ScienceCBSE Class 10 · Chapter 9

Light – Reflection and Refraction Class 10: Notes and Solved Questions

Q: Do I need to derive the mirror formula or the lens formula for the board exam?

No. The 2026-27 syllabus says the derivation is not required for either formula. You must know both formulas, apply the sign convention correctly and solve numericals with them.

Q: Do I have to memorise the refractive index table?

No. NCERT says you need not memorise the data. You should be able to read the table, pick the optically densest or rarest medium, and compare speeds of light in different media.

Q: Why is there a minus sign in mirror magnification but not in lens magnification?

For a mirror, a real image forms in front, on the same side as the object, so v and u are both negative; the minus sign makes m negative for real images. For a lens, a real image forms on the other side, so v is positive and u negative, and v divided by u is already negative. Both rules give negative m for real, inverted images.

Q: Is the formula for the power of lenses in contact part of the syllabus?

The idea that powers of lenses in contact add up appears in a More to Know box in NCERT, and CBSE says information in text boxes is not assessed in the board exam. It is simple to understand, but focus your preparation on the main text: power, focal length and the sign of power.

Q: How do I remember which rays to use in a ray diagram?

Choose the two rays whose paths you know for sure. For mirrors, the ray parallel to the axis and the ray through or towards C are usually easiest; for lenses, the ray parallel to the axis and the ray through the optical centre. Always add arrowheads and use dotted lines for virtual rays and images.

Updated 14 October 2026 · Fact-checked

Light – Reflection and Refraction is the Class 10 Physics chapter on how spherical mirrors and lenses form images, how light bends when it changes medium, and what refractive index and lens power mean. The board asks image-nature MCQs, assertion-reason items, ray diagrams, mirror and lens formula numericals with sign convention, and practical-based questions.

What Light – Reflection and Refraction is about

This chapter treats light as rays that travel in straight lines. You first see how curved (spherical) mirrors reflect light and form images. Then you see how light bends, or refracts, when it passes from one transparent medium into another: through a glass slab and then through convex and concave lenses. Two formulas, the mirror formula and the lens formula, let you calculate where an image forms and how big it is. Refractive index and the power of a lens put numbers on how strongly light is bent.

The chapter builds on the laws of reflection and plane mirrors from earlier classes. It uses simple coordinate signs (left is negative, right is positive) and fractions with reciprocals, so your maths must be steady. The idea of lens power, measured in dioptres, is how opticians describe spectacle lenses, so it comes back whenever corrective lenses are discussed.

Board questions take several shapes. There are 1-mark MCQs on the nature and position of images and assertion-reason items that test a reason, not just a fact. There are 2- and 3-mark questions asking for reasons, definitions or a sign-convention numerical, and ray diagrams to draw and label. Case-based and data-based questions give you a passage, a ray figure in words or a table of object and image distances to analyse. Practical-based questions draw on finding focal length using a distant object and tracing rays through a glass slab.

Weightage: This chapter is part of Unit III, Natural Phenomena, which carries 12 of the 80 theory marks together with The Human Eye and the Colourful World.Source: CBSE Science syllabus 2026-27, Unit III Natural Phenomena

In real board papers: asked in 21 of the 21 board papers we hold (2023 to 2026), on average 7.6 marks per paper. See the questions

Key concepts

Reflection and the language of spherical mirrors

NCERT section 9.2 Spherical Mirrors

The laws of reflection you learnt for plane mirrors still hold on a curved surface: the angle of incidence equals the angle of reflection, and the incident ray, the normal and the reflected ray lie in one plane. A spherical mirror is a piece of the surface of a sphere. If the shiny side faces the centre of that sphere (curved inwards, like the inside of a spoon) it is a concave mirror. If it bulges outwards it is a convex mirror.

Learn the labels, because every ray rule uses them. The pole P is the centre of the mirror's surface. The centre of curvature C is the centre of the sphere. It is not on the mirror: it lies in front of a concave mirror and behind a convex one. PC is the radius of curvature R. The principal axis joins P and C and is normal to the mirror at P. Rays parallel to the axis meet at (concave) or seem to come from (convex) the principal focus F. PF is the focal length f. For small apertures $R = 2 f$ , so F sits midway between P and C.

Worked example

The centre of curvature of a spherical mirror lies behind its reflecting surface, and its radius of curvature is 36 cm. Using the New Cartesian Sign Convention, its focal length is:
(a) −18 cm
(b) +18 cm
(c) +36 cm
(d) −72 cm

Try it first, then show the solutionSolution

C lies behind the mirror, so the mirror bulges outwards: it is a convex mirror.
Focal length is half the radius of curvature: $f = \frac{R}{2} = \frac{36}{2} = 18$ cm.
The focus of a convex mirror is behind it, to the right of the pole, so f is positive.

Answer: (b) +18 cm

Images formed by a concave mirror and ray diagrams

NCERT section 9.2.1 Image Formation by Spherical Mirrors

A concave mirror makes reflected rays converge, so the image depends on where the object stands relative to P, F and C. Learn the pattern, not six loose facts:

At infinity: image at F, point-sized, real, inverted.
Beyond C: image between F and C, diminished, real, inverted.
At C: image at C, same size, real, inverted.
Between C and F: image beyond C, enlarged, real, inverted.
At F: image at infinity.
Between P and F: image behind the mirror, enlarged, virtual, erect.

To draw any case, use two of four rays: parallel to the axis reflects through F; through F reflects parallel; through C comes straight back, because it strikes along the normal; aimed at P reflects making equal angles with the axis. The image is where two reflected rays cross.

This explains the uses: torches and headlights (source at F gives a parallel beam), shaving and dentists' mirrors (object inside F gives an enlarged erect image) and solar furnaces (sunlight concentrated to produce heat).

Worked example

A concave mirror has a focal length of 12 cm.
(a) Where must an object be placed to get a real, inverted image of the same size?
(b) The object is then moved to 8 cm in front of the mirror. Use the mirror formula to find the image position and magnification.
(c) Can the image in (b) be obtained on a screen? Give a reason. (3 marks)

Try it first, then show the solutionSolution

(a) A same-size real image forms when the object is at C. (½)
C is at $2 f = 24$ cm, so the object is 24 cm in front of the mirror, $u = - 24$ cm. (½)
(b) $u = - 8$ cm, $f = - 12$ cm; $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = - \frac{1}{12} + \frac{1}{8}$ (½)
$\frac{1}{v} = \frac{- 2 + 3}{24} = \frac{1}{24}$ , so $v = + 24$ cm, behind the mirror. (½)
$m = - \frac{v}{u} = - \frac{24}{- 8} = + 3$ (½)
(c) Positive v and m mean the image is virtual, erect and three times enlarged, so it cannot be obtained on a screen. (½)

Answer: (a) At C, 24 cm in front of the mirror. (b) v = +24 cm (24 cm behind the mirror), m = +3. (c) No, the image is virtual and erect.

Images formed by a convex mirror and its uses

NCERT section 9.2.2 Representation of Images Formed by Spherical Mirrors Using Ray Diagrams

A convex mirror spreads reflected rays apart. They never meet in front of the mirror, so you extend them backwards with dotted lines; they appear to meet behind the mirror. So the image is always virtual and erect, and it always lies between P and F behind the mirror.

There are only two cases. An object at infinity gives a point-sized image at F. An object anywhere between infinity and P gives a diminished, virtual, erect image between P and F. As the object moves away, the image becomes smaller and moves towards F (Activity 9.5), but it never goes past F.

For ray diagrams: a ray parallel to the axis reflects as if coming from F; a ray aimed at F reflects parallel to the axis; a ray aimed at C comes back along itself.

Because the surface bulges outwards, it shows a much wider area than a plane mirror, and you can see the full image of a tall tree or building in a small convex mirror. An always-upright image plus a wider field of view is why vehicles use convex mirrors as rear-view (wing) mirrors.

Worked example

A driver notices that the image of a truck in her side mirror is always upright and smaller than the truck, and that it gets smaller still as the truck falls further behind.
(a) Name the mirror and state where the image is formed.
(b) Give two reasons why this mirror is preferred as a rear-view mirror.
(c) As the truck moves away, towards which point does the image move? (3 marks)

Try it first, then show the solutionSolution

(a) Convex mirror. (½)
The image is virtual, formed behind the mirror between the pole P and the focus F. (½)
(b) It always gives an erect image, whatever the distance of the vehicle behind. (½)
It is curved outwards, so it has a wider field of view and shows a much larger area than a plane mirror. (½)
(c) As the object moves away, the image becomes smaller. (½)
The image moves closer to the principal focus F, reaching F only when the object is at infinity. (½)

Answer: (a) Convex mirror; image behind the mirror between P and F. (b) Always erect image and wider field of view. (c) The image moves towards F.

New Cartesian sign convention, mirror formula and magnification

NCERT section 9.2.4 Mirror Formula and Magnification

Numericals need a fixed rule for signs, and CBSE uses the New Cartesian Sign Convention. Put the pole at the origin and the principal axis along the x-axis. The object is always on the left, so light travels left to right. Measure every distance from the pole: to the right is positive, to the left negative. Heights above the axis are positive, below negative.

Consequences to know before you start: $u$ is always negative; $f$ is negative for a concave mirror (F in front) and positive for a convex mirror (F behind); a real image in front has negative $v$ , a virtual image behind has positive $v$ .

The mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ links object distance, image distance and focal length for every spherical mirror and every object position. You use it; you do not derive it. Magnification $m = \frac{h ^{'}}{h} = - \frac{v}{u}$ gives size and nature together: negative $m$ means real and inverted, positive means virtual and erect, and a value above 1 in size means enlarged. Always finish by stating position, nature and size in words.

Worked example

An object 2 cm tall is placed 15 cm in front of a concave mirror of focal length 10 cm. At what distance from the mirror should a screen be placed to get a sharp image? Find the size and nature of the image. (3 marks)

Try it first, then show the solutionSolution

Data with signs: $u = - 15$ cm, $f = - 10$ cm, $h = + 2$ cm. (½)
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = - \frac{1}{10} + \frac{1}{15}$ (½)
$\frac{1}{v} = \frac{- 3 + 2}{30} = - \frac{1}{30}$ , so $v = - 30$ cm: screen 30 cm in front of the mirror. (½)
$m = - \frac{v}{u} = - \frac{- 30}{- 15} = - 2$ (½)
$h^{'} = m \times h = - 2 \times 2 = - 4$ cm (½)
Negative v and m: the image is real, inverted and twice the size of the object. (½)

Answer: Screen at 30 cm in front of the mirror (v = −30 cm); image 4 cm tall (h′ = −4 cm), real, inverted and enlarged.

Refraction of light and the rectangular glass slab

NCERT section 9.3.1 Refraction through a Rectangular Glass Slab

Inside one transparent medium light goes in straight lines, but when it passes obliquely into another medium its direction changes. This is refraction, and it happens because light travels at different speeds in different media. That is why a pencil in water looks displaced at the surface, a pond looks shallower and letters under a glass slab look raised.

The bending follows the speed. From a rarer to a denser medium (air to glass) light slows and bends towards the normal. From denser to rarer it speeds up and bends away from the normal.

In the glass slab activity the ray bends towards the normal at the first face and away from it at the second. The faces are parallel, so the two bendings are equal and opposite: the emergent ray is parallel to the incident ray but shifted slightly sideways. So the angle of emergence equals the angle of incidence.

The laws of refraction: the incident ray, refracted ray and normal lie in one plane; and for a given colour and pair of media, $\frac{s i n i}{s i n r}$ is constant (Snell's law), for $0 < i < 9 0^{\circ}$ .

Worked example

In a practical, a student traces a ray through a rectangular glass slab. For an angle of incidence of 45°, she measures the angle of refraction as 28° and the angle of emergence as 45°.
(a) Why is the angle of refraction smaller than the angle of incidence?
(b) Why is the angle of emergence equal to the angle of incidence?
(c) Does the emergent ray lie along the same straight line as the incident ray? (3 marks)

Try it first, then show the solutionSolution

(a) At the first face light goes from air (rarer) into glass (denser), slows down and bends towards the normal, so r is less than i. (1)
(b) The bending at the air-glass face and at the glass-air face is equal and opposite because the two faces are parallel, so the emergent ray is parallel to the incident ray and e = i. (1)
(c) No. The emergent ray is parallel to the incident ray but shifted sideways from it. (1)

Answer: (a) Light slows in glass and bends towards the normal. (b) Equal and opposite bending at parallel faces makes the emergent ray parallel to the incident ray. (c) No, it is parallel but laterally shifted.

Refractive index and optical density

NCERT section 9.3.2 The Refractive Index

The constant in Snell's law is the refractive index of the second medium with respect to the first. It measures how much a pair of media changes the direction of light, and it is linked to speed: $n_{21} = \frac{v _{1}}{v _{2}}$ , the speed in medium 1 divided by the speed in medium 2.

When the first medium is vacuum (or air, which is almost the same) you get the absolute refractive index $n = \frac{c}{v}$ , with $c = 3 \times 1 0^{8} m s^{- 1}$ . Light is fastest in vacuum, so a material's refractive index is more than 1. Water has 1.33: light in air is 1.33 times as fast as in water. Diamond, at 2.42, has the highest value in the NCERT table. You need not memorise the table, but you must be able to read it.

The medium with the larger refractive index is optically denser: light is slower in it, and a ray entering it bends towards the normal. Optical density is not mass density. Kerosene (1.44) is optically denser than water (1.33) although its mass density is less.

Worked example

The speed of light is $3 \times 1 0^{8} m s^{- 1}$ in vacuum, $2.25 \times 1 0^{8} m s^{- 1}$ in water and $2 \times 1 0^{8} m s^{- 1}$ in a type of glass.
(a) Find the absolute refractive index of the glass and of water.
(b) Find the refractive index of the glass with respect to water.
(c) A ray passes obliquely from water into this glass. Does it bend towards or away from the normal? Why? (3 marks)

Try it first, then show the solutionSolution

(a) $n_{g} = \frac{c}{v _{g}} = \frac{3 \times 1 0 ^{8}}{2 \times 1 0 ^{8}} = 1.5$ ; $n_{w} = \frac{3 \times 1 0 ^{8}}{2.25 \times 1 0 ^{8}} = \frac{4}{3} \approx 1.33$ (1)
(b) $n_{g w} = \frac{v _{w}}{v _{g}} = \frac{2.25 \times 1 0 ^{8}}{2 \times 1 0 ^{8}} = 1.125$ (1)
(c) Glass has the larger refractive index, so it is optically denser; light slows on entering it and bends towards the normal. (1)

Answer: (a) Glass 1.5, water about 1.33. (b) 1.125 (no unit). (c) Towards the normal, because glass is optically denser than water.

Spherical lenses and the images they form

NCERT section 9.3.4 Image Formation by Lenses

A lens is a transparent material bounded by two surfaces, at least one of them spherical. A convex lens is thicker in the middle and converges parallel rays. A concave lens is thicker at the edges and diverges them. Each lens has two centres of curvature, a principal axis through them, an optical centre O, and two principal foci $F_{1}$ and $F_{2}$ , one on each side.

Three rays build every lens diagram: a ray parallel to the axis passes through F on the far side (convex) or appears to come from F on the same side (concave); a ray through F, or heading towards F for a concave lens, emerges parallel; a ray through O goes straight on.

The convex lens table matches the concave mirror table with $2 F$ in place of C. Beyond $2 F_{1}$ : diminished real image between $F_{2}$ and $2 F_{2}$ . At $2 F_{1}$ : same-size real image at $2 F_{2}$ . Between $F_{1}$ and $2 F_{1}$ : enlarged real image beyond $2 F_{2}$ . At $F_{1}$ : image at infinity. Between $F_{1}$ and O: enlarged, virtual, erect image on the object's side. A concave lens always gives a virtual, erect, diminished image between F and O.

Worked example

A student has a convex lens of focal length 15 cm.
(a) Where should she place a candle to get a real image of the same size on a screen, and where will the screen be?
(b) Where should the candle be to see an enlarged, erect image through the lens?
(c) Her friend says a concave lens could also give a real image if the candle were placed close enough. Is he right? (3 marks)

Try it first, then show the solutionSolution

(a) Object at $2 F_{1}$ , that is 30 cm from the lens; the same-size real, inverted image forms at $2 F_{2}$ , 30 cm on the other side, where the screen goes. (1)
(b) Object between $F_{1}$ and O, that is less than 15 cm from the lens; the image is virtual, erect, enlarged and on the same side as the candle. (1)
(c) No. A concave lens always forms a virtual, erect, diminished image, whatever the position of the object. (1)

Answer: (a) Candle 30 cm from the lens, screen 30 cm on the other side. (b) Candle closer than 15 cm. (c) No, a concave lens never gives a real image.

Lens sign convention, lens formula and magnification

NCERT section 9.3.7 Lens Formula and Magnification

Lenses use the same sign convention as mirrors, but every distance is measured from the optical centre O. The object is on the left, so $u$ is negative. A convex lens has positive $f$ and a concave lens has negative $f$ .

The lens formula is $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ . Notice the minus sign, unlike the mirror formula. The reason is the geometry: a lens forms a real image on the far side (the right, positive $v$ ), while a mirror forms a real image in front (the left, negative $v$ ).

Magnification for a lens is $m = \frac{h ^{'}}{h} = \frac{v}{u}$ , with no minus sign. Again negative $m$ means real and inverted, positive means virtual and erect. Check: a real image has positive $v$ , and with negative $u$ that gives negative $m$ , as it should.

In a numerical, write the data with signs, rearrange to $\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$ (or solve for $u$ ), substitute in brackets, then use the signs of $v$ and $m$ to say where the image is and what it is like. You use the formula; no derivation is asked.

Worked example

An object 3 cm tall is placed 30 cm from a convex lens of focal length 20 cm, perpendicular to its principal axis. Find the position, size and nature of the image. (3 marks)

Try it first, then show the solutionSolution

Data with signs: $u = - 30$ cm, $f = + 20$ cm, $h = + 3$ cm. (½)
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{20} + \frac{1}{- 30}$ (½)
$\frac{1}{v} = \frac{3 - 2}{60} = \frac{1}{60}$ , so $v = + 60$ cm, on the other side of the lens. (½)
$m = \frac{v}{u} = \frac{60}{- 30} = - 2$ (½)
$h^{'} = m \times h = - 2 \times 3 = - 6$ cm (½)
Negative m and h′: the image is real, inverted and twice as large, formed below the axis. (½)

Answer: Image 60 cm on the other side of the lens (v = +60 cm), 6 cm tall (h′ = −6 cm), real, inverted and enlarged (m = −2).

