PhysicsICSE Class 9 · Spherical Mirrors

Images Formed by Concave and Convex Mirrors: Ray Rules, Object Positions and Uses (ICSE Class 9)

Updated 17 October 2026 · Fact-checked

A concave mirror gives a real, inverted image for any object beyond F and a virtual, erect, magnified image when the object is between P and F. A convex mirror gives only a virtual, erect, diminished image between P and F, whatever the object position. Locate images with two rays: parallel to the axis through F, through F parallel to the axis, through C straight back.

Understand Images Formed by Concave and Convex Mirrors: Ray Rules and Nature of Images

Terms. Pole $P$ , centre of curvature $C$ , radius of curvature $R = P C$ , principal axis, principal focus $F$ , focal length $f = P F = \frac{R}{2}$ (small aperture).

Concave mirror: reflects from the inner surface; converges parallel rays to a real focus in front.
Convex mirror: reflects from the outer surface; diverges parallel rays so that they appear to come from a virtual focus behind it.

Rules for rays (concave).
1. A ray parallel to the axis passes through $F$ after reflection.
2. A ray through $F$ emerges parallel to the axis.
3. A ray through $C$ is reflected straight back (it strikes along the normal).
4. A ray striking the pole is reflected with equal angle to the axis on the other side.

For a convex mirror use the same rules with "towards $F$ " and "towards $C$ " (behind the mirror); the reflected rays appear to come from $F$ . Two of these rays are enough to locate the image.

Concave mirror: object position and image.

Object	Image position	Nature
At infinity	At $F$	Real, inverted, highly 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, magnified
At $F$	At infinity	Real, inverted, highly magnified
Between $F$ and $P$	Behind the mirror	Virtual, erect, magnified

Only the last case is virtual. Real images can be received on a screen; virtual images cannot.

Convex mirror. For every position of a real object the image is behind the mirror, between $P$ and $F$ , virtual, erect and diminished. A convex mirror gives a wide field of view.

Uses.

Concave: shaving and make-up mirrors (face within $f$ ), dentist’s mirror, torch and headlight reflectors (source at $F$ gives a parallel beam), solar cookers (sunlight converges at $F$ ), doctor’s head mirror.
Convex: rear-view mirrors, security mirrors, street-light reflectors.

Identifying the mirror. Bring it close to your face: a magnified erect image means concave; a small erect image means convex. Concave mirrors also focus sunlight to a bright spot on paper.

Checking with the formula. If $f$ and $u$ (both with Cartesian signs) are known, $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ tells where the image lies; its position relative to $F$ and $C$ gives the same table.

Images Formed by Concave and Convex Mirrors: Ray Rules and Nature of Images: formulas and key facts

Focal length: $f = \frac{R}{2}$; Small aperture; concave f negative, convex f positive (Cartesian).
Mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$; Used to check ray-diagram results.
Magnification: $m = - \frac{v}{u}$; Negative: real, inverted; positive: virtual, erect.

How to solve Images Formed by Concave and Convex Mirrors: Ray Rules and Nature of Images questions

Place the object relative to F and C first, then read the image from the table or confirm with the formula.

1Find $f = R /2$ and mark $F$ and $C$ on the axis.
2Decide the region of the object: beyond $C$ , at $C$ , between $C$ and $F$ , at $F$ , or inside $F$ .
3Draw two of the four ray rules to locate the image.
4State position, nature (real or virtual), orientation and size.
5If numbers are needed, apply $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ with signs.

Images Formed by Concave and Convex Mirrors: Ray Rules and Nature of Images in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A torch reflector

Suppose a torch has a concave reflector of radius of curvature 12 cm and the bulb is at its principal focus.

Where must the bulb be placed, and what happens to the beam if the bulb is moved closer to the mirror by 2 cm?

Try it first, then show the solutionSolution

$f = \frac{12}{2} = 6$ cm: the bulb is 6 cm from the pole.
At 4 cm the bulb is inside $F$ : $\frac{1}{v} = \frac{1}{- 6} - \frac{1}{- 4} = \frac{1}{12}$ , $v = + 12$ cm; the image is virtual, so the reflected beam spreads out.

Answer: Bulb 6 cm from the pole; moving it in makes the beam diverge.

Memory hook: At F: parallel beam. Inside F: spreading beam.

Tricks to remember Images Formed by Concave and Convex Mirrors: Ray Rules and Nature of Images

Short memory aids. Say each one aloud once, then use it on a question the same day.

