In short

The two laws of reflection. When a ray of light strikes a surface:

  1. The angle of incidence i equals the angle of reflection r, both measured from the normal (the line perpendicular to the surface at the point of incidence).
  2. The incident ray, the reflected ray, and the normal all lie in the same plane.
\boxed{\; i = r \;}

For a plane mirror. The image of a point object is as far behind the mirror as the object is in front of it. The image is virtual (light only appears to come from it — no light actually reaches it), erect (right way up), same size as the object, and laterally inverted (left becomes right).

Multiple images from two mirrors. Two plane mirrors hinged at an angle \theta produce

\boxed{\; N \;=\; \frac{360^\circ}{\theta} - 1 \;}

images — provided 360^\circ/\theta is an integer. Set \theta = 60^\circ and you see 5 images; set \theta = 0^\circ (parallel mirrors) and the number blows up — that is the barber's-shop infinity.

Real vs virtual images. A real image forms where reflected rays actually meet — you can catch it on a screen. A virtual image forms where reflected rays appear to diverge from, behind the mirror — no screen catches it. Plane mirrors always give virtual images; concave mirrors can give real or virtual depending on the object position.

Stand on the edge of the Yamuna at dawn and the Taj Mahal doubles. The white marble dome rises above the water; a mirror-image dome hangs below it, slightly rippled. If the river were perfectly still, you could not tell the real dome from the reflected one by geometry alone — they would be reflections of each other through the water's surface, equally tall, equally sharp. What keeps them separable is that reflections carry no actual light from where they appear to be: if you dived into the Yamuna and tried to touch the reflected dome, your hand would go through empty water. The image is virtual — it exists in the geometry, not in the world.

Every reflection you have ever seen — your face in a shaving mirror at 6 a.m., the headlights of a truck bouncing off a wet Mumbai road, the orange of a sunset repeated in the waters of the Hooghly in Kolkata, the fifteen copies of yourself in the barber's two facing mirrors — obeys a single rule. Light hits a surface and bounces off at the same angle it arrived, measured from the line perpendicular to that surface. That one sentence is the whole of reflection. Everything else in this chapter — the geometry of images in plane mirrors, the counting of multiple reflections, the distinction between real and virtual images, the preview of what curved mirrors will do — is consequences.

The two laws of reflection

Place a torch near a polished steel plate in a darkened room. Aim the beam obliquely at the plate. The beam bounces off and strikes the wall. Now rotate the plate a little: the reflected spot on the wall moves twice as fast as the plate did. That factor of two is a clue — and it is the first fact about reflection that any careful observer notices.

The full rule is cleaner than it looks. Draw the normal at the point where the ray hits — the line perpendicular to the surface. Measure both the incoming and the outgoing ray from this normal, not from the surface itself. Call these angles the angle of incidence i and the angle of reflection r.

Laws of reflection

First law. The angle of incidence equals the angle of reflection:

i = r

Both measured from the normal to the surface at the point of incidence.

Second law. The incident ray, the reflected ray, and the normal all lie in the same plane (the plane of incidence).

The law of reflection: incident ray, normal, reflected ray A horizontal mirror surface. A normal (dashed) rises vertically from the point of incidence. An incident ray comes down from the upper left at angle i from the normal. A reflected ray departs to the upper right at equal angle r from the normal. mirror normal incident reflected i r point of incidence i = r
The normal is the reference line — never measure angles from the mirror surface itself. When light hits at $40°$ from the normal, it leaves at $40°$ from the normal, on the other side.

Why the rotating-mirror trick doubles the angle

Go back to the rotating polished plate. Call the original angle of incidence i. You rotate the mirror by \alpha. What happens to the reflected beam?

Step 1. Before the rotation, the normal points in some direction; the reflected ray makes angle i on the opposite side.

Step 2. Rotate the mirror by \alpha. The normal, which is perpendicular to the mirror, also rotates by \alpha. The incident ray direction has not changed — it still comes from the same torch. So the new angle of incidence is i + \alpha (measured from the new normal).

Why: when you tilt the mirror by \alpha, the angle between the incoming ray and the normal opens up by exactly \alpha. The ray did not move; the reference line did.

Step 3. By the law of reflection, the new angle of reflection is also i + \alpha — but on the other side of the new normal, which is itself tilted by \alpha.