Power of a lens

NCERT section 9.3.8 Power of a Lens

Some lenses bend light more sharply than others. A convex lens of short focal length brings rays to a focus close to the lens, so it converges strongly; a concave lens of short focal length diverges strongly. Power expresses this degree of convergence or divergence: $P = \frac{1}{f}$ .

The SI unit is the dioptre (D), and it works only when $f$ is in metres: $1 D = 1 m^{- 1}$ , the power of a lens of focal length 1 m. If $f$ is in centimetres, use $P = \frac{100}{f}$ with $f$ in cm, or convert to metres first.

Power carries the sign of the focal length: convex lenses have positive power, concave lenses negative. So the sign tells you the type of lens straight away. Opticians write prescriptions in power. For example, +4.0 D means a convex lens of focal length +0.25 m, and −2.0 D means a concave lens of focal length −0.50 m. The bigger the size of the power, the shorter the focal length and the stronger the bending.

Worked example

Lens A has a power of +4.0 D and lens B has a power of −2.5 D.
(a) Find the focal length of each lens in cm.
(b) Name the type of each lens.
(c) Which lens bends light more strongly? Give a reason. (3 marks)

Try it first, then show the solutionSolution

(a) $f_{A} = \frac{1}{P} = \frac{1}{4.0} = + 0.25$ m $= + 25$ cm (½)
$f_{B} = \frac{1}{- 2.5} = - 0.40$ m $= - 40$ cm (½)
(b) Positive power, so A is convex (converging). (½)
Negative power, so B is concave (diverging). (½)
(c) Lens A, because its power has the larger magnitude (4.0 D against 2.5 D). (½)
Larger power means shorter focal length, so the rays are bent through larger angles. (½)

Answer: (a) f(A) = +25 cm, f(B) = −40 cm. (b) A convex, B concave. (c) Lens A, as it has greater power and shorter focal length.

Science in real life: Light – Reflection and Refraction

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.

Images formed by a concave mirror and ray diagrams

Suppose you open an old torch and find a small shiny curved mirror behind the bulb, curved inwards towards the bulb. You measure its radius of curvature as 6 cm. You want to refit the bulb so the torch throws a strong straight beam.

Name the mirror, state where the bulb must be placed to get a parallel beam, and find its distance from the pole. What would happen to the beam if the bulb were placed a little beyond the focus instead?

Try it first, then show the solutionSolution

The mirror curves inwards towards the bulb, so it is a concave mirror.
A ray passing through the principal focus of a concave mirror reflects parallel to the principal axis.
So the bulb must be at the principal focus F to send out a parallel beam.
Focal length $f = \frac{R}{2} = \frac{6}{2} = 3$ cm, so the bulb is 3 cm in front of the pole.
If the bulb is between F and C, the reflected rays converge to form an image beyond C, so the beam is no longer parallel.

Answer: Concave mirror; bulb at the principal focus, 3 cm from the pole. Beyond F the rays would converge instead of forming a parallel beam.

Memory hook: Bulb at F, beam goes parallel: the torch is a concave mirror run backwards.

Images formed by a convex mirror and its uses

Imagine your family car has a side mirror that curves outwards, with a focal length of 2 m. A car 1.6 m tall is 6 m behind it on the road.

Find the position, nature and height of the image of the car in the mirror. Why is this mirror used as a rear-view mirror?

Try it first, then show the solutionSolution

Convex mirror: $f = + 2$ m, $u = - 6$ m, $h = + 1.6$ m.
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{2} + \frac{1}{6} = \frac{3 + 1}{6} = \frac{2}{3}$
$v = + 1.5$ m, so the image is 1.5 m behind the mirror, between P and F.
$m = - \frac{v}{u} = - \frac{1.5}{- 6} = + 0.25$
$h^{'} = m \times h = 0.25 \times 1.6 = + 0.4$ m
Positive m: the image is virtual and erect, and only a quarter of the car's height, so it is diminished.
A convex mirror always gives an erect, diminished image, and because it is curved outwards it has a wider field of view, so the driver sees a much larger area behind than a plane mirror would show.

Answer: Image 1.5 m behind the mirror, virtual, erect and diminished, 0.4 m tall (m = +0.25). It is used as a rear-view mirror because it always gives an erect image and has a wider field of view.

Memory hook: Wing mirror: always upright, smaller, and shows a wide area.

New Cartesian sign convention, mirror formula and magnification

Suppose your father uses a concave shaving mirror of focal length 20 cm. He holds his face 10 cm in front of it.

Find the position of the image and the magnification. State the nature of the image and explain why this mirror suits shaving.

Try it first, then show the solutionSolution

Data: $u = - 10$ cm, $f = - 20$ cm (concave mirror).
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = - \frac{1}{20} + \frac{1}{10} = \frac{- 1 + 2}{20} = \frac{1}{20}$
$v = + 20$ cm, so the image is 20 cm behind the mirror.
$m = - \frac{v}{u} = - \frac{20}{- 10} = + 2$
Positive m: virtual, erect and twice the size of the face.
The face is between P and F, which gives an enlarged erect image, so fine details are easy to see.

Answer: Image 20 cm behind the mirror, m = +2: virtual, erect and enlarged two times.

Memory hook: Face inside F of a concave mirror: big and upright, perfect for shaving.

Refraction of light and the rectangular glass slab

Imagine a coin lying at the bottom of a bucket full of water in your courtyard. You look at it from one side, above the water, and try to pick it up in one go, but your fingers land in the wrong place.

Explain, using refraction, why you miss the coin and how you should change your reach. Would the error be exactly the same if the bucket held a different clear liquid?

Try it first, then show the solutionSolution

Light from the coin travels obliquely from water (optically denser) into air (rarer).
At the surface it speeds up and bends away from the normal.
Your eye assumes light travels in a straight line, so it traces the ray back to a point higher than the real coin.
The coin therefore appears raised; it is actually deeper than it looks, so you should reach lower than where it seems to be.
The amount of bending depends on the pair of media, so with a different liquid the coin would appear raised by a different amount.

Answer: The coin appears raised because light bends away from the normal as it leaves water; reach deeper than it appears. The shift would differ for a different liquid.

Memory hook: Water lifts the coin to your eye: reach below what you see.

Lens sign convention, lens formula and magnification

Suppose your grandmother uses a convex magnifying glass of focal length 10 cm to read tiny print on a medicine strip. She holds the lens 5 cm above the print.

Find where the image of the print is formed and its magnification. State its nature and explain why she must hold the lens closer than 10 cm.

Try it first, then show the solutionSolution

Data: $u = - 5$ cm, $f = + 10$ cm (convex lens).
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} - \frac{1}{5} = \frac{1 - 2}{10} = - \frac{1}{10}$
$v = - 10$ cm: the image is 10 cm from the lens on the same side as the print.
$m = \frac{v}{u} = \frac{- 10}{- 5} = + 2$
Positive m: virtual, erect and twice the size.
Only an object between F and O of a convex lens gives an enlarged erect image, so the print must be within 10 cm of the lens.

Answer: Image 10 cm from the lens on the same side as the print (v = −10 cm), m = +2, virtual, erect and enlarged.

Memory hook: Magnifier rule: print inside F, letters big and upright.

Power of a lens

Imagine you and a friend compare spectacle prescriptions after an eye test. Yours reads −2.0 D and your friend's reads +2.5 D.

Find the focal length of each lens in cm, state which lens is converging and which is diverging, and say which lens bends light more strongly.

Try it first, then show the solutionSolution

Your lens: $f = \frac{1}{P} = \frac{1}{- 2.0} = - 0.50$ m $= - 50$ cm.
Friend's lens: $f = \frac{1}{+ 2.5} = + 0.40$ m $= + 40$ cm.
Negative power and focal length: your lens is concave, a diverging lens.
Positive power: your friend's lens is convex, a converging lens.
Your friend's lens has the larger power in size (2.5 D against 2.0 D) and a shorter focal length, so it bends light more strongly.

Answer: Yours: −50 cm, concave (diverging). Friend's: +40 cm, convex (converging). The +2.5 D lens bends light more strongly.

Memory hook: Sign on the prescription tells the lens: plus convex, minus concave.

