Drag the object arrow along the principal axis
Controls
Mirror type
Place the object (concave mirror)
Readings
- Object distance, u
- -30.0cm
- Image distance, v
- -15.0cm
- Focal length, f
- -10.0cm
- Radius of curvature, R
- -20.0cm
- Magnification, m = −v/u
- -0.50
- Image height, h′ = m × h
- -2.5cm
- Nature of the image
- Real, inverted, diminished
How to use this simulation
- Pick a concave, convex or plane mirror under "Mirror type". The pole P, principal focus F and centre of curvature C are marked on the principal axis.
- Drag the blue object arrow along the axis with a mouse or finger, or use the "Object distance" slider.
- For the concave mirror, the "Place the object" buttons jump to the six textbook positions: very far, beyond C, at C, between F and C, at F and between F and P.
- Move the "Focal length" slider and the radius of curvature follows as R = 2f, so F and C slide together.
- Orange lines are incident rays and red lines are reflected rays. Where reflected rays really cross in front of the mirror, the green image is drawn solid (real); where only their dashed extensions meet behind the mirror, it is dashed (virtual).
- Read u, v, f, R, the magnification m and the nature of the image in the readings panel, and compare them with the solved problems below.
A spoon, a bus mirror and a torch: curved mirrors are everywhere
Pick up a shiny steel spoon and look into its hollow side. Your face is upside down. Turn the spoon over and look at the back: now your face is the right way up, but smaller and a little stretched. It is the same spoon, yet the two sides give completely different pictures, because the hollow side behaves like a concave mirror and the back behaves like a convex mirror.
Now bring the hollow side slowly towards your eye. At some point the upside-down face turns into a blur, and when the spoon is very close you see a big, upright picture instead. Changing only the distance flipped the image from inverted to erect. That single observation is what this whole page explains, and the simulation above lets you repeat it step by step.
Curved mirrors are all around you. The mirror on a bus or a motorbike that shows the road behind is convex. A dentist's little mirror, a shaving or make-up mirror, and the shiny bowl behind a torch bulb are concave. Once you know a few simple rules for how light bounces, you can predict with a ruler and a pencil where any of these mirrors puts its image, whether it is upright or upside down, and how big it is.
Starting from zero: the two laws of reflection
When light hits a smooth, shiny surface and bounces back into the same medium, we call it reflection. A household mirror is a sheet of glass with a thin coat of silver or aluminium on the back; the coat stops the light and sends it back.
Reflection follows two laws, and curved mirrors obey them exactly like flat ones. First, the incident ray, the reflected ray and the normal at the point of incidence all lie in the same plane. Second, the angle of incidence equals the angle of reflection, i = r.
The only extra difficulty with a curved mirror is that the normal points in a different direction at every point. Here geometry helps: on a sphere, the normal at any point always passes through the centre of the sphere. So a line drawn from the centre of curvature to any point on the mirror is the normal at that point. Every rule on this page grows out of that one fact.
Imagine cutting a piece out of a hollow rubber ball. If the inside of the piece is polished, you have a concave mirror; if the outside is polished, you have a convex mirror. A concave mirror makes parallel rays meet at a point, so it is called a converging mirror. A convex mirror makes parallel rays spread out, so it is called a diverging mirror.
Key words for spherical mirrors
Before drawing a single ray, get the vocabulary straight. Most one-mark definition questions come straight from this table, and every mark on the simulation uses these names.
| Term | Symbol | What it means in plain words |
|---|---|---|
| Pole | P | The centre of the reflecting surface; every distance is measured from here |
| Centre of curvature | C | The centre of the hollow sphere the mirror is a piece of |
| Radius of curvature | R | The radius of that sphere, i.e. the distance PC |
| Principal axis | — | The straight line through P and C (the horizontal line in the diagram) |
| Principal focus | F | Where rays parallel to the axis meet after reflection (concave), or seem to come from (convex) |
| Focal length | f | The distance PF from the pole to the principal focus; f = R/2 |
| Aperture | — | The width of the reflecting surface, edge to edge |
| Real image | — | Formed where reflected rays really meet, in front of the mirror; can be caught on a screen |
| Virtual image | — | Where reflected rays only appear to meet, behind the mirror; cannot be caught on a screen |
| Magnification | m | The ratio of image height to object height, m = h′/h |
Why the focal length is half the radius: f = R/2
Take a concave mirror and a ray travelling parallel to the principal axis that strikes the mirror at a point M. The line CM is the normal at M, because a radius is always perpendicular to the surface of a sphere. Call the angle of incidence θ.