Step 4. Measure the new reflected ray against the old normal direction. It makes angle (i + \alpha) + \alpha = i + 2\alpha.

Why: the reflected ray is at angle i + \alpha from the new normal, and the new normal is at \alpha from the old normal, on the same side. Angles add.

Step 5. The reflected ray has moved by (i + 2\alpha) - i = 2\alpha.

So when the mirror turns by \alpha, the reflected ray turns by 2\alpha. This is the principle behind the mirror galvanometer — the sensitive current detector used in early physics labs, where a tiny coil twist was amplified into a large spot motion on a distant scale.

Regular vs diffuse reflection

A mirror reflects a clean beam. A sheet of paper does not — hit it with a torch and the beam disappears. Both surfaces obey the same law of reflection at every point. The difference is that the paper is, on a microscopic scale, a jumble of tiny flat facets facing every direction. Each facet reflects cleanly, but the facets reflect into random directions, and the result is a diffuse scatter that lights up the whole room.

Regular (specular) reflection: smooth surface, all normals point the same way, a parallel beam stays a parallel beam. Mirrors, still water, polished metal.

Diffuse reflection: rough surface on the wavelength scale, normals point every which way, a parallel beam becomes a halo. Paper, cloth, unpolished stone, whitewashed walls.

This is why you can read a book illuminated by a bulb but not by a mirror: the book needs diffuse reflection to send light from every part of the page to your eye, and a mirror would only reflect the bulb's own image into one specific direction.

Image formation by a plane mirror

A plane mirror is a flat reflective surface. Stand a lit candle in front of one and there seems to be another candle behind the glass. What is going on geometrically?

Image formation in a plane mirror A point object O at distance d in front of a vertical mirror. Two rays from O strike the mirror and reflect. Extended backwards, the reflected rays meet at point I, distance d behind the mirror. mirror O object I image d d eye
Two rays from the object $O$ strike the mirror and reflect toward the eye. Traced backwards (dashed), the reflected rays appear to come from a point $I$ the same distance behind the mirror as $O$ is in front. No light actually reaches $I$ — which is what makes the image virtual.

Proving the image is equidistant

Pick any point object O at distance d from a plane mirror. Take any ray from O that strikes the mirror at point P. The normal at P is perpendicular to the mirror. Call the angle of incidence i. The reflected ray leaves P at angle i on the other side of the normal.

Extend the reflected ray backwards, through the mirror, into the geometry behind. Do this for a second ray from O striking a different point Q. The two backward-extended reflected rays meet at some point I. The claim is: I is on the perpendicular from O to the mirror, the same distance d behind it.

Step 1. Drop a perpendicular from O to the mirror. Call the foot M. The segment OM has length d.

Step 2. Take a general ray OP with angle of incidence i. The normal at P is perpendicular to the mirror, so it is parallel to OM. The triangle OPM is a right triangle at M, with angle \angle OPM = 90° - i (because the angle between OP and the mirror is 90° - i when the angle from the normal is i).

Why: OP makes angle i with the normal and the normal is perpendicular to the mirror, so OP makes angle 90° - i with the mirror. In triangle OPM, the angle at P between the mirror (the side PM) and OP is exactly this 90° - i.

Step 3. Now reflect. The reflected ray leaves P at angle i on the other side of the normal, which means at angle 90° - i with the mirror, on the other side of PM. Extend this reflected ray backwards — it continues behind the mirror at angle 90° - i on the far side.

Step 4. The backward extension, the perpendicular PM, and the line through M perpendicular to the mirror define a right triangle congruent to OPM. Call the point where the backward ray meets the perpendicular line from M the image point I.

Step 5. In the triangle IPM (on the mirror's back side), \angle IPM = 90° - i and \angle IMP = 90°. This is the same triangle as OPM, just reflected through the mirror. So IM = OM = d.

Why: the two right triangles OPM and IPM share the side PM and have equal angles at P (both 90° - i) and at M (both 90°). By ASA, they are congruent. Therefore IM = OM.

Step 6. This argument worked for any angle of incidence i. Pick a second ray from O — say, OQ — and the same reasoning puts its backward extension through the same point I at distance d. So all reflected rays, traced backwards, converge to I.