Topics in this chapter

Reflection of Light and Spherical MirrorsComing soon
Image Formation by Concave and Convex MirrorsComing soon
Mirror Formula, Magnification and Sign Convention
Refraction of Light and Refractive Index
Image Formation by Spherical LensesComing soon
Lens Formula, Magnification and Power of a Lens

Light – Reflection and Refraction: formulas and key facts

Laws of reflection: $A n g l eo f in c i d e n ce = an g l eo f r e f l ec t i o n, an d t h e in c i d e n t r a y, n or ma l a tt h e p o in t o f in c i d e n ce an d r e f l ec t e d r a y l i e in t h es am e pl an e .$; Hold for plane and spherical mirrors alike. Every ray-diagram rule for mirrors follows from them.
Radius of curvature and focal length: $R = 2 f$; For small-aperture spherical mirrors. F lies midway between P and C. Use it to get f when R is given.
Mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$; For all spherical mirrors and all object positions. Substitute with New Cartesian signs. Derivation not required.
Magnification by a mirror: $m = \frac{h ^{'}}{h} = - \frac{v}{u}$; Note the minus sign for mirrors. Negative m: real and inverted; positive m: virtual and erect.
Plane mirror image: Virtual, erect, same size, as far behind as the object is in front, laterally inverted; $m = + 1$ .; Used for comparison MCQs. m = +1 means same size and erect (virtual).
New Cartesian Sign Convention: Measured from pole or optical centre
Right of origin positive, left negative
Above axis positive, below negative
Object always on the left; So u is always negative. f: concave mirror −, convex mirror +, convex lens +, concave lens −.
Snell's law (second law of refraction): $\frac{s i n i}{s i n r} = constant$; For a given colour and pair of media, with 0 < i < 90°. The constant is the refractive index of the second medium with respect to the first.
First law of refraction: The incident ray, the refracted ray and the normal at the point of incidence lie in the same plane.; Quote with Snell's law when asked to state the laws of refraction.
Relative refractive index: $n_{21} = \frac{v _{1}}{v _{2}}$; Refractive index of medium 2 with respect to medium 1 = speed in medium 1 divided by speed in medium 2. No unit.
Absolute refractive index: $n_{m} = \frac{c}{v}$; c is speed in vacuum or air, v is speed in the medium. Larger n means optically denser and slower light.
Speed of light in vacuum: $c = 3 \times 1 0^{8} m s^{- 1}$; Light is fastest in vacuum; only slightly slower in air; much slower in glass or water.
Lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$; For all spherical lenses. Distances from the optical centre. Note the minus sign. Derivation not required.
Magnification by a lens: $m = \frac{h ^{'}}{h} = \frac{v}{u}$; No minus sign for lenses. Negative m: real, inverted; positive m: virtual, erect.
Power of a lens: $P = \frac{1}{f}$; f must be in metres to get dioptres. Convex lens: positive power; concave lens: negative power.
Unit of power: $1 D = 1 m^{- 1}$; One dioptre is the power of a lens of focal length 1 m. With f in cm, P = 100/f.
Bending rule: Rarer to denser: slows, bends towards the normal. Denser to rarer: speeds up, bends away from the normal.; Use for reasoning questions: glass slab, coin in water, pencil in water, apparent raising.

Measured from	pole or optical centre
Right of origin	positive, left negative
Above axis	positive, below negative
Object	always on the left

Light – Reflection and Refraction: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Image formation by a concave mirror (six object positions)

2- or 3-mark ray diagram questions, often part of a 5-mark answer: 'Draw a ray diagram to show the image formed when the object is placed between F and C' or 'to get a virtual, enlarged image'.

Draw

Draw the concave mirror as a curve bulging away from the object, shaded on the back (non-reflecting) side, with the principal axis as a horizontal line through the pole P.
Mark F and C on the principal axis in front of the mirror, with F midway between P and C.
Draw the object as an upright arrow standing on the axis at the position asked (beyond C, at C, between C and F, at F, or between P and F).
Ray 1: from the top of the object, parallel to the principal axis; after reflection it passes through F.
Ray 2: from the top of the object through F (or through C); after reflection it goes parallel to the axis (or comes straight back along the same path).
Where the reflected rays meet in front of the mirror, draw the real image as an inverted arrow.
For an object between P and F, extend the reflected rays behind the mirror as dotted lines; where they meet, draw the virtual, erect, enlarged image behind the mirror.

Label

P (pole)
F (principal focus)
C (centre of curvature)
Principal axis
Object AB
Image A'B'
Arrows on every incident and reflected ray
Dotted lines for rays extended behind the mirror

Watch out: Rays must obey the laws of reflection. Put arrows on every ray. Draw virtual rays (behind the mirror) as dotted lines only. A very common slip is drawing the ray through F reflecting back through F instead of going parallel to the axis.

Image formation by a convex mirror

2- or 3-mark questions on rear-view mirrors, or 'Draw a ray diagram for an object placed in front of a convex mirror'.

Draw

Draw the convex mirror bulging towards the object, shaded on the side away from the object.
Mark P on the mirror, and F and C on the principal axis behind the mirror (F midway between P and C).
Draw the object as an upright arrow in front of the mirror.
Ray 1: from the top of the object parallel to the axis; after reflection it goes away from the axis. Extend it backwards as a dotted line so that it appears to come from F.
Ray 2: from the top of the object aimed towards C; it reflects straight back along the same path. Extend it as a dotted line to C.
Where the two dotted lines meet behind the mirror, between P and F, draw a small, erect image.

Label

P
F (behind the mirror)
C (behind the mirror)
Principal axis
Object
Virtual, erect, diminished image
Arrows on incident and reflected rays
Dotted extensions

Watch out: F and C of a convex mirror are behind the mirror. The image is always between P and F, erect and smaller. Do not draw the image beyond F or show it inverted.

Refraction through a rectangular glass slab

2- or 3-mark questions and practical-based questions on Activity 9.10 / Practical 11: 'Draw the path of a ray through a glass slab and mark i, r and e'.

Draw

Draw the rectangle ABCD for the slab outline.
Draw the incident ray EO striking the upper face AB obliquely at O, with an arrow towards the slab.
Draw the normal NN' at O as a dashed line perpendicular to AB.
Draw the refracted ray OO' inside the slab, bent towards the normal, meeting the lower face CD at O'.
Draw the normal MM' at O' perpendicular to CD.
Draw the emergent ray O'H bent away from the normal, parallel to the incident ray.
Extend the incident ray EO as a dotted line beyond O to show that the emergent ray is shifted sideways.

Label

A, B, C, D (slab corners)
Incident ray EO
Refracted ray OO'
Emergent ray O'H
Normals NN' and MM'
Angle of incidence i
Angle of refraction r
Angle of emergence e
Lateral shift between the dotted line and O'H

Watch out: Measure all angles from the normal, not from the surface. The emergent ray must be parallel to the incident ray, not along the same line. A common slip is bending the ray away from the normal on entering the glass.

Image formation by a convex lens (six object positions)

2- or 3-mark ray diagrams, and part of 5-mark lens answers: 'Draw a ray diagram for an object between F1 and 2F1' or 'to use the lens as a magnifying glass'.

Draw

Draw the convex lens as a thin double-bulging outline (or a vertical line with outward arrowheads) with the principal axis through its optical centre O.
Mark $F_{1}$ and $2 F_{1}$ on the object side and $F_{2}$ and $2 F_{2}$ on the other side, equally spaced.
Draw the upright object at the asked position.
Ray 1: from the top of the object parallel to the axis; after refraction it passes through $F_{2}$ .
Ray 2: from the top of the object through O; it goes straight on without deviation.
Where these meet on the far side, draw the real, inverted image.
For an object between $F_{1}$ and O, the refracted rays spread out; extend them backwards as dotted lines to meet on the object side and draw the virtual, erect, enlarged image there.

Label

O (optical centre)
$F_{1}$ , $F_{2}$
$2 F_{1}$ , $2 F_{2}$
Principal axis
Object
Image
Arrows on all rays
Dotted backward extensions for a virtual image

Watch out: The ray through O must not bend. The parallel ray bends through the focus on the opposite side ( $F_{2}$ ), not the same side. Do not forget to mark 2F, which loses the position mark.

Image formation by a concave lens

2-mark ray diagrams and numericals that ask you to draw the ray diagram for a concave lens.

Draw

Draw the concave lens (thin in the middle, thick at the edges, or a vertical line with inward arrowheads) and the principal axis through O.
Mark $F_{1}$ on the object side and $F_{2}$ on the other side.
Draw the upright object anywhere on the left.
Ray 1: from the top of the object parallel to the axis; after refraction it bends away from the axis. Extend it backwards as a dotted line so it appears to come from $F_{1}$ .
Ray 2: from the top of the object through O, undeviated.
Where the dotted line meets ray 2, between $F_{1}$ and O, draw a small, erect image.

Label

O
$F_{1}$ , $F_{2}$
Principal axis
Object
Virtual, erect, diminished image
Arrows on rays
Dotted extension to $F_{1}$

Watch out: The diverging ray must be traced back to the focus on the same side as the object. The image is always between $F_{1}$ and O, erect and diminished. Never show it on the far side or inverted.