The incident ray is parallel to the axis and CM cuts across both, so the alternate angle ∠MCP is also θ. By the law of reflection the reflected ray also makes θ with CM, and it crosses the axis at F. Triangle FMC therefore has two equal angles, so it is isosceles and FM = FC.
For a mirror with a small aperture, M is very close to the pole, so FM is almost equal to FP. Then FP = FC, which puts F exactly halfway between P and C. So PF = PC/2, or f = R/2. The same argument works for a convex mirror, with F and C both behind the mirror.
f = R/2 or R = 2ffor a spherical mirror of small aperture
|f| = 10 cm gives |R| = 20 cmthe simulation's default mirror
The four principal rays
Countless rays leave the tip of the object, but any two of them are enough to find its image: wherever two reflected rays cross, or appear to cross, the image of the tip sits. These four rays are the easy ones, because we know in advance exactly where each goes.
Ray 1: parallel to the axis, then through the focus
For a concave mirror, a ray parallel to the principal axis passes through the principal focus after reflection. For a convex mirror, it bounces off as if it were coming from the focus behind the mirror. In the simulation it is the ray leaving the object tip horizontally.
Ray 2: through the focus, then parallel to the axis
For a concave mirror, a ray passing through the principal focus comes back parallel to the axis. This is Ray 1 run backwards, because light paths are reversible. For a convex mirror, a ray heading towards the focus behind the mirror comes back parallel.
Ray 3: through the centre of curvature, then straight back
A ray passing through C strikes the mirror along the normal, so its angle of incidence is zero and it reflects straight back along the same path. For a convex mirror, a ray aimed at the C behind the mirror also returns along itself.
Ray 4: to the pole, then out at an equal angle
At the pole the principal axis itself is the normal, so a ray hitting P obliquely reflects to the other side of the axis at the same angle, like a ball bouncing off the floor. This ray is the quickest way to prove that m = h′/h = −v/u.
Concave mirror ray diagram: all six cases
This table is the most-asked table in the chapter. The last column is computed from the mirror formula for a concave mirror with f = -10 cm (R = -20 cm) in the New Cartesian sign convention. Press the "Place the object" buttons in the simulation to check each row.
| Position of object | Position of image | Size and nature of image | Example (f = -10 cm) |
|---|---|---|---|
| At infinity | At F | Real, inverted, highly diminished (point-sized) | u → −∞, v = f = -10 |
| Beyond C | Between F and C | Real, inverted, diminished | u = -30, v = -15, m = -0.50 |
| At C | At C | Real, inverted, same size | u = -20, v = -20, m = -1 |
| Between F and C | Beyond C | Real, inverted, magnified | u = -15, v = -30, m = -2 |
| At F | At infinity | Highly magnified (reflected rays leave parallel) | Image at infinity |
| Between F and P | Behind the mirror | Virtual, erect, magnified | u = -6, v = 15, m = 2.50 |
Convex and plane mirrors: always the same answer
Put the object anywhere in front of a convex mirror and the image is always behind the mirror, between P and F, virtual, erect and diminished. The further the object goes, the smaller the image and the closer it creeps to F, but it never gets past F. The table below is computed for a convex mirror with f = +15 cm, in the Cartesian convention.
That is why vehicles use convex mirrors as rear-view and wing mirrors: a smaller image means a much wider field of view fits in a small mirror, and the image is always upright, so the driver is never confused. The catch is that things look further away than they are, which is why many wing mirrors carry the warning "objects in mirror are closer than they appear".
A plane mirror can be treated as a spherical mirror whose radius of curvature is infinite, so its focal length is infinite too. Putting 1/f = 0 into the mirror formula gives v = −u: the image is exactly as far behind the mirror as the object is in front, and m = +1, so it is virtual, erect and the same size. Choose "Plane mirror" in the simulation and watch the dashed extensions meet at the object's mirror twin.
| Object distance u (cm) | Image distance v (cm) | Magnification m = −v/u |
|---|---|---|
| -10 | 6 | 0.60 |
| -30 | 10 | 0.33 |
| -60 | 12 | 0.20 |
The mirror formula: 1/v + 1/u = 1/f
A ray diagram tells you roughly where the image is; the mirror formula tells you exactly. Here u is the object distance, v the image distance and f the focal length, all measured from the pole.
