CBSE Class 10 Science · 2026–27
A mirror can make an image appear behind glass, turn it upside down, or show more of a busy corridor. Follow the rays and each result becomes predictable. Work through these short topics, try the checks, and finish with signed numerical examples and a chapter revision.
For CBSE Class 10 Science, 2026–27: spherical mirrors, image formation, mirror formula and magnification. Plane-mirror ideas build the foundation. Formula derivation is not needed.
- Trace reflected light using the normal and i=r.
- Distinguish real and virtual images and locate a plane-mirror image.
- Label P, F, C, R and f for concave and convex mirrors.
- Construct and interpret the six concave-image cases and the convex-image pattern.
- Apply Cartesian signs, the mirror formula and signed magnification with consistent units.
- Choose a mirror for a purpose and interpret a distant-object focal-length practical.
01Follow the light to your eye
Why can you read a page beside a lamp, yet lose it in a dark room? Your eyes need light arriving from the page. The lamp emits light; the page reflects some of that light towards you. The useful path is source → object → eye. Your eyes do not send out a searching beam.
A glowing source is luminous. A page, a tree and the Moon are examples of objects usually seen by reflected light. For a luminous object, light can travel directly from the source to your eye.
In a uniform transparent medium, we use straight lines called rays to represent light’s direction. Put an arrow on every ray. A bundle of rays is a beam. This ray model is a useful tool for mirrors; it does not mean light has no wave properties.
Try the idea: imagine a book under an opaque box. A lamp outside the box cannot make the book visible unless light reaches the book and some reflected light reaches the viewer.
Pause. Think. Try.
02Measure from the normal
A ray arrives at a mirror. Which way will it leave? First mark the point where it hits. Draw a normal: a construction line perpendicular to the surface at that point. The incoming ray is the incident ray; the departing ray is the reflected ray.
- The incident ray, reflected ray and normal lie in the same plane.
- The angle of incidence equals the angle of reflection: i = r.
Both angles are measured from the normal, not from the mirror surface. If an incident ray makes 62° with the surface, it makes 90° − 62° = 28° with the normal, so r = 28°. A ray along the normal has i = r = 0° and returns along its incoming path.
A smooth mirror sends an initially parallel bundle in an orderly direction. A rough wall has differently tilted tiny surface patches, so it sends rays in many directions. This diffuse reflection still obeys i = r at each patch. It helps a page remain visible from different seats; it does not form a sharp mirror image.
Sketch habit: draw the normal first, then the two equal angles. Give the reflected ray an arrow pointing away from the mirror.
Pause. Think. Try.
03Locate a plane-mirror image
There is no object behind the bathroom mirror, so why does your image seem to be there? Rays from one point on your face reflect into your eye. Extend those reflected rays backwards with dashed lines: their extensions meet behind the mirror. That apparent origin is the image point.
The image is virtual: the actual reflected rays do not meet behind the mirror, so a bare screen placed there cannot receive the image. Virtual does not mean invisible; you can see it, and a camera looking into the mirror can photograph it.
A plane-mirror image is upright, the same size as the object, and equally far behind the mirror as the object is in front. It shows mirror writing, usually called lateral inversion. The geometry reverses the direction perpendicular to the mirror; comparing your image with a person facing you makes the sideways reversal familiar.
If a card is 1.8 m in front of a fixed mirror, its image is 1.8 m behind it: card-to-image separation is 3.6 m. Move the card 0.3 m towards the mirror and the separation becomes 1.5 + 1.5 = 3.0 m.
A real image is different: actual rays meet at the image position and can produce a focused image on a screen there. In our single-mirror examples with a real upright object, real images are inverted and virtual images are upright.
Pause. Think. Try.
04Meet P, F and C
A spoon offers a quick clue: one side bends in, the other bulges out. A spherical mirror is a small part of a spherical reflecting surface. A concave mirror reflects from its inward-curved side; a convex mirror reflects from its outward-curved side. A spoon is a useful shape analogy, not a precision spherical mirror.
- P, pole: the middle point of the reflecting surface.
- C, centre of curvature: the centre of the sphere of which the mirror is a part.
- Principal axis: the line through P and C.
- R, radius of curvature: the signed distance PC.
- F, principal focus: where rays close and parallel to the principal axis meet after reflection, or appear to come from.
- f, focal length: the signed distance PF.
A concave mirror brings these parallel rays together at a real focus in front. A convex mirror makes them diverge as if they came from a virtual focus behind.
For the small-aperture, near-axis spherical mirrors used in this chapter, f = R/2. Thus a radius magnitude of 48 cm gives a focal-length magnitude of 24 cm. We will attach signs after fixing the coordinate convention. The near-axis condition matters: a wide spherical mirror need not focus every incoming ray at exactly one point.
Pause. Think. Try.