How the board asks Light – Reflection and Refraction

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
MCQ on image position and nature	A 1-mark question gives the nature of an image (for example virtual, erect, enlarged) and asks where the object is, or gives a focal length sign and asks which mirror and lens it could be.	Picture the tables: for a concave mirror or convex lens only one position (between the pole or optical centre and F) gives a virtual erect image. Convex mirrors and concave lenses always give virtual, erect, diminished images. Check sign: f negative means concave mirror or concave lens.
Assertion-reason	Two statements, for example about a convex mirror, apparent raising of a coin in water, or kerosene being optically denser than water. You choose (a) to (d).	Judge A and R separately as true or false first. If a number is involved, run the formula quickly (for a convex mirror with u = −f, the image is at v = +f/2, not at infinity). Only if both are true, ask whether R actually explains A.
Mirror or lens numerical (2-3 marks)	Object height, object distance and focal length (or R) are given; find image position, size and nature, or find u or f from given data such as magnification.	Write a data line with signs. Write the correct formula (mirror: plus; lens: minus). Substitute in brackets, solve with a common denominator, then find m (mirror −v/u, lens v/u) and h′. End with a sentence on position, nature and size.
Ray diagram (2-3 marks, sometimes part of a 5-mark answer)	Draw the image formed by a concave mirror or convex lens for a stated object position, or by a convex mirror or concave lens, and state the nature of the image.	Draw and label checklist: principal axis as a straight line; mirror or lens with P or O, F and C (or F1, F2, 2F1, 2F2) marked at correct spacing; object as an upright arrow; two standard rays with arrowheads; real rays solid, virtual extensions dotted; image drawn where rays meet (dotted for virtual). Common slip: missing arrowheads or drawing a virtual image with solid lines.
Refractive index and data tables	Speeds of light or a table of refractive indices are given; find refractive index, speed in a medium, the optically densest medium, or the direction of bending.	Use n = c/v or n21 = v1/v2. Larger n means optically denser and slower light. Remember optical density is not mass density. Refractive index has no unit.
Case-based or data-based (4 marks)	A passage on lenses or mirrors, or a table of object and image distances for a convex lens, with sub-questions on focal length, image nature and power.	In a u-v table, look for the row where the image and object distances are equal in size (for example u = −50 cm, v = +50 cm): the object is at 2F, so f is half that, here 25 cm. Confirm with the lens formula, then find power in dioptres with f in metres.
Practical-based	Questions on finding focal length of a concave mirror or convex lens by focusing a distant object, or on tracing a ray through a rectangular glass slab and comparing angles.	For focal length: get the sharpest, smallest image of a distant object on a screen and measure its distance from the mirror or lens. For the slab: the angle of emergence equals the angle of incidence, the angle of refraction is smaller, and the emergent ray is parallel to but shifted from the incident ray.

Tricks to remember Light – Reflection and Refraction

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

The converging cross pair
Concave MIRROR and convex LENS both converge light; convex mirror and concave lens both diverge. The names swap between mirrors and lenses.
Use it when: Quick check in MCQs: the diverging pair (convex mirror, concave lens) always gives virtual, erect, diminished images.
Mirror plus, lens minus
Mirror: plus in the formula (1/v + 1/u), minus in magnification (−v/u). Lens: minus in the formula (1/v − 1/u), plus in magnification (v/u). Each one has exactly one minus sign.
Use it when: Stops you mixing the two formulas in numericals.
Left is negative
The object is always on the left, so u is always negative. f is positive only when the focus is on the right: convex mirror and convex lens.
Use it when: Fill in the data line with signs before you touch the formula.
Negative m = Real = Inverted
Think 'NRI': Negative, Real, Inverted. Positive m means virtual and erect.
Use it when: Lets you state the nature of the image straight from the sign of m.
Only one escape
For a concave mirror or convex lens, every object position gives a real inverted image except one: between F and the pole or optical centre, which gives a virtual, erect, enlarged image.
Use it when: Answers 'erect and enlarged' questions and shaving mirror or magnifier items instantly.
Centimetres to dioptres
P = 100 divided by f in cm. A 25 cm convex lens is 100/25 = +4 D.
Use it when: Avoids the 0.04 D mistake when f is given in cm.

Light – Reflection and Refraction: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Object between F1 and 2F1 of a convex lens · 1 mark · MCQBoard level

An object is placed between $F_{1}$ and $2 F_{1}$ of a convex lens. The image formed is:
(a) between $F_{2}$ and $2 F_{2}$ , real and diminished
(b) beyond $2 F_{2}$ , real, inverted and enlarged
(c) on the same side as the object, virtual and enlarged
(d) at $2 F_{2}$ , real and of the same size

Try it first, then show the solutionSolution

Between $F_{1}$ and $2 F_{1}$ is the row just before 'at $F_{1}$ ' in the convex lens table.
The image moves beyond $2 F_{2}$ , real and inverted, and is enlarged.

Answer: (b) beyond $2 F_{2}$ , real, inverted and enlarged.

Example 2: Ray through the centre of curvature · 1 mark · Assertion-reasonBoard level

Assertion (A): A ray of light passing through the centre of curvature of a concave mirror is reflected back along the same path.
Reason (R): Such a ray falls on the mirror along the normal to the reflecting surface, so the angle of incidence is zero.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

Any line from C to the mirror is a radius, so it is normal to the surface.
Angle of incidence is 0°, so angle of reflection is 0° and the ray retraces its path. R explains A.

Answer: (a) Both A and R are true and R is the correct explanation of A.

Example 3: Concave lens: position, size and nature · 3 marks · Short answerBoard level

An object 6 cm tall is placed 20 cm in front of a concave lens of focal length 20 cm. Find the position, size and nature of the image. Show that the image lies between the focus and the optical centre. (3 marks)

Try it first, then show the solutionSolution

Data: $h = + 6 cm$ , $u = - 20 cm$ , $f = - 20 cm$ (concave lens). (½)
Lens formula: $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = - \frac{1}{20} - \frac{1}{20} = - \frac{2}{20}$ (½)
$v = - 10 cm$ : image 10 cm from the lens on the object side. (½)
$m = \frac{v}{u} = \frac{- 10}{- 20} = + 0.5$ (½)
$h^{'} = mh = 0.5 \times 6 = + 3 cm$ ; virtual, erect, diminished. (½)
The focus is 20 cm from O and the image is 10 cm from O on the same side, so it lies between $F_{1}$ and O. (½)

Answer: Image 10 cm from the lens on the same side as the object ( $v = - 10 cm$ ), 3 cm tall, virtual, erect and diminished, between $F_{1}$ and O.

Example 4: Concave mirror: diagram, numerical and use · 5 marks · Long answerHarder board level

(a) Draw a ray diagram (describe it in words) for an object placed between F and C of a concave mirror, and state the nature of the image.
(b) An object 3 cm tall is placed 18 cm in front of a concave mirror of focal length 12 cm. Find the position and size of the image and the magnification.
(c) Name one use of a concave mirror and give the reason it suits that use. (5 marks)

Try it first, then show the solutionSolution

(a) Ray 1 from the top of the object parallel to the axis reflects through F; ray 2 through F reflects parallel to the axis (or ray through C returns along itself). (1)
They meet beyond C: the image is real, inverted, enlarged and beyond C. (1)
(b) Data: $u = - 18 cm$ , $f = - 12 cm$ , $h = + 3 cm$ . (½)
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = - \frac{1}{12} + \frac{1}{18} = \frac{- 3 + 2}{36} = - \frac{1}{36}$ , so $v = - 36 cm$ . (1)
$m = - \frac{v}{u} = - \frac{- 36}{- 18} = - 2$ ; $h^{'} = - 2 \times 3 = - 6 cm$ . (1)
(c) Use with reason, for example: shaving mirror, because the face held between P and F gives an enlarged, erect image (or torch or headlight: source at F gives a parallel beam; or solar furnace: sunlight is concentrated to produce heat). (½)

Answer: (a) Real, inverted, enlarged image beyond C. (b) Image 36 cm in front of the mirror, 6 cm tall, inverted; $m = - 2$ . (c) For example, shaving mirror: the face within F gives an enlarged, erect image.

Example 5: Practical: tracing a ray through a glass slab · 3 marks · Short answerHarder board level

In Practical 11, a student traces a ray through a rectangular glass slab. For an angle of incidence of 30°, she measures the angle of refraction as 19° and the angle of emergence as 30°.
(a) Why is the angle of refraction less than the angle of incidence?
(b) Explain why the angle of emergence equals the angle of incidence.
(c) She repeats the experiment with an angle of incidence of 45°. What will stay the same between the two readings, by the laws of refraction? (3 marks)

Try it first, then show the solutionSolution

(a) Light goes from air (optically rarer) into glass (optically denser), slows down and bends towards the normal, so r < i. (1)
(b) The bending at face AB (air to glass) and at face CD (glass to air) is equal and opposite because the faces are parallel, so the emergent ray is parallel to the incident ray and e = i. (1)
(c) The ratio $\frac{s i n i}{s i n r}$ stays the same (Snell's law) because the colour of light and the pair of media are unchanged; this constant is the refractive index of glass with respect to air. (1)

Answer: (a) The ray enters a denser medium and bends towards the normal. (b) Equal and opposite bending at parallel faces makes e = i. (c) $\frac{s i n i}{s i n r}$ remains constant.