Result. The image I sits on the perpendicular from O to the mirror, at distance d behind it.

Properties of the plane-mirror image

From the geometry just proved, five properties follow:

  1. Virtual. No light actually reaches I. The reflected rays only appear to diverge from I; they really leave the mirror and travel forward. You could not catch the image on a screen placed at I — the screen would have no light to register.

  2. Erect. Stand upright in front of the mirror and the image stands upright too. The top of your head maps to the top of the image; your feet map to the feet.

  3. Same size. Every point of an extended object maps to a point at the same perpendicular distance behind the mirror. The geometry is a rigid reflection, which preserves all distances. Your image in a full-length mirror is exactly your height.

  4. Laterally inverted. Your right hand appears as the image's left hand. This is not a "the mirror flips left and right" mystery — it is pure geometry. When you look at yourself, you are facing the mirror, and the image is facing back at you. Your right hand, on your right side, is on the image's left as you view the image — because the image has turned around to face you. The mirror does not flip left and right; it flips front and back, and your brain interprets the flipped-through image as someone who turned around, which swaps their left and right from your point of view.

  5. Distance to image equals distance to object. Walk toward the mirror at 1 m/s and your image walks toward you at 1 m/s — from your frame, it closes at 2 m/s. This is why a mirror on your wardrobe seems to double the depth of the room.

The minimum size of a full-length mirror

Here is a consequence that the geometry settles cleanly. You want a mirror in which you can see yourself from head to toe. How tall does the mirror need to be?

Let your height be H. Let E be the position of your eyes. Let the distance from the top of your head to your eyes be a, and from your eyes to your feet be b, so a + b = H.

Step 1. A ray from the top of your head reaches your eyes after bouncing off the mirror. The law of reflection forces the mirror point to be exactly halfway (in height) between the head-top and the eyes — because the normal at that point is horizontal, and the angles above and below it must match.

Why: the head-top is at height a above the eyes. For the reflected ray from head-top to arrive at the eyes, the strike point on the mirror must be at height a/2 above the eyes — then the incoming ray drops a/2 and the outgoing ray also drops a/2, putting the ray at eye level.

Step 2. Similarly, a ray from your feet reaches your eyes after striking the mirror at height b/2 below the eyes.

Step 3. So the mirror must extend from a/2 above your eye level to b/2 below. Its minimum height is

h_{\min} = \frac{a}{2} + \frac{b}{2} = \frac{a+b}{2} = \frac{H}{2}

Result. You only need a mirror half your height to see yourself from head to toe. And — surprisingly — this is independent of how far you stand from the mirror. Walk closer or farther; a mirror half your height still works, as long as its top edge is at the right height (at H/2 plus half the distance from eyes to feet from the floor).

This fact is why mirror designers at showrooms like Hindware Lacasa can put a 90 cm mirror on a bathroom door and let a 180 cm customer see the whole of themselves. Half the height, full the view.

Multiple reflections — two mirrors at an angle

Stand between two parallel mirrors in a barber's shop and you see an infinite row of selves, each smaller than the last (because each round-trip through the mirrors loses a little light and perspective compresses). Tilt the mirrors away from parallel and the row closes into a ring of discrete images. How many?

The formula

Two plane mirrors are hinged at an angle \theta. An object is placed between them. The number of images formed is:

\boxed{\; N = \frac{360°}{\theta} - 1 \;}

provided 360°/\theta is an even integer, or an integer with special conditions described below.

Multiple images from two mirrors at 60 degreesTwo mirrors meet at vertex V at 60 degrees. An object between them produces 5 images arranged around V on a hexagon. M₁ M₂ 60° O I₂ I₄ I₅ I₃ I₁ Two mirrors at 60° → 5 images (hexagonal pattern)
With two mirrors hinged at $\theta = 60°$, the formula gives $N = 360°/60° - 1 = 5$ images. Together with the original object, six equally spaced points lie on a circle around the hinge — a regular hexagon. This is the optics behind the kaleidoscope, a toy whose cardboard tube typically houses three mirrors at $60°$ each.

Why that formula

The proof is a tidy piece of reasoning about reflections as rotations.

Step 1. Two successive reflections in two lines that meet at an angle \theta is the same as a rotation by angle 2\theta about the intersection point.