Real is inverted, virtual is erect
Real images are always inverted; virtual images are erect.
Use it when: Nature of image.
Only inside F is virtual (concave)
Between F and P: virtual, erect, magnified; everywhere else real and inverted.
Use it when: Table of positions.
Convex is always the same
Behind, between P and F, virtual, erect, diminished.
Use it when: Convex mirror questions.

Images Formed by Concave and Convex Mirrors: Ray Rules and Nature of Images: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Magnified virtual image · 1 mark · MCQBoard level

A concave mirror forms a virtual, erect, magnified image when the object is placed:
(a) beyond C
(b) at C
(c) between C and F
(d) between F and P

Try it first, then show the solutionSolution

Only an object between $F$ and the pole gives a virtual, erect, magnified image.

Answer: (d)

Example 2: Convex mirror image · 2 marks · Short answerBoard level

State the position and nature of the image formed by a convex mirror of an object placed (a) far away, (b) close to the mirror. Give one use of the mirror based on this.

Try it first, then show the solutionSolution

In both cases the image is behind the mirror between $P$ and $F$ (near $F$ for a far object, near $P$ for a close one), virtual, erect and diminished. (1)
Use: rear-view mirror (erect image, wide field of view). (1)

Answer: Behind the mirror, between P and F, virtual, erect, diminished

Example 3: Locating the region · 3 marks · Short answerHarder board level

A concave mirror has radius of curvature 24 cm. An object is placed 30 cm in front of it. State in which region the image lies, then find it with the mirror formula, and give its nature and magnification.

Try it first, then show the solutionSolution

$f = 12$ cm and $C$ is at 24 cm; the object is beyond $C$ , so the image lies between $F$ and $C$ , real, inverted, diminished. (1)
$\frac{1}{v} = \frac{1}{- 12} - \frac{1}{- 30} = - \frac{3}{60}$ , so $v = - 20$ cm (between 12 cm and 24 cm). (1)
$m = - \frac{v}{u} = - \frac{- 20}{- 30} = - \frac{2}{3}$ : inverted, diminished. (1)

Answer: Between F and C; v = −20 cm; m = −2/3

Example 4: Object moving towards a mirror · 3 marks · Short answerHarder board level

A concave mirror has R = 40 cm. Find the image distance and magnification for objects at 60, 40, 30, 20 and 10 cm in front of it, and state the nature of each.

Try it first, then show the solutionSolution

$f = - 20$ cm. (½)
$u = - 60$ : $v = - 30$ , $m = - 0.5$ (real, inverted, diminished). $u = - 40$ : $v = - 40$ , $m = - 1$ (real, inverted, same size). (1)
$u = - 30$ : $v = - 60$ , $m = - 2$ (real, inverted, magnified). $u = - 20$ : image at infinity. $u = - 10$ : $v = + 20$ , $m = + 2$ (virtual, erect, magnified). (1½)

Answer: v = −30, −40, −60, ∞, +20 cm; m = −0.5, −1, −2, ∞, +2

Example 5: A torch reflector off its focus · 4 marks · Case-basedHOTS / competency

A torch has a concave reflector of radius of curvature 12 cm.
(a) Find its focal length and the position of the bulb for a parallel beam.
(b) Find the image distance when the bulb is 4 cm from the pole and say what happens to the beam.
(c) Find the image distance when the bulb is 9 cm from the pole and say what happens to the beam.
(d) Which of the three positions would you choose for a long-range searchlight, and why?

Try it first, then show the solutionSolution

(a) $f = 6$ cm; bulb at 6 cm. (½)
(b) $\frac{1}{v} = - \frac{1}{6} + \frac{1}{4} = \frac{1}{12}$ , $v = + 12$ cm: virtual image behind; beam diverges. (1½)
(c) $\frac{1}{v} = - \frac{1}{6} + \frac{1}{9} = - \frac{1}{18}$ , $v = - 18$ cm: real image in front; beam converges to a point then spreads. (1½)
(d) At the focus: the beam is parallel, so it travels far without spreading. (½)

Answer: (a) 6 cm (b) +12 cm, diverging (c) −18 cm, converging then spreading (d) at the focus

Where students lose marks in Images Formed by Concave and Convex Mirrors: Ray Rules and Nature of Images

The slips examiners see most often, and what to write instead.