Example 6: Convex lens: real and virtual images from one lens · 5 marks · Long answerHarder board level

An object 4 cm tall is placed 12 cm in front of a convex lens of focal length 8 cm.
(a) Find the position, size and nature of the image.
(b) Find the power of the lens.
(c) The object is now moved to 6 cm from the lens. Find the new image position and magnification, and state the nature of the image. (5 marks)

Try it first, then show the solutionSolution

(a) Data: $u = - 12 cm$ , $f = + 8 cm$ , $h = + 4 cm$ . (½)
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{8} - \frac{1}{12} = \frac{3 - 2}{24} = \frac{1}{24}$ , so $v = + 24 cm$ . (1)
$m = \frac{v}{u} = \frac{24}{- 12} = - 2$ ; $h^{'} = - 8 cm$ : real, inverted, enlarged, 24 cm on the other side. (1)
(b) $P = \frac{1}{f} = \frac{1}{0.08 m} = + 12.5 D$ (½)
(c) $u = - 6 cm$ : $\frac{1}{v} = \frac{1}{8} - \frac{1}{6} = \frac{3 - 4}{24} = - \frac{1}{24}$ , so $v = - 24 cm$ . (1)
$m = \frac{- 24}{- 6} = + 4$ : virtual, erect, enlarged, on the same side as the object. (1)

Answer: (a) $v = + 24 cm$ , $h^{'} = - 8 cm$ , real, inverted, enlarged. (b) +12.5 D. (c) $v = - 24 cm$ , $m = + 4$ , virtual, erect, enlarged.

Example 7: Analysing a convex lens data table · 3 marks · Short answerHOTS / competency

A student records these readings for a convex lens (u in cm, v in cm):

u = −60, v = +30
u = −40, v = +40
u = −30, v = +60
u = −25, v = +100
u = −15, v = +30

(a) Using the most suitable reading, find the focal length without using the lens formula.
(b) Which reading is wrong? Give a reason.
(c) Find the magnification for u = −25 cm. (3 marks)

Try it first, then show the solutionSolution

(a) When u and v are equal in size (40 cm), the object is at $2 F_{1}$ and the image at $2 F_{2}$ , so $2 f = 40 cm$ and $f = 20 cm$ . (1)
(b) For u = −15 cm the object is between $F_{1}$ and O (15 cm < 20 cm). The image must then be virtual, on the object side, so v cannot be +30 cm. Check: $\frac{1}{v} = \frac{1}{20} - \frac{1}{15} = - \frac{1}{60}$ , so $v = - 60 cm$ . (1)
(c) $m = \frac{v}{u} = \frac{100}{- 25} = - 4$ (real, inverted, 4 times enlarged). (1)

Answer: (a) f = 20 cm. (b) The last reading; the true image is virtual at $v = - 60 cm$ . (c) $m = - 4$ .

Example 8: Identify the mirror from its image · 3 marks · Short answerBoard level

A mirror forms an erect image one-third the size of the object, 10 cm behind the mirror. Identify the mirror, and find the object distance and the focal length. Justify your identification in two ways. (3 marks)

Try it first, then show the solutionSolution

An erect image smaller than the object can come only from a convex mirror (a concave mirror's erect image is always enlarged; a plane mirror's is the same size). (1)
Image behind the mirror: $v = + 10 cm$ ; $m = + \frac{1}{3} = - \frac{v}{u}$ , so $u = - 3 v = - 30 cm$ . (1)
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{10} - \frac{1}{30} = \frac{2}{30}$ , so $f = + 15 cm$ ; the positive focal length confirms a convex mirror. (1)

Answer: Convex mirror; object 30 cm in front ( $u = - 30 cm$ ); focal length +15 cm.

Example 9: Convex lens followed by a plane mirror · 3 marks · Short answerHarder board level

An object 2 cm tall is placed 15 cm in front of a convex lens of focal length 10 cm. A plane mirror is placed perpendicular to the principal axis, 40 cm from the lens on the other side. The light from the lens first forms an image and then reaches the mirror.
(a) Find the position and size of the image formed by the lens.
(b) Where does the plane mirror form the final image, and how far is it from the lens?
(c) State the nature and size of the final image compared with the original object. (3 marks)

Try it first, then show the solutionSolution

(a) Data: $u = - 15$ cm, $f = + 10$ cm, $h = + 2$ cm; $\frac{1}{v} = \frac{1}{10} - \frac{1}{15} = \frac{3 - 2}{30} = \frac{1}{30}$ , so $v = + 30$ cm. (½)
$m = \frac{v}{u} = \frac{30}{- 15} = - 2$ , so $h^{'} = - 4$ cm: real, inverted, 30 cm from the lens. (½)
(b) This image is $40 - 30 = 10$ cm in front of the plane mirror and acts as the object for the mirror. (½)
A plane mirror forms its image as far behind it as the object is in front, so the final image is 10 cm behind the mirror, $40 + 10 = 50$ cm from the lens. (½)
(c) A plane mirror gives an erect image of the same size as its own object, so the final image stays 4 cm tall and inverted compared with the original object. (½)
It is formed behind the mirror, so it is virtual. (½)

Answer: (a) 30 cm beyond the lens, 4 cm tall, real and inverted. (b) 10 cm behind the mirror, 50 cm from the lens. (c) Virtual, 4 cm tall and inverted relative to the original object (twice its size).

Example 10: Fixed object and screen: two lens positions · 3 marks · Short answerHOTS / competency

A candle and a screen are fixed 90 cm apart. A convex lens of focal length 20 cm is placed between them, 30 cm from the candle.
(a) Show that a sharp image forms on the screen and find the magnification.
(b) Without moving the candle or the screen, the student slides the lens towards the screen and finds a second position where the image is sharp again. Find this position and the new magnification.
(c) How far did she move the lens? (3 marks)

Try it first, then show the solutionSolution

(a) $u = - 30$ cm, $f = + 20$ cm: $\frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{1}{60}$ , so $v = + 60$ cm; $30 + 60 = 90$ cm, exactly where the screen is. (½)
$m = \frac{v}{u} = \frac{60}{- 30} = - 2$ : real, inverted, twice enlarged. (½)
(b) For a real image the formula reads $\frac{1}{∣ v ∣} + \frac{1}{∣ u ∣} = \frac{1}{f}$ , which is unchanged if the two distances swap, so try the lens 60 cm from the candle: $u = - 60$ cm. (½)
$\frac{1}{v} = \frac{1}{20} - \frac{1}{60} = \frac{2}{60}$ , so $v = + 30$ cm, and $60 + 30 = 90$ cm: the image is again on the screen. (½)
$m = \frac{30}{- 60} = - 0.5$ : real, inverted, half the size of the candle. (½)
(c) The lens moved from 30 cm to 60 cm from the candle, a shift of 30 cm towards the screen. (½)

Answer: (a) v = +60 cm, on the screen; m = −2. (b) Lens 60 cm from the candle (30 cm from the screen); m = −0.5. (c) 30 cm.

Example 11: From power to image: a converging lens · 5 marks · Long answerHarder board level

A lens has a power of +2.5 D.
(a) Find its focal length and name the type of lens.
(b) An object 1.5 cm tall is placed 60 cm from this lens. Find the position, size and nature of the image.
(c) The object is moved to 20 cm from the lens. Find the new image position and magnification. Can this image be obtained on a screen? Give a reason. (5 marks)

Try it first, then show the solutionSolution

(a) $f = \frac{1}{P} = \frac{1}{2.5} = + 0.40$ m $= + 40$ cm; positive power, so it is a convex (converging) lens. (1)
(b) $u = - 60$ cm: $\frac{1}{v} = \frac{1}{40} - \frac{1}{60} = \frac{3 - 2}{120} = \frac{1}{120}$ , so $v = + 120$ cm. (1)
$m = \frac{120}{- 60} = - 2$ ; $h^{'} = - 2 \times 1.5 = - 3$ cm: real, inverted, enlarged, 120 cm on the other side. (1)
(c) $u = - 20$ cm: $\frac{1}{v} = \frac{1}{40} - \frac{1}{20} = - \frac{1}{40}$ , so $v = - 40$ cm. (½)
$m = \frac{- 40}{- 20} = + 2$ : virtual, erect, enlarged, on the same side as the object. (½)
No. The object is between $F_{1}$ and O, so the refracted rays do not actually meet; the image is virtual and cannot be caught on a screen. (1)

Answer: (a) +40 cm, convex lens. (b) v = +120 cm, h′ = −3 cm, real, inverted, enlarged. (c) v = −40 cm, m = +2; no, the image is virtual.

Example 12: Concave mirror throwing an image on a wall · 3 marks · Short answerHarder board level

A concave mirror of focal length 15 cm is used to throw the image of a 2 cm tall bulb filament on a wall 60 cm in front of the mirror.
(a) Where must the filament be placed?
(b) Find the size of the image.
(c) Between which two points of the mirror does the filament lie? Is this consistent with the image you found? (3 marks)

Try it first, then show the solutionSolution

(a) The image on the wall is real and in front: $v = - 60$ cm, $f = - 15$ cm. (½)
$\frac{1}{u} = \frac{1}{f} - \frac{1}{v} = - \frac{1}{15} + \frac{1}{60} = \frac{- 4 + 1}{60} = - \frac{3}{60}$ , so $u = - 20$ cm. (½)
(b) $m = - \frac{v}{u} = - \frac{- 60}{- 20} = - 3$ (½)
$h^{'} = - 3 \times 2 = - 6$ cm: real, inverted, three times enlarged. (½)
(c) C is at $2 f = 30$ cm, so the filament at 20 cm lies between F (15 cm) and C (30 cm). (½)
For an object between F and C a concave mirror gives a real, inverted, enlarged image beyond C; the image at 60 cm is beyond C, so the results agree. (½)

Answer: (a) 20 cm in front of the mirror (u = −20 cm). (b) 6 cm tall and inverted (h′ = −6 cm). (c) Between F and C; yes, this gives an enlarged real image beyond C.