This page uses the New Cartesian sign convention, the one in NCERT, most international textbooks and JEE/NEET. The pole is the origin and the incident light travels in the positive direction (left to right in the simulation). Distances measured along the incident light are positive, against it negative; heights above the axis are positive, below negative.
Because the object always sits in front of the mirror, against the incident light, u is always negative. The focus of a concave mirror is in front, so f is negative; the focus of a convex mirror is behind, so f is positive. A real image in front of the mirror has negative v, and a virtual image behind it has positive v.
Magnification is m = h′/h = −v/u. A negative m means a real, inverted image; a positive m means a virtual, erect image. If |m| > 1 the image is magnified, if |m| < 1 it is diminished. The readings panel above prints exactly these signs.
1/v + 1/u = 1/fmirror formula (New Cartesian convention)
m = h′/h = −v/umagnification; negative → real and inverted
v = uf/(u − f)the formula rearranged to give v directly
Where the formula comes from (sketch)
Put an object AB beyond C with its image A′B′. The ray to the pole gives a pair of similar triangles, so A′B′/AB = PB′/PB. The ray through C gives another pair, so A′B′/AB = CB′/CB. With the distances written as sizes, PB′/PB = (PC − PB′)/(PB − PC).
Now put in the Cartesian signs, PB = −u, PB′ = −v and PC = −R, cross-multiply and tidy up: you get 2uv = R(u + v). Divide both sides by uvR and use R = 2f to reach 1/v + 1/u = 2/R = 1/f. Because every quantity carries its own sign, this one formula covers concave and convex mirrors, real and virtual images.
The other convention: real-is-positive (1/u + 1/v = 1/f)
Some syllabi, including Bangladesh's NCTB, use the "real-is-positive" convention instead: real objects and real images have positive distances, virtual images negative, a concave mirror has positive f and a convex mirror negative f, and magnification is written m = v/u.
In that convention, Problem 1 below reads u = 30 cm, f = 10 cm, v = 15 cm and m = 0.50. The sizes are identical; only the signs, and the meaning of the sign of m, change. Pick the convention your exam expects and never switch in the middle of a problem.
Things to try in the simulation
Do these one at a time, and after each one ask yourself "why did that happen?"
Experiment 1: chase the image
With a concave mirror of focal length 10 cm, drag the object from the far left slowly towards the mirror. The green inverted image starts near F, runs out past C and grows. When the object reaches F the image vanishes to infinity, and once the object is inside F a dashed, upright image appears behind the mirror.
Experiment 2: the same-size point
Press "At C". The panel shows u = -20 cm, v = -20 cm and m = -1: object and image share the same spot, but the image is upside down. Change the focal length, put the object back at C, and check that the result does not change.
Experiment 3: the stubborn convex mirror
Switch to "Convex mirror" and drag the object anywhere. The image never comes in front of the mirror, never flips, never grows bigger than the object. The magnification stays positive and below 1 however hard you try.
Experiment 4: the plane mirror
Choose "Plane mirror" and move the object. The panel always shows v = −u and m = 1: the image is as far behind the mirror as the object is in front, and it never changes size.
Experiment 5: design a shaving mirror
Put the object between F and P of a concave mirror, then increase the focal length. At the same object distance a longer focal length gives less magnification, and the closer the object sits to F, the bigger the upright image. That is why a make-up mirror works best held fairly close to the face.
Solved problems (New Cartesian convention)
In each problem, write down what is given with signs, then the formula, then substitute. Problem 1 is the simulation's default state, so the panel shows the same answer when the page loads.
Problem 1: object 30 cm in front of a concave mirror of focal length 10 cm
Given u = -30 cm and f = -10 cm (concave). Then 1/v = 1/f − 1/u = −1/10 + 1/30 = −1/15, so v = -15 cm.
Magnification m = −v/u = −(-15)/(-30) = -0.50. v is negative, so the image is real, in front of the mirror; m is negative, so it is inverted; |m| < 1, so it is diminished. The object is beyond C and the image lies between F and C, just as the table says.
Problem 2: object at the centre of curvature (20 cm)
R = 2f, so C is 20 cm from the pole. With u = -20 cm: 1/v = −1/10 + 1/20, giving v = -20 cm and m = -1. The image is at C, real, inverted and exactly the same size.
Problem 3: object between F and C (15 cm)
With u = -15 cm: 1/v = −1/10 + 1/15, giving v = -30 cm and m = -2. The image is beyond C, real, inverted and magnified. To throw a large upside-down picture onto a wall, this is where the object has to be.