05Build a concave-mirror ray diagram
You do not need to guess an image position. Let two rays find it. Draw the principal axis, mirror, P, F and C. Place the base of an upright object arrow on the axis. Trace two different rays from the same point, usually the arrow tip.
- A ray parallel to the axis reflects through F.
- A ray passing through F reflects parallel to the axis.
- A ray directed through C meets the surface normally and returns along its path.
- A ray striking P reflects at an equal angle to the principal axis, which is the normal at P.
Choose two convenient rules. Where actual reflected rays cross, mark a real image point. If reflected rays diverge, extend them backwards with dashed construction lines to locate a virtual image point. Draw the image arrow from the axis to that point.
Keep the diagram honest: rays need arrowheads, P is the distance origin, and F must lie halfway between P and C. Dashed extensions are not light travelling through the mirror. If your chosen two rays overlap, use another independent ray.
Example: place an object between F and C. The parallel ray reflects through F; the pole ray reflects below the axis. They cross beyond C, so the image is real, inverted and enlarged.
Pause. Think. Try.
06Move the object, predict the image
A concave mirror can enlarge your face or turn a distant object upside down. The difference is the object’s position relative to F and C. Read the pattern from far away towards the mirror.
| Object position | Image position | Nature and size |
|---|---|---|
| Very far away | Near F, in front | Real, inverted, strongly diminished |
| Beyond C | Between F and C | Real, inverted, diminished |
| At C | At C | Real, inverted, same size |
| Between C and F | Beyond C | Real, inverted, enlarged |
| At F | No finite image position; often described as at infinity | Reflected rays from each object point are parallel; no sharp image on a finite screen |
| Between F and P | Behind the mirror | Virtual, upright, enlarged |
At the focus is a limiting case. Exactly parallel rays do not eventually meet somewhere far away. As an object outside F approaches F, its real image moves farther away and grows. At F itself, the ideal model gives 1/v = 0.
For a mirror with |f| = 12 cm, C is 24 cm in front. An object at 30 cm gives a diminished real image; at 18 cm it gives an enlarged real image; at 8 cm it gives an enlarged virtual image. A screen test distinguishes the two enlarged cases.
Memory anchor: inside F, the image is upright and virtual; outside F, the finite image is inverted and real. A ray diagram tells you the size.
Pause. Think. Try.
07A convex mirror keeps the view wide
Why can a small convex safety mirror show a broad stretch of a corridor? Its outward-curved surface spreads reflected rays. The upright images are smaller, allowing a wider field of view than a comparable plane mirror in a similar viewing arrangement.
Start with two rays from the same object tip. A parallel ray reflects as if it came from F behind the mirror. A ray directed towards F reflects parallel to the axis. You can also use a ray directed towards C, which returns along its incoming line, or the equal-angle pole ray.
The reflected rays diverge. Extend them backwards: their extensions meet behind the mirror, between P and F. For every finite real object in front, the image is virtual, upright and diminished. Move the object farther away and the image moves towards F while becoming smaller. A very distant object gives the limiting image near F.
A convex mirror cannot project its usual real-object image onto a bare screen behind it. Nor does a smaller image mean the object is physically small or safely far away; image size depends on mirror geometry and distance.
Compare carefully: an upright same-size image suggests a plane mirror; an upright enlarged image can come from a concave mirror with the object inside F; an upright diminished image is the convex pattern in these single-mirror cases.
Pause. Think. Try.
08Give each distance its sign
A minus sign is a direction, not a mistake. Fix the diagram before substituting numbers. We put the real object on the left and let the incident light travel left to right. P is the origin.
- Right of P is positive; left of P is negative.
- Above the axis is positive height; below it is negative height.
- Measure u, v, f and R from P, not from F or C.
| Quantity | Concave mirror | Convex mirror |
|---|---|---|
| Object distance u | Negative | Negative |
| Focal length f and radius R | Negative | Positive |
| Image distance v | Negative for real image; positive for virtual image | Positive |
For an upright object, ho is positive. An inverted image has hi negative; an upright image has hi positive. So a concave mirror of focal-length magnitude 18 cm uses f = −18 cm. An object 12 cm in front uses u = −12 cm.
Leave an unknown signed quantity as a symbol and let the calculation find its sign. Then translate the result back into a location: v = +36 cm means 36 cm behind the mirror. If a diagram uses a different direction for incident light, apply the coordinate rule again rather than memorising left/right without context.
Pause. Think. Try.
09Solve with the mirror formula
Can a diagram’s prediction survive a calculation? For a near-axis spherical-mirror image, use 1/f = 1/v + 1/u. It applies to concave and convex mirrors when all distances use the same sign convention and length unit. You need to use the relation; its derivation is not required here.
- Write the known signed values and units.
- Rearrange for the unknown reciprocal.
- Substitute, calculate, then take the reciprocal.
- State the physical position and check it against a ray diagram.