Why: a reflection in line \ell_1 followed by a reflection in line \ell_2 (meeting \ell_1 at angle \theta at point V) sends any point P to a new point P'' such that \angle PVP'' = 2\theta and |VP''| = |VP|. This is a standard result in plane isometries — compose two reflections about intersecting lines, get a rotation by twice the angle between them.

Step 2. Start with the object O at some position. Reflect it in mirror M_1 to get image I_1. Reflect I_1 in mirror M_2 to get I_2. Then reflect I_2 in M_1 to get I_3, and so on. Each pair of successive reflections rotates by 2\theta. So the objects-and-images form a ring around the hinge point V, spaced by 2\theta per pair.

Step 3. How many distinct images fit in a full 360°? The answer is 360°/\theta total positions (object plus images), because each reflection step advances the position by \theta around the circle — a reflection in one mirror, then the other, then the first, and so on, all rotate through multiples of \theta when viewed as points on the circle.

Step 4. Subtract 1 for the object itself: N = 360°/\theta - 1 images.

Step 5. When 360°/\theta is an odd integer. An odd count means one of the images sits exactly opposite O along the hinge bisector, where two reflection chains (object → M_1 → … and object → M_2 → …) land at the same point. That shared image gets counted once, not twice, so the formula still gives the right total.

Step 6. When 360°/\theta is not an integer. If \theta = 50°, then 360°/50° = 7.2. Take \lfloor 7.2 \rfloor = 7. One common convention gives N = 7 images when 360°/\theta is not a whole number — the reflections do not close into a clean ring, and the last image falls inside one of the mirrors rather than completing the cycle.

For JEE-standard problems, the formula N = 360°/\theta - 1 applied when the division is an even integer is the safe default. Examples:

\theta 360°/\theta N (images)
30° 12 11
45° 8 7
60° 6 5
72° 5 4
90° 4 3
120° 3 2
180° 2 1

The last row is just a single mirror (\theta = 180° means the mirrors have unfolded into one plane), giving one image, as expected.

Parallel mirrors — why infinity, qualitatively

Set \theta \to 0° and the formula blows up: N \to \infty. Two parallel mirrors do indeed, in principle, produce an infinite sequence of images. Each round trip of a ray between them creates another image, each one further behind the next mirror by 2d (where d is the mirror separation).

In practice, every reflection loses a few percent of the light to absorption in the silvering. After 10 bounces only (0.95)^{10} \approx 60\% of the original light remains; after 50 bounces it is 8\%; after 100 bounces it is well under 1\%. The barber's-shop infinity fades to darkness after a dozen or so visible reflections. But the geometry is genuinely infinite — only the photons run out.

Real vs virtual images — the distinction

A real image is formed where reflected (or refracted) rays actually converge and cross. If you place a screen at that location, a bright, focused spot appears on it. A concave mirror focusing sunlight to a point on a sheet of paper — that point is a real image of the Sun. A cinema projector's beam converging onto the screen in a PVR auditorium — that is a real image of the film frame.

A virtual image is formed where rays only appear to converge, after you extend them backwards. Place a screen there and nothing happens — no light has reached the screen, it only seems to originate from that point when you extend the rays along straight lines. Every image in a plane mirror is virtual.

The distinction is important because:

For plane mirrors, the answer is always the same: virtual, erect, same size, laterally inverted. For curved mirrors, the answer depends on where you put the object — which is the subject of the next article, Spherical Mirrors.

Preview: curved mirrors in one sentence each

The full geometry of curved mirrors, the mirror formula 1/v + 1/u = 1/f, and the magnification formula m = -v/u are all developed in Spherical Mirrors.

Worked examples

Example 1: A ray bouncing off a wall mirror

A ray of light travels along the direction making an angle of 35° with a vertical wall and strikes a plane mirror fixed flush against the wall. Find the angle the reflected ray makes with the wall, and the total angle between the incident and reflected rays.

Ray striking a vertical wall mirror at 35 degrees from the wall A vertical wall-mirror. An incident ray arrives from the upper left at 35 degrees from the vertical wall (i.e. 55 degrees from the normal). The reflected ray leaves at the same angle on the other side. wall mirror normal incident reflected 55° 55° 35°
The ray arrives at $35°$ from the wall, which is $55°$ from the normal. It leaves at $55°$ from the normal on the other side, i.e. $35°$ from the wall again — but now heading back outward.