Mistake: Image called real for an object between F and P
Between F and P the image is virtual.
Wrong working
Real, inverted, magnified
Right working
Virtual, erect, magnified, behind the mirror
Do this: Learn the single virtual case.
Usually costs: 1 mark.
Mistake: Convex mirror image said to be magnified
It is always diminished.
Wrong working
Virtual, erect, magnified
Right working
Virtual, erect, diminished
Do this: Convex = small.
Usually costs: 1 mark.
Mistake: Reflected ray drawn without the ray rules
Rules fix the path.
Wrong working
Freehand rays.
Right working
Ray parallel to axis through F; ray through C back on itself.
Do this: Use two standard rays with arrowheads.
Usually costs: Diagram marks.

Exam tips

Draw the principal axis, mark P, F and C before the rays.
State the nature of the image as: position, real or virtual, erect or inverted, size.
Draw arrows on all rays and show backward extensions dotted for virtual images.
Give a use together with the property that makes it suitable.

Practice questions on Images Formed by Concave and Convex Mirrors: Ray Rules and Nature of Images

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
A convex mirror is preferred as a rear-view mirror mainly because it:
(a) forms a real image
(b) magnifies the traffic behind
(c) gives a wide field of view
(d) focuses the light of headlamps
Hint
Think of how much road you see.
Answer
(c) gives a wide field of view
Q2 · 1 mark · Very short answerBoard level
A concave mirror has a radius of curvature 50 cm. State its focal length.
Hint
f = R/2.
Answer
25 cm (−25 cm with Cartesian signs).
Q3 · 2 marks · Short answerHarder board level
Where must an object be placed so that a concave mirror forms an image at infinity? State the nature of the image.
Hint
Rays parallel after reflection.
Answer
At the principal focus F; the image is real, inverted and highly magnified (formed at infinity).
Q4 · 3 marks · Short answerHarder board level
A convex mirror has a focal length of 20 cm. An object is placed 60 cm in front of it. Find the image distance and the magnification, and describe the image.
Hint
f = +20, u = −60.
Answer
$\frac{1}{v} = \frac{1}{20} + \frac{1}{60} = \frac{4}{60}$ , so $v = + 15$ cm; $m = - \frac{15}{- 60} = + 0.25$ . The image is virtual, erect, diminished (a quarter of the object size), 15 cm behind the mirror.
Q5 · 2 marks · Assertion-reasonHarder board level
Assertion (A): A convex mirror can never form a real image of a real object.
Reason (R): A convex mirror makes parallel rays diverge, so the reflected rays never actually meet.
(a) Both true, R explains A.
(b) Both true, R does not explain A.
(c) A true, R false.
(d) A false, R true.
Hint
Divergence.
Answer
(a)
Q6 · 3 marks · Short answerHOTS / competency
A student looks at her face in a concave mirror of focal length 20 cm. Show with the mirror formula what she sees when her face is (a) 10 cm and (b) 30 cm from the mirror, and say which position suits a shaving mirror.
Hint
Use u = −10 and −30 with f = −20.
Answer
(a) $\frac{1}{v} = - \frac{1}{20} + \frac{1}{10} = \frac{1}{20}$ , $v = + 20$ cm, $m = + 2$ : erect, twice as large. (b) $\frac{1}{v} = - \frac{1}{20} + \frac{1}{30} = - \frac{1}{60}$ , $v = - 60$ cm, $m = - 2$ : real, inverted. The shaving mirror needs position (a), inside $F$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Focal length and radius of curvature?Show answer
f = R/2.
2.Image of an object at C in a concave mirror?Show answer
At C, real, inverted, same size.
3.When does a concave mirror give a virtual image?Show answer
When the object is between F and P.
4.Convex mirror image?Show answer
Virtual, erect, diminished, between P and F.
5.Why is a convex mirror used as a rear-view mirror?Show answer
Erect image and wide field of view.

Images Formed by Concave and Convex Mirrors: Ray Rules and Nature of Images: frequently asked questions

What images are formed by a concave mirror?

Real and inverted for objects beyond F (diminished beyond C, same size at C, magnified between C and F); virtual, erect and magnified for objects between F and P.

Why does a convex mirror always give a virtual image?

It diverges parallel rays, so the reflected rays never actually meet in front of the mirror; they only appear to come from a point behind it.

Which mirror is used in a torch and why?

A concave mirror, because a bulb at its focus gives a parallel beam.