For a concave mirror with f = −12 cm and u = −18 cm:
1/v = 1/f − 1/u = −1/12 + 1/18 = −1/36 cm−1, so v = −36 cm. The image is 36 cm in front of the mirror. The object lies between F and C, so an image beyond C is sensible.
For a convex mirror with f = +16 cm and u = −48 cm, 1/v = 1/16 + 1/48 = 1/12 cm−1; hence v = +12 cm. This is behind the mirror and between P and F, as the ray picture predicts.
Common slip: 1/v = −1/36 cm−1 does not mean v = −1/36 cm. Inverting the reciprocal is the final algebra step. Keep centimetres throughout, or convert every distance to metres first.
Pause. Think. Try.
10Read size and orientation separately
An image can be larger and still have a negative magnification. Use m = hi/ho = −v/u for mirrors.
- The sign tells orientation: positive means upright, negative means inverted relative to the upright object.
- The magnitude |m| tells relative size: greater than 1 means enlarged, equal to 1 means same size, less than 1 means diminished.
Thus m = −2 means an inverted image twice the object’s height, whereas m = +0.25 means an upright image one-quarter as tall. It is wrong to call every numerical value below 1 a diminished image: −2 is below 1, but |−2| is 2.
Return to u = −18 cm, v = −36 cm. Then m = −(−36)/(−18) = −2. If the upright object is 3 cm tall, hi = m ho = −6 cm. The image is 6 cm tall and inverted.
Final answer check: report position, real or virtual nature, orientation and size. For a finite real object, a convex mirror should give v > 0 and 0 < m < 1. If your result violates that pattern, revisit the signs or arithmetic.
Pause. Think. Try.
11Choose a mirror, then test your prediction
Different jobs need different images. A close inspection mirror needs an enlarged upright view: use a concave mirror with the object inside F. A broad corridor view needs upright diminished images: use a convex mirror. An ideal concave reflector sends an approximately parallel beam out when a small source is at F. Real headlamps and instruments may use more elaborate shapes and optics.
For a teacher-guided focal-length check, use a secure concave mirror, a proper white screen on a holder, and a ruler. Choose a brightly lit distant non-solar object, such as a building edge or tree. Keep the object, mirror and screen suitably aligned; move the screen until the image is sharp. Measure pole-to-screen distance, not object-to-screen distance. Repeat, recording units and the spread of readings. Use no direct sunlight or laser beam.
A distant object is only approximately at infinity. For a finite real object beyond C, the screen is slightly farther from P than |f|; increasing object distance brings the image closer to F. That is why a distant target improves the approximation.
| Trial | Sharp-image screen reading | Pole-to-screen distance |
|---|---|---|
| 1 | 34.1 cm | 23.9 cm |
| 2 | 34.3 cm | 23.7 cm |
| 3 | 34.2 cm | 23.8 cm |
The mean distance is 23.8 cm, an approximate focal-length magnitude. In our signed diagram, f ≈ −23.8 cm. These are supplied classroom practice readings, not a claim that an experiment has already been performed.
Finish the chapter: redraw one real concave image, one virtual concave image and one convex image. Solve a numerical for each, then check that signs, orientation and relative size agree.
Pause. Think. Try.
Your formula shelf
| Idea | Relation | Remember |
|---|---|---|
| Reflection law | i = r | Both angles measured from the normal at the point of incidence. |
| Radius and focal length | f = R/2 | Small-aperture spherical mirror with near-axis rays; keep signed f and R consistent. |
| Mirror formula | 1/f = 1/v + 1/u | Distances from P; one length unit; real object and near-axis spherical-mirror model. |
| Magnification | m = hi/ho = −v/u | Sign gives orientation; |m| gives size ratio. Magnification has no unit. |
| Image height | hi = m ho | Use signed heights relative to the principal axis. |
Work through the numbers
A 3.0 cm upright object is 18 cm in front of a concave mirror with focal length magnitude 12 cm. Find image position, magnification, height and nature.
- Use u=−18 cm, f=−12 cm and hₒ=+3.0 cm.
- 1/v=1/f−1/u=−1/12+1/18=−1/36 cm⁻¹, so v=−36 cm.
- m=−v/u=−(−36)/(−18)=−2.
- hᵢ=m hₒ=−2×3.0=−6.0 cm.
The image is 36 cm in front, real, inverted and twice as tall. F is 12 cm and C is 24 cm in front; the predicted image beyond C matches the calculation.
A 2.0 cm upright object is 12 cm in front of a concave mirror with focal length magnitude 18 cm. Find image position and height.
- u=−12 cm, f=−18 cm, hₒ=+2.0 cm.
- 1/v=−1/18+1/12=+1/36 cm⁻¹, giving v=+36 cm.
- m=−36/(−12)=+3; hᵢ=+3×2.0=+6.0 cm.