Step 1. Identify the angle from the normal. The wall is the mirror surface; the normal is perpendicular to the wall (horizontal in the figure). The ray makes 35° with the wall, so it makes 90° - 35° = 55° with the normal.

i = 55°

Why: always measure angles from the normal, not from the surface. A ray "grazing" the surface has a small angle from the surface but a large angle from the normal.

Step 2. Apply the law of reflection.

r = i = 55°

The reflected ray leaves the mirror at 55° from the normal, on the opposite side of the normal from the incident ray.

Step 3. Convert back to angle from the wall.

\text{angle from wall} = 90° - 55° = 35°

Step 4. Total angle between the incident and reflected rays. They lie on opposite sides of the normal, each at 55° from it.

\text{total angle} = i + r = 55° + 55° = 110°

Why: the angle between the two rays (measured on the far side of the mirror from the normal) is the sum of the individual angles from the normal, because those angles are adjacent and open in opposite directions.

Result. The reflected ray makes 35° with the wall, and the total angle between the incident and reflected rays is 110°.

What this shows. Always translate to the normal before applying the law. Angles quoted from the surface must be subtracted from 90° first. This pattern — surface angle → normal angle → apply law → normal angle → back to surface — is the standard workflow for every reflection problem.

Example 2: Two mirrors at 45 degrees and a torch flash

In a physics lab at IIT Madras, two plane mirrors are hinged at an angle of 45°. A small LED is placed on the bisector of the angle between them. How many images does a student see? Sketch the symmetric pattern and verify with the formula.

Two mirrors at 45 degrees producing 7 imagesTwo mirrors meet at vertex V at 45 degrees. An object on the bisector. 7 images and the object lie at 8 equally spaced points around V. M₁ M₂ 45° O I₅ I₂ I₃ I₄ I₆ I₇ I₁ bisector
A $45°$ hinge gives $360°/45° - 1 = 7$ images. The object $O$ (solid red) and its 7 images $I_1, \ldots, I_7$ (faded red) lie on a circle around the hinge, equally spaced by $45°$. This is the optical pattern inside a kaleidoscope with two mirrors at $45°$.

Step 1. Apply the formula.

N = \frac{360°}{\theta} - 1 = \frac{360°}{45°} - 1 = 8 - 1 = 7

Why: the formula counts the number of images other than the original object. The total number of bright points — object plus images — is 360°/\theta = 8, equally spaced around the hinge.

Step 2. Verify by constructing the images one at a time.

  • Start with object O on the bisector at 22.5° from each mirror.
  • Reflect O in mirror M_1 → I_1 at -22.5° (below M_1, so outside the physical region but the geometry still tracks).
  • Reflect I_1 in M_2 → I_2.
  • Continue alternating reflections until the chain closes.
  • A second chain starts with reflecting O in M_2 → I_1', then in M_1, and so on.
  • The two chains close after enough steps such that the total number of distinct points around the hinge is 8, giving 7 images plus the original.

Step 3. Check for the "shared image" case. For \theta = 45°, 360°/\theta = 8 is even, so no image is shared between the two chains — every image is distinct.

Why: a shared image appears only when 360°/\theta is odd, because then the two reflection chains both reach the single image exactly opposite the hinge bisector. For even 360°/\theta, the chains meet at a point that is itself a reflection pair, not a single shared image.

Step 4. Arrange the 8 points (object plus 7 images) on a circle centred at the hinge. They are equally spaced at 45° each. The kaleidoscope you held as a child was showing exactly this pattern — a rotationally symmetric polygon of images, growing into a snowflake as you twisted the tube.

Result. The student sees \mathbf{7} images. Together with the original LED, 8 bright points form a perfect octagon around the hinge.

What this shows. The formula N = 360°/\theta - 1 reduces a tricky geometry problem to a single division. The assumption behind it — that 360°/\theta is a whole number — is why only certain hinge angles give clean image counts; \theta = 30°, 40°, 45°, 60°, 72°, 90° all work nicely. Intermediate angles like 50° give partial chains that do not close cleanly.