Problem 4: a shaving mirror, face 6 cm away
Now u = -6 cm, closer than the focus. 1/v = −1/10 + 1/6 is positive, so v = +15 cm.
A positive v means the image is behind the mirror, so it is virtual. m = −v/u = 2.50: positive, so erect, and greater than 1, so magnified. That is exactly why your face looks big and upright in a shaving mirror.
Problem 5: a convex wing mirror, f = +15 cm, object 30 cm away
Given u = -30 cm and f = +15 cm (convex). 1/v = 1/15 + 1/30, so v = +10 cm.
The image is virtual, behind the mirror, closer than F. m = 0.33: erect and diminished. Set the simulation to a convex mirror with |f| = 15 cm and the object at 30 cm to check.
Problem 6: focal length and radius from an image on a screen
In the lab, a candle 36 cm in front of a concave mirror gives a sharp image on a screen 18 cm in front of the mirror. The image is on a screen, so it is real and in front: u = -36 cm, v = -18 cm.
1/f = 1/v + 1/u = −1/18 − 1/36, so f = -12 cm and R = 2f = -24 cm. The negative sign confirms that the mirror is concave.
Problem 7: how tall is the image?
If the object in Problem 3 is 2 cm tall, the image height is h′ = m × h = -2 × 2 = -4 cm. The minus sign means it is inverted, below the axis. Set the object height to 2 cm and the distance to 15 cm in the simulation and read the image-height box.
Problem 8: a dentist wants an erect image 3 times as large
With a concave mirror of f = -10 cm, the dentist wants an upright image 3 times the size of the tooth. Upright means virtual, so m = +3, which gives v = −3u.
Substitute: 1/(−3u) + 1/u = −1/10. Simplifying gives u = -6.67 cm and v = +20 cm, so the image is 20 cm behind the mirror. The mirror must be held a little inside the focal length from the tooth.
Problem 9: standing 1.50 m from a plane mirror
The image is 1.50 m behind the mirror, so you and your image are 3 m apart. Step 0.50 m forward and the gap becomes 2 m: your image came 1 m closer to you, twice the distance you walked.
Concave mirror vs convex mirror
For every "state the differences" question, here is the whole comparison:
| Feature | Concave mirror | Convex mirror |
|---|---|---|
| Reflecting surface | Inner (caved-in) side of the sphere | Outer (bulging) side of the sphere |
| Effect on light | Converging | Diverging |
| F and C | In front of the mirror, real | Behind the mirror, virtual |
| Sign of f (Cartesian) | Negative | Positive |
| Images | Real or virtual depending on the object position | Always virtual, erect, diminished |
| Uses | Shaving and make-up mirrors, dentist's mirror, torch and headlight reflectors, solar cookers | Rear-view and wing mirrors, road-bend mirrors, shop security mirrors |
Common mistakes
Examiners see these again and again. Check your own answers for them.
- Forgetting the sign of u. In the Cartesian convention the object distance is always negative, because the object is in front of the mirror.
- Giving a concave mirror a positive focal length in Cartesian problems. Concave f is negative, convex f is positive.
- Mixing up R and f: if a question gives the radius of curvature, halve it first.
- Drawing the part behind the mirror with solid lines. Every virtual ray extension and every virtual image is dashed.
- Leaving arrowheads off the rays, or pointing them the wrong way. Light travels from the object to the mirror and then away from it.
- Saying the image is "at F" when the object is at F. The reflected rays are parallel and the image is at infinity.
- Writing m = v/u in a Cartesian problem. For mirrors the Cartesian magnification is m = −v/u, unlike a lens.
Spherical mirrors in real life
Shaving and make-up mirrors: concave mirrors with a long focal length. Your face is inside F, so you see an upright, enlarged image where every small detail stands out.
Dentist's mirror: a small concave mirror on a handle, held close to a tooth, gives an erect, magnified view of it. The round mirror an ENT doctor wears on the forehead is concave too; it gathers light onto a spot to light up the inside of an ear or throat.
Torches, car headlights and searchlights: the bulb sits at the principal focus of a concave reflector. Rays from F come back parallel to the axis (Ray 2), so the light leaves as a strong, narrow beam that reaches a long way instead of spreading out.
Solar cookers and solar furnaces: large concave mirrors collect parallel sunlight at the focus, where a black pot gets hot enough to cook.