The image is 36 cm behind the mirror, virtual, upright and enlarged. The object lies inside F, consistent with this result. A bare screen behind the mirror cannot receive it.
An 8.0 cm upright object is 48 cm in front of a convex mirror of focal length 16 cm. Find the image position, magnification and height.
- u=−48 cm, f=+16 cm, hₒ=+8.0 cm.
- 1/v=1/16+1/48=1/12 cm⁻¹; v=+12 cm.
- m=−12/(−48)=+0.25; hᵢ=+0.25×8.0=+2.0 cm.
The image is 12 cm behind the mirror, between P and F. It is virtual, upright and one-quarter as tall.
A mirror forms a real image half as tall as an upright object 45 cm in front. Find v, f and the mirror type.
- The image is inverted in this real-object single-mirror setup, so m=−0.5 and u=−45 cm.
- From m=−v/u, v=−m u=−(−0.5)(−45)=−22.5 cm.
- 1/f=1/v+1/u=−1/22.5−1/45=−1/15 cm⁻¹; f=−15 cm.
The mirror is concave. C is 30 cm in front, so the object beyond C produces an image between C and F, as required.
A concave mirror has radius-of-curvature magnitude 40 cm. A real object is 60 cm in front. Find image position and relative size.
- R=−40 cm, so f=R/2=−20 cm; u=−60 cm.
- 1/v=−1/20+1/60=−1/30 cm⁻¹; v=−30 cm.
- m=−(−30)/(−60)=−0.5.
The image is 30 cm in front, real, inverted and half as tall. It lies between F at 20 cm and C at 40 cm.
Bring it all together
Mix the ideas. Try each question before opening its explanation.
Write it. Explain it.
Use paper for a calculation, a labelled diagram or a short explanation. Then compare the reasoning. Suggested marks help you check how complete your answer is.
A ray makes 64° with a plane-mirror surface. Find its angle of reflection and state the reference line used to measure it. · 2 marks
The normal is perpendicular to the surface. Therefore i=90°−64°=26°, and r=i=26°. Both angles are measured from the normal at the point of incidence.
- 1 marks: Correct conversion to 26° from the normal.
- 1 marks: r=i with correct reference line.
A concave mirror with f=−16 cm has an upright object at u=−24 cm. Find v and m, then describe the image. · 3 marks
1/v=1/f−1/u=−1/16+1/24=−1/48 cm⁻¹, so v=−48 cm. Then m=−(−48)/(−24)=−2. The image is 48 cm in front of the mirror, real, inverted and twice as tall.
- 1 marks: Correct mirror relation and signed substitution.
- 1 marks: v=−48 cm and physical position.
- 1 marks: m=−2 and consistent image description.
A school compares two mirrors using an upright object. Mirror A gives an upright same-size image at every tested distance. Mirror B gives an upright diminished image at every tested distance. Identify each mirror, explain the advantage of B for a corridor, and say whether a bare screen behind B can receive its image. · 4 marks
A is a plane mirror; B is a convex mirror in this real-object setup. The convex surface gives a wider field of view through upright diminished images, useful for observing a corridor junction. Its image is virtual: actual reflected rays do not meet behind it, so a bare screen there cannot receive the image.
- 1 marks: A identified as plane.
- 1 marks: B identified as convex.
- 1 marks: Wider field of view connected to upright diminished images.
- 1 marks: No screen image, explained using virtual ray geometry.
Describe or draw how a concave mirror forms an enlarged upright image. State the object region and show how two rays locate the image. A fully labelled text description may be used in place of a drawing. · 3 marks
Place the object between P and F. From its tip, draw a ray parallel to the axis; it reflects through F. Draw a second ray to P and reflect it at an equal angle to the principal axis. The reflected rays diverge in front. Their dashed backward extensions meet behind the mirror at a taller upright image. It is virtual.
- 1 marks: Object between P and F, with P/F/C placed consistently.
- 1 marks: Two correct independent ray paths or equivalent text description.
- 1 marks: Dashed-extension intersection behind mirror; enlarged upright virtual image.
A concave mirror is at ruler position 72.0 cm. A distant non-solar object gives sharp screen images at 53.4, 53.6 and 53.5 cm. Estimate the focal-length magnitude. Explain one reason the result is approximate and name the point from which distance is measured. · 3 marks
The separations are 18.6, 18.4 and 18.5 cm. Their mean is 18.5 cm, so |f|≈18.5 cm (f≈−18.5 cm with this chapter’s convention). Measure from the pole P. A finite object distance, limited sharpness judgement or ruler/position uncertainty makes the estimate approximate; a finite distant object forms an image slightly beyond F.
- 1 marks: Correct separations or equivalent mean calculation.
- 1 marks: |f|≈18.5 cm, measured from P.
- 1 marks: One valid approximation/measurement reason.