Case-study practice (Section E style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Finding a lens's focal length in the school lab

Suppose in your school physics lab, a student holds a convex lens near an open window and focuses the image of a distant tree on the opposite wall. The sharpest, smallest image appears when the lens is 20 cm from the wall. She then fixes the lens on a stand and places a candle 30 cm in front of it. On the other side she moves a screen until a sharp image of the flame appears. The flame is 2 cm tall. Finally, she slides the candle to 10 cm from the lens and finds that no image appears on the screen, wherever she places it, though she can see an image when she looks through the lens.

(i) · 1 mark
What is the focal length of the lens, and what is its power?
Answer
Answer: f = 20 cm; P = +5 D
Rays from the distant tree are nearly parallel and meet at the focus, so $f = + 20 cm = + 0.20 m$ . $P = \frac{1}{0.20} = + 5 D$ .
(ii) · 1 mark
Where must the screen be placed to get a sharp image when the candle is 30 cm from the lens?
Answer
Answer: 60 cm from the lens on the other side
$u = - 30 cm$ , $f = + 20 cm$ : $\frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{1}{60}$ , so $v = + 60 cm$ .
(iii) · 2 marks
Find the size and nature of the flame's image in the second step. Why does no image form on the screen in the last step?
Answer
Answer: 4 cm, real, inverted, enlarged; at 10 cm the image is virtual
$m = \frac{v}{u} = \frac{60}{- 30} = - 2$ , so $h^{'} = - 2 \times 2 = - 4 cm$ : real, inverted, twice enlarged. At 10 cm the candle is between $F_{1}$ and O, so the lens gives a virtual, erect, enlarged image on the same side as the candle; a virtual image cannot be caught on a screen.

A student's u-v table for a convex lens

Suppose a student uses a convex lens on an optical bench and records the object distance u and the image distance v (both in cm) for several candle positions. Her readings are: u = −30, v = +15; u = −20, v = +20; u = −15, v = +30; u = −12, v = +60; u = −5, v = +10. She also took a reading at u = −60 but forgot to note the image distance. For the reading at u = −5 cm she wrote down a value even though she could not find any sharp image on the screen.

(i) · 1 mark
Using the most suitable reading, find the focal length and the power of the lens.
Answer
Answer: f = +10 cm; P = +10 D
At u = −20 cm, v = +20 cm: the object and image are equally far from the lens, so the object is at $2 F_{1}$ and the image at $2 F_{2}$ . So $2 f = 20$ cm, $f = + 10$ cm $= + 0.10$ m, and $P = \frac{1}{0.10} = + 10$ D.
(ii) · 1 mark
Complete the missing reading: find v for u = −60 cm, and state the nature and relative size of the image.
Answer
Answer: v = +12 cm; real, inverted, diminished
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} - \frac{1}{60} = \frac{6 - 1}{60} = \frac{5}{60}$ , so $v = + 12$ cm. $m = \frac{12}{- 60} = - 0.2$ : real, inverted and diminished, as expected for an object beyond $2 F_{1}$ .
(iii) · 2 marks
Which reading cannot be correct? Give a reason, and find the correct image distance and magnification for it.
Answer
Answer: u = −5 cm; correct v = −10 cm, m = +2
At u = −5 cm the object is between $F_{1}$ and O (5 cm is less than f = 10 cm), so the image is virtual, on the same side as the object, and cannot form on a screen; a positive v is impossible. $\frac{1}{v} = \frac{1}{10} - \frac{1}{5} = - \frac{1}{10}$ , so $v = - 10$ cm. $m = \frac{- 10}{- 5} = + 2$ : virtual, erect, twice enlarged.

Lenses and a mirror at an optician's shop

Imagine an optician's shop. Two lenses lie on the counter, one marked +4.0 D and the other marked −2.0 D. The optician explains that the power of a lens tells how strongly it converges or diverges light, that its sign tells the type of lens, and that power in dioptres needs the focal length in metres. On the wall hangs a concave mirror of radius of curvature 60 cm, in which customers check how their new frames look.

(i) · 1 mark
Find the focal length of each lens in cm and name its type.
Answer
Answer: +25 cm, convex; −50 cm, concave
+4.0 D: $f = \frac{1}{4.0} = + 0.25$ m $= + 25$ cm, a convex (converging) lens. −2.0 D: $f = \frac{1}{- 2.0} = - 0.50$ m $= - 50$ cm, a concave (diverging) lens.
(ii) · 2 marks
A customer stands with her face 20 cm in front of the wall mirror. Find the position of the image and the magnification, and state the nature of the image.
Answer
Answer: v = +60 cm (behind the mirror), m = +3; virtual, erect, enlarged
$f = \frac{R}{2} = 30$ cm, and for a concave mirror $f = - 30$ cm; $u = - 20$ cm. $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = - \frac{1}{30} + \frac{1}{20} = \frac{- 2 + 3}{60} = \frac{1}{60}$ , so $v = + 60$ cm, behind the mirror. $m = - \frac{v}{u} = - \frac{60}{- 20} = + 3$ : virtual, erect and three times enlarged, because the face is between P and F.
(iii) · 1 mark
A customer asks whether the −2.0 D lens could be used as a magnifying glass for small print. What should the optician say, and why?
Answer
Answer: No; a concave lens always gives a diminished image
No. Negative power means a concave lens, and a concave lens always gives a virtual, erect, diminished image whatever the position of the object, so it can never enlarge print. The +4.0 D convex lens, with the print held closer than 25 cm (between $F_{1}$ and O), would give an enlarged, erect image.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

A convex lens has a focal length of 50 cm. Find its power and state what the sign of the power tells you. (2 marks) (2 marks)

Step	Marks
Convert the focal length to metres with its sign: $f = + 50 cm = + 0.50 m$	0.5
Write the formula: $P = \frac{1}{f}$ (with f in metres)	0.5
Substitute and give the unit: $P = \frac{1}{+ 0.50} = + 2 D$	0.5
Interpret the sign: positive power means a converging (convex) lens	0.5

State the two laws of refraction of light. A ray of light passes obliquely from air into a transparent medium of refractive index 1.6. Find the speed of light in the medium. (Speed of light in air = $3 \times 1 0^{8} m s^{- 1}$ ) (3 marks) (3 marks)

Step	Marks
First law: the incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane	1
Second law (Snell's law): for light of a given colour and a given pair of media, $\frac{s i n i}{s i n r}$ is a constant	1
Use $n_{m} = \frac{c}{v}$ , so $v = \frac{c}{n _{m}} = \frac{3 \times 1 0 ^{8}}{1.6}$	0.5
$v = 1.875 \times 1 0^{8} m s^{- 1}$	0.5

(a) With the help of a ray diagram, show the image formed by a convex lens when the object is placed at $2 F_{1}$ .
(b) An object 5 cm tall is placed perpendicular to the principal axis, 50 cm from a convex lens of focal length 25 cm. Using the lens formula, find the position and size of the image and its magnification. State the nature of the image.
(c) Find the power of the lens. (5 marks) (5 marks)

Step	Marks
Ray diagram: one ray from the top of the object parallel to the principal axis, refracted through $F_{2}$ ; a second ray through the optical centre O going undeviated; arrows on both rays	1
Image drawn at $2 F_{2}$ , inverted, same size as the object, with $F_{1}$ , $2 F_{1}$ , $F_{2}$ , $2 F_{2}$ and O labelled	0.5
Data with signs: $h = + 5 cm$ , $u = - 50 cm$ , $f = + 25 cm$	0.5
Lens formula: $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{25} - \frac{1}{50} = \frac{1}{50}$	0.5
$v = + 50 cm$ : image 50 cm on the other side of the lens	0.5
$m = \frac{v}{u} = \frac{+ 50}{- 50} = - 1$	0.5
$h^{'} = m \times h = - 1 \times 5 = - 5 cm$	0.5
Nature: real, inverted, same size as the object	0.5
$P = \frac{1}{f} = \frac{1}{+ 0.25 m} = + 4 D$	0.5