Rear-view and wing mirrors: convex mirrors give an upright, smaller image, so a small mirror shows a wide stretch of road behind.
Road-bend and shop security mirrors: big convex mirrors at blind corners on hill roads and in the corners of supermarkets let you see round the bend or the whole shop floor at once.
Spherical aberration and parabolic mirrors
The rule f = R/2 is only exact for mirrors with a small aperture. On a wide spherical mirror, parallel rays that strike near the edge cross the axis a little in front of F, closer to the mirror. The rays no longer meet at one point but smear over a small region, and the image blurs. This defect is called spherical aberration.
To avoid it, torches, headlights, satellite dishes and large telescopes use parabolic mirrors instead of spherical ones. A parabola has the special property that every ray parallel to its axis, near the edge or near the middle, reflects through exactly the same focus. Like the textbooks, this simulation assumes a small aperture, so its rays meet perfectly.
Exam corner
Spherical mirrors appear every year in Grade 10 physics under "Light: reflection and refraction", return in Grade 12 ray optics, and turn up in JEE and NEET as quick sign-convention numericals. The questions follow a few patterns:
- Definitions: pole, centre of curvature, radius of curvature, principal focus, focal length, aperture, magnification.
- Proofs: show that f = R/2 for a mirror of small aperture, using the law of reflection and an isosceles triangle.
- Ray diagrams: draw the image for an object at a given position, usually between F and C or between F and P, and state its nature.
- Numericals: given u and f (or R), find v, m and the image height with correct signs, as in Problems 1 to 8.
- Applications: which mirror is used in headlights, as a shaving mirror or as a rear-view mirror, and why.
Quick revision
If you only read one section the night before the exam, read this one.
- A concave mirror converges light; a convex mirror diverges it; a plane mirror has infinite focal length.
- f = R/2: the focal length is half the radius of curvature.
- Four rays: parallel → through F; through F → parallel; through C → straight back; to P → equal angle on the other side.
- Mirror formula 1/v + 1/u = 1/f (Cartesian, u negative); magnification m = h′/h = −v/u.
- Concave mirror: object beyond F → real and inverted; between F and P → virtual, erect, magnified.
- Object at C → same-size image at C; object at F → image at infinity.
- Convex mirror → image always virtual, erect, diminished, between P and F.
- Simulation default: f = -10 cm, u = -30 cm → v = -15 cm, m = -0.50.
Frequently asked questions
What is a concave mirror?
A concave mirror is a spherical mirror whose reflecting surface is the inner, caved-in side of the sphere. It makes rays parallel to the principal axis meet at the principal focus after reflection, so it is also called a converging mirror.
What is a convex mirror?
A convex mirror is a spherical mirror whose reflecting surface is the outer, bulging side of the sphere. It spreads parallel rays out as if they came from a focus behind the mirror, so it is also called a diverging mirror.
What is the relation between focal length and radius of curvature?
For a spherical mirror of small aperture, the focal length is half the radius of curvature: f = R/2. A mirror with R = 20 cm has f = 10 cm.
When does a concave mirror form a virtual image?
Only when the object is between the principal focus and the pole. The image is then behind the mirror, erect and magnified. Shaving mirrors and dentist's mirrors work this way.
Can a convex mirror form a real image?
Not of a real object. Wherever the object is placed, a convex mirror forms a virtual, erect, diminished image behind the mirror, between the pole and the focus.
Why are convex mirrors used as rear-view mirrors?
Because they always give an erect, diminished image, which lets a small mirror show a much wider field of view than a plane mirror of the same size.
What does a negative magnification mean for a mirror?
In the New Cartesian convention, a negative m = −v/u means the image is real and inverted; a positive m means virtual and erect. In the real-is-positive convention (m = v/u) the meaning of the sign is the other way round.
Where is the image when the object is at the centre of curvature?
At the centre of curvature itself: real, inverted and the same size. For f = -10 cm and u = -20 cm, v = -20 cm and m = -1.
Which mirror is used in torches and car headlights?
A concave mirror, usually shaped as a parabola. The bulb sits at its focus, so the reflected light leaves as a parallel beam that travels far without spreading.
Why is the object distance negative in the mirror formula?
In the New Cartesian convention distances are measured from the pole, and the direction of the incident light is positive. The object is always in front of the mirror, against that direction, so u comes out negative.
Keep studying this topic
The animation made the idea click; now turn it into marks. Syllabus, suggestions, textbooks and admission-test guides are below.
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