same object is placed between two plane mirrors inclined at 60 degrees. How many images are formed?

Step 1. For part (a), apply the plane-mirror symmetry: the image is as far behind the mirror as the object is in front, so it is $\boxed{15\ \mathrm{cm}}$ behind the mirror. The object-to-image distance is therefore $15 + 15 = \boxed{30\ \mathrm{cm}}$.

Step 2. State the nature fully — boards mark each property: the image is virtual (behind the mirror, cannot be caught on a screen), erect, the same size as the object (magnification $+1$), and laterally inverted. Giving only "virtual" throws away easy marks.

Step 3. For part (b), use the inclined-mirror count with $\theta = 60^\circ$: since $360/60 = 6$ is an even whole number, $N = 360^\circ/60^\circ - 1 = 6 - 1 = \boxed{5}$ images, all virtual. Sense-check with the kaleidoscope picture: object plus five images make six points evenly spaced by $60^\circ$ around the hinge — a full circle.

Competitive level — the minimum wall mirror (JEE and NEET pattern)

A person of height 1.72 m, with eyes 1.60 m above the floor, wants to buy the shortest possible plane wall mirror in which they can see themselves from head to toe. Find the minimum length of the mirror and the height of its lower edge above the floor. Does stepping closer let them use a shorter mirror?

Step 1. Locate the top edge. The ray from the top of the head (height $1.72\ \mathrm{m}$) to the eye (height $1.60\ \mathrm{m}$) reflects at the average height, $(1.72 + 1.60)/2 = 1.66\ \mathrm{m}$. That is where the mirror's top edge must be.

Step 2. Locate the bottom edge. The ray from the feet (height $0$) to the eye reflects at $(0 + 1.60)/2 = 0.80\ \mathrm{m}$. That is the mirror's lower edge, so its lower edge sits $\boxed{0.80\ \mathrm{m}}$ above the floor.

Step 3. The length is the difference: $L_{\min} = 1.66 - 0.80 = \boxed{0.86\ \mathrm{m}}$, which is exactly half the height, $1.72/2 = 0.86\ \mathrm{m}$.

Step 4. Does distance matter? No. Neither edge calculation used the distance to the mirror — both reflection points depend only on heights — so stepping closer or further changes nothing. The answer is $0.86\ \mathrm{m}$ from any distance, and a shorter mirror can never show the whole person. Timing note: about 75 seconds. The whole question is the two averages, $(H+E)/2$ and $E/2$; write them down first and the arithmetic is trivial. The trap is to waste time hunting for the distance to the mirror — it is deliberately not needed.

Calling the image real. Covered in the warning, and it is the single most common error on plane mirrors across CBSE, ICSE, IGCSE and A-Level. Reflected rays from a flat mirror always diverge, so the image is always virtual and behind the glass. If you ever write "real" for a plane mirror, it is wrong.

Thinking the minimum mirror length depends on distance. It does not — $L_{\min} = H/2$ at every distance, because the eye height and the distance both cancel in the geometry. A favourite JEE distractor offers a length that grows or shrinks with distance; it is always wrong. Only the position of the mirror on the wall matters, not its length.

Ignoring the even/odd caveat in image counting. $N = 360/\theta - 1$ holds cleanly only when $360/\theta$ is an even whole number. When it is odd, an object off the bisector gives one more image; when $360/\theta$ is not whole, take the integer part. State which case applies before quoting a number.

Explaining lateral inversion as a genuine left-right flip. The mirror reverses front-to-back along the line of sight, not left-to-right. The "left-right" appearance is your own mental rotation. Boards that ask why mirror writing is reversed want the front-to-back answer.

Board scope. Plane mirrors and image formation are on CBSE, ICSE, JEE Main, NEET, IGCSE and A-Level — this article's boards. Note that IB (both SL and HL) dropped geometric optics from its guide, so if you are an IB candidate, plane and curved mirrors are enrichment for you, not exam material. Everyone else needs the five properties, the inclined-mirror count and the half-height rule cold.

Texts. NCERT Science Class 10, Ch. 10 ("Light — Reflection and Refraction"), plane mirrors and image characteristics · NCERT Physics Class 12, Ch. 9 ("Ray Optics"), reflection by plane surfaces · H. C. Verma, Concepts of Physics Vol. 1, "Geometrical Optics" · Cambridge IGCSE 0625 and A-Level 9702, "Reflection and plane mirrors".

Provenance. Standard results. Recompute: object 15 cm in front → image 15 cm behind, separation $15 + 15 = 30$ cm, magnification $+1$; inclined-mirror counts $360/90 - 1 = 3$, $360/60 - 1 = 5$, $360/45 - 1 = 7$, $360/30 - 1 = 11$, and the odd case $360/72 = 5$; the minimum-mirror length $1.72/2 = 0.86$ m with edges $(1.72 + 1.60)/2 = 1.66$ m and $1.60/2 = 0.80$ m; the image-approach speed $2 \times 1 = 2$ m/s.

Prerequisites. The Ray Model of Light and Laws of Reflection — this article applies its single law, $\theta_i = \theta_r$, to a flat surface. Leads to. Spherical Mirrors: Concave and Convex, where the surface curves and the image can at last become real, magnified or diminished; then The Mirror Formula and Sign Convention, which turns the ray trace into an equation.

Go deeper (UG). The wave-optics account of why a mirror image is perfectly sharp, and how the phase of the reflected wave behaves, is developed in the university optics track. It is enrichment only; every result on this page follows from ray tracing alone, and no exam here will ask for the wave treatment.