Example 3: The minimum wardrobe mirror

Ananya is 165 cm tall. Her eyes are 10 cm below the top of her head. She wants the smallest possible mirror on her wardrobe door that still lets her see herself from the crown of her head to the soles of her feet. Find the height of the mirror and where its top edge should be placed.

Minimum mirror height to see full body A person of height 165 cm with eyes 10 cm below the crown. A mirror on a wall to the right. Rays from head and feet bounce off the mirror to the eyes. The mirror extends from 5 cm above eye level to 77.5 cm below eye level. head (crown) eye mirror (82.5 cm) floor ray from head ray from feet
The mirror needs to cover only from halfway-up-the-head (at $5$ cm above eye level) to halfway-down-to-the-floor ($77.5$ cm below eye level). Total: $82.5$ cm — exactly half of Ananya's $165$ cm height.

Step 1. Identify the geometry. Ananya's total height H = 165 cm. Crown to eyes: a = 10 cm. Eyes to feet: b = H - a = 165 - 10 = 155 cm.

Step 2. Find the mirror-strike point for a ray from the crown to the eyes.

The ray leaves the crown (which is 10 cm above the eyes) and must arrive at the eyes after one reflection. By the law of reflection, the strike point is at height a/2 = 5 cm above eye level.

Why: the mirror is vertical, so its normal is horizontal. For the reflected ray to reach the eyes at their height, the ray from the crown (at +a above eye level) must strike the mirror at height +a/2 — the midpoint. Then both the incoming and outgoing ray descend by a/2 in height.

Step 3. Find the mirror-strike point for a ray from the feet.

The feet are b = 155 cm below the eyes. By the same argument, the strike point is at height -b/2 = -77.5 cm (i.e., 77.5 cm below eye level).

Step 4. Compute the total mirror height.

h_{\min} = \frac{a}{2} + \frac{b}{2} = \frac{a+b}{2} = \frac{H}{2} = \frac{165}{2} = 82.5 \text{ cm}

Step 5. Find where the top edge of the mirror should sit above the floor.

Ananya's eyes are at b = 155 cm above the floor. The top of the mirror is a/2 = 5 cm above her eyes:

\text{top of mirror} = 155 + 5 = 160 \text{ cm above the floor}

Why: the top edge only needs to be as high as the ray from the crown strikes. Anything above that is wasted glass.

Result. A mirror 82.5 cm tall, with its top edge 160 cm above the floor, is the minimum needed for Ananya to see herself head-to-toe. The bottom edge sits at 160 - 82.5 = 77.5 cm above the floor — nowhere near the ground. The visible region below the mirror is taken care of because the reflection includes the floor near her feet.

What this shows. Full-length does not mean full-height. A half-height mirror is enough. Shopkeepers who sell tall floor mirrors are, geometrically speaking, selling more glass than you need — but a taller mirror gives you slack if you stand closer or if another person of a different height uses it.

Common confusions

You have the two laws of reflection, the proof that the plane-mirror image is equidistant, the multiple-image formula, and the real-vs-virtual distinction. Stop here if that is what you came for. What follows is the wave-optics justification of the law of reflection (which you otherwise accept as a postulate of ray optics), and the exact reason the law fails at grazing angles or near wavelength-scale structures.

Deriving the law of reflection from Huygens' principle

Ray optics treats light as straight-line rays that bounce at equal angles. This is a useful fiction: light is really an electromagnetic wave. At every point on a wavefront, Huygens said, treat the point as a source of tiny secondary wavelets; the new wavefront a moment later is the envelope of all these wavelets. When a plane wavefront strikes a flat mirror, the law of reflection emerges from Huygens' construction.

Setup. A plane wavefront AB approaches a flat mirror. Point A is just touching the mirror; point B is still above it at distance AB \cdot \sin\theta_i measured perpendicular, where \theta_i is the angle of incidence.

Step 1. At time t = 0, point A starts emitting a secondary wavelet from the mirror surface. Point B still has a distance d = AB \cdot \sin\theta_i to travel before it reaches the mirror.

Step 2. Let the time for B to reach the mirror be \tau = d/c = AB\sin\theta_i / c. During this time, the wavelet from A spreads as a hemisphere of radius c\tau = AB\sin\theta_i.

Why: Huygens' wavelets travel at the same speed c as the incident wave. If B takes time \tau to reach its landing point B' on the mirror, then during the same \tau the wavelet from A has grown to a sphere of radius c\tau.

Step 3. Now construct the new wavefront. From the endpoint B' on the mirror (where B just landed), draw the tangent to the sphere centred at A with radius c\tau. This tangent is the new wavefront, call it A'B'. By geometry, A'A = c\tau = AB\sin\theta_i.

Step 4. The new wavefront makes an angle \theta_r with the mirror. In the right triangle formed by A, A', B': \sin\theta_r = AA'/AB' = c\tau / AB.

Step 5. Compare with Step 2: \sin\theta_i = d/AB = AB\sin\theta_i / AB, which just gives back \sin\theta_i — the setup statement. The key is that both \sin\theta_i and \sin\theta_r are ratios of the same pair: c\tau / AB and d/AB. And we constructed c\tau = d = AB\sin\theta_i. So

\sin\theta_r = \frac{c\tau}{AB} = \frac{AB\sin\theta_i}{AB} = \sin\theta_i

Therefore \theta_r = \theta_i.

Why: the Huygens construction forces the reflected wavefront to be the mirror image of the incident wavefront through the mirror plane. The equality of angles is not an assumption — it is forced by the requirement that wavelets from different points of the wavefront arrive at the new wavefront in phase.

This is the deeper reason the law of reflection holds. It is not a contingent fact about light; it is a consequence of any wave reflecting off a flat boundary at a speed independent of direction.

When the law fails — grazing angles and wavelength-scale roughness

Two limits break the simple picture.

Grazing angles (\theta_i \to 90°). At very oblique incidence, the reflectivity of even a modestly smooth surface approaches 100%. This is why a polished floor at the Chhatrapati Shivaji International Airport terminal looks mirror-like when you view it at a shallow angle but dull when you look straight down — at grazing angles, even micro-roughness cannot scatter the wave efficiently, because the component of surface height perpendicular to the wavefront is much less than a wavelength. The formal criterion (the Rayleigh roughness criterion) is that a surface looks specular when its roughness h satisfies h\cos\theta_i \lesssim \lambda/8. For a 500 nm wave, this condition is met by a 0.5 micrometre-rough surface viewed at \theta_i = 89°, but not viewed head-on.

Wavelength-scale roughness. If the surface has bumps comparable to the wavelength of the light, the simple law of reflection gives way to diffraction. Each bump acts as a secondary source; the total scattered wave has a complicated angular distribution. This is the physics of antireflection coatings (deliberate sub-wavelength roughness to suppress specular reflection) and of structural colour in butterfly wings (sub-wavelength periodic structure that reflects selected wavelengths strongly). Ray optics simply does not apply — you have to compute the full electromagnetic diffraction problem.

These corrections do not change anything about JEE-level reflection: for smooth, macroscopic surfaces and ordinary angles, \theta_i = \theta_r holds to better than one part in a million. But they are where the ray-optics approximation meets its limits.

The Fermat principle view

There is a third way to understand the law of reflection, which physicists often find the most elegant. Fermat's principle of least time says that the actual path a light ray takes between two points is the one that makes the total travel time stationary — usually minimum.

Put a light source S and a receiver R on the same side of a flat mirror. You insist the ray must bounce off the mirror somewhere. Where is the bounce point?

Reflect R through the mirror to get its mirror image R'. The distance from S to a point P on the mirror and then to R equals the distance from S to P and then to R' (because PR = PR' by the plane-mirror isometry). Minimising this sum is the same as finding the straight line from S to R' through P — which forces P to lie on the line SR', and this geometry is exactly what gives \theta_i = \theta_r.

Why: the straight line from S to R' makes equal angles with the mirror on the two sides of P by symmetry (the mirror image reflects angles through itself). These equal angles are the angle of incidence and angle of reflection, measured from the normal.

Fermat's principle is more general than Huygens' construction — it works for refraction, for light in a medium of varying index, and (in its generalised form as the principle of stationary action) it reappears as Hamilton's principle in mechanics and as the path integral in quantum field theory. For reflection, it gives the fastest-to-a-schoolchild derivation: "the ray takes the shortest path, and the shortest path bounces at equal angles."

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