Where students lose marks in Light – Reflection and Refraction

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

Mistake: Taking the focal length of a concave mirror as positive
Students copy the number from the question and forget that F of a concave mirror is in front of it, on the negative side.
Wrong working
Concave mirror, f = +15 cm, u = −30 cm: $\frac{1}{v} = \frac{1}{15} + \frac{1}{30} = \frac{3}{30}$ , so v = +10 cm, image behind the mirror.
Right working
f = −15 cm, u = −30 cm: $\frac{1}{v} = - \frac{1}{15} + \frac{1}{30} = - \frac{1}{30}$ , so v = −30 cm: real image at C, 30 cm in front.
Do this: Write a data line with every sign first and ask: is F in front (negative) or behind (positive)?
Usually costs: Usually 1 to 2 marks of a 3-mark numerical, since every later step goes wrong.
Mistake: Using the mirror magnification formula for a lens
The two formulas look alike, and students carry the minus sign over from mirrors.
Wrong working
Convex lens, u = −30 cm, v = +60 cm: m = −v/u = −60/−30 = +2, so the image is virtual and erect.
Right working
For a lens m = v/u = 60/−30 = −2, so the image is real, inverted and twice the size.
Do this: Remember 'mirror plus, lens minus': the mirror's minus sign sits in m, the lens's minus sign sits in the formula.
Usually costs: About 1 mark: the magnification and the stated nature are both wrong.
Mistake: Finding power with the focal length in centimetres
The dioptre is defined for focal length in metres, but students substitute the cm value directly.
Wrong working
f = +25 cm, so P = 1/25 = 0.04 D.
Right working
f = +25 cm = +0.25 m, so P = 1/0.25 = +4 D (convex lens).
Do this: Convert to metres before using P = 1/f, or use P = 100/f with f in cm, and keep the sign.
Usually costs: ½ to 1 mark, plus the type of lens if the sign is dropped.
Mistake: Leaving the answer as a bare number
Students stop once v is found, without reading the sign as a position and nature.
Wrong working
v = 30 cm.
Right working
v = −30 cm: the image is 30 cm in front of the mirror, real and inverted; m = −2, so it is enlarged two times.
Do this: Finish every numerical with one sentence on position, nature and size, using the signs of v and m.
Usually costs: ½ to 1 mark in most 3-mark numericals.
Mistake: Careless ray diagrams
Students rush, forget arrowheads and draw virtual rays and images as solid lines.
Wrong working
Convex mirror diagram: reflected rays extended behind the mirror as solid lines, no arrowheads, image drawn as a solid arrow, F and C not labelled.
Right working
Arrowheads on every ray; extensions behind the mirror dotted; virtual image drawn dotted between P and F; P, F and C labelled with F midway between P and C.
Do this: Use the Draw and label checklist: axis, labelled points, two standard rays with arrows, dotted virtual lines, image clearly marked.
Usually costs: ½ to 1 mark per diagram.
Mistake: Saying the emergent ray from a glass slab is in line with the incident ray, or bends towards the normal
Students know the emergent ray is parallel and confuse 'parallel' with 'same line', or forget the glass-air bend is away from the normal.
Wrong working
The ray bends towards the normal at both faces and comes out along the same line as the incident ray.
Right working
It bends towards the normal at the air-glass face and away from it at the glass-air face; the emergent ray is parallel to the incident ray but shifted sideways.
Do this: Say both facts: equal and opposite bending at parallel faces, so parallel but laterally shifted.
Usually costs: 1 mark in a reasoning or practical-based question.
Mistake: Treating optical density as mass density
The words 'denser' and 'rarer' make students think of how heavy a liquid is.
Wrong working
Water is denser than kerosene, so light bends more in water and water is optically denser.
Right working
Kerosene has refractive index 1.44, water 1.33, so kerosene is optically denser even though its mass density is less.
Do this: Decide optical density only from refractive index: larger n means optically denser.
Usually costs: 1 mark in an MCQ or assertion-reason item.

NCERT exercise guide: Light – Reflection and Refraction

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

In-text questions after section 9.2.2

What it tests: Basic terms of spherical mirrors: principal focus, the link $R = 2 f$ , which mirror gives an erect enlarged image, and why convex mirrors are used as rear-view mirrors.

How to approach it: Learn each definition in one clear sentence using the words 'parallel to the principal axis'. For $R = 2 f$ questions, halve R and give the sign by the type of mirror. For 'name the mirror' questions, recall Table 9.1 and Table 9.2 and give one reason.

Mirrors the questions on R and f and on which mirror gives a particular image

The radius of curvature of a concave mirror is 24 cm. Find its focal length. Where must an object be placed in front of it to get an erect, enlarged image? (2 marks)

Show the solutionSolution

For a spherical mirror of small aperture, $f = \frac{R}{2} = \frac{24}{2} = 12 cm$ .
For a concave mirror, F is in front of the mirror, so $f = - 12 cm$ in the New Cartesian Sign Convention.
A concave mirror gives an erect, enlarged (virtual) image only when the object is between P and F.
So the object must be placed less than 12 cm from the mirror.

Answer: Focal length is 12 cm ( $f = - 12 cm$ ). Place the object between the pole and the focus, at less than 12 cm from the mirror.

In-text questions after section 9.2.4

What it tests: Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ and magnification $m = - \frac{v}{u}$ with correct signs.

How to approach it: Write the data with signs first: u is always negative, f is negative for concave and positive for convex. For a real image, m is negative; for a virtual image, m is positive. End with position, nature and size in words.

Mirrors the question on a magnified real image from a concave mirror

A concave mirror forms a real image twice the size of an object placed 15 cm in front of it. Find the image position and the focal length of the mirror. (3 marks)

Show the solutionSolution

The image is real, so it is inverted: $m = - 2$ . Object distance $u = - 15 cm$ .
$m = - \frac{v}{u}$ gives $- 2 = - \frac{v}{- 15}$ , so $v = - 30 cm$ .
Mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{- 30} + \frac{1}{- 15} = \frac{- 1 - 2}{30} = - \frac{3}{30}$
$f = - 10 cm$ .

Answer: The image is 30 cm in front of the mirror ( $v = - 30 cm$ ), and the focal length is 10 cm ( $f = - 10 cm$ ).

In-text questions after section 9.3.2

What it tests: Direction of bending at a boundary, speed of light in a medium from its refractive index, and reading Table 9.3 for optical density.

How to approach it: Remember: going into the medium of higher refractive index (optically denser), light slows and bends towards the normal. Use $n_{m} = \frac{c}{v}$ for speed. To compare media, compare refractive indices, not mass densities.

Mirrors the questions on speed of light in a medium and the meaning of a refractive index value

The absolute refractive index of a transparent liquid is 1.25. (a) What does this value mean? (b) Find the speed of light in the liquid. Speed of light in vacuum = $3 \times 1 0^{8} m s^{- 1}$ . (2 marks)

Show the solutionSolution

(a) It means that the ratio of the speed of light in vacuum (or air) to the speed of light in the liquid is 1.25.
(b) $n = \frac{c}{v}$ , so $v = \frac{c}{n} = \frac{3 \times 1 0 ^{8}}{1.25}$
$v = 2.4 \times 1 0^{8} m s^{- 1}$

Answer: Light travels 1.25 times faster in vacuum than in the liquid; speed in the liquid is $2.4 \times 1 0^{8} m s^{- 1}$ .

In-text questions after section 9.3.8

What it tests: Definition of 1 dioptre, finding power from focal length, and using the same-size image condition for a convex lens.

How to approach it: Always convert focal length to metres before using $P = \frac{1}{f}$ . Give the sign: positive for convex, negative for concave. For a same-size real image from a convex lens, object and image are both at 2F.

Mirrors the question on a same-size image and the power of a convex lens

A convex lens forms a real, inverted image of a pin, of the same size as the pin, at 80 cm from the lens. Where is the pin placed? Find the power of the lens. (3 marks)

Show the solutionSolution

A same-size real image from a convex lens forms at $2 F_{2}$ when the object is at $2 F_{1}$ .
So the pin is 80 cm in front of the lens: $u = - 80 cm$ , $v = + 80 cm$ .
$2 f = 80 cm$ , so $f = + 40 cm = + 0.40 m$ (check: $\frac{1}{80} + \frac{1}{80} = \frac{1}{40}$ ).
$P = \frac{1}{f} = \frac{1}{0.40} = + 2.5 D$

Answer: The pin is 80 cm in front of the lens, and the power of the lens is +2.5 D.

End-of-chapter exercises

What it tests: MCQs on image position and lens material, naming mirrors for uses with reasons, ray diagrams, mirror and lens numericals, the meaning of magnification, and power with lens type.

How to approach it: For MCQs, picture the matching row of Tables 9.1, 9.2, 9.4 or 9.5. For numericals, list data with signs, use the correct formula (mirror: plus sign; lens: minus sign), and state position, size and nature. Draw ray diagrams with two standard rays and arrows.

Mirrors the convex mirror numericals on position, nature and size

An object 4 cm tall is placed 20 cm in front of a convex mirror of radius of curvature 40 cm. Find the position, nature and size of the image. (3 marks)

Show the solutionSolution

$f = \frac{R}{2} = + 20 cm$ (convex: F behind the mirror); $u = - 20 cm$ ; $h = + 4 cm$ .
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20}$ , so $v = + 10 cm$ .
$m = - \frac{v}{u} = - \frac{10}{- 20} = + 0.5$
$h^{'} = m \times h = 0.5 \times 4 = + 2 cm$

Answer: The image is 10 cm behind the mirror, virtual, erect and diminished, 2 cm tall.

Activity 9.10 and Activities 9.2, 9.11, 9.12 (practicals on the glass slab and focal length)

What it tests: Tracing a ray through a rectangular glass slab and measuring i, r and e; finding focal length by focusing a distant object.

How to approach it: For the slab, measure every angle from the normal and expect $e = i$ . For focal length, focus a distant object (the Sun only through the stated caution, or a distant tree or window) to the sharpest, smallest image; the lens or mirror to screen distance is the approximate focal length.

Mirrors practical questions on finding focal length with a distant object

A student focuses the image of a distant building on a screen using a convex lens. The sharpest image is obtained with the screen 12.5 cm from the lens. (a) What is the approximate focal length? (b) Find the power of the lens. (c) Why must the object be distant? (3 marks)

Show the solutionSolution

(a) Rays from a distant object are nearly parallel, so they converge at the focus: $f \approx 12.5 cm$ .
(b) $P = \frac{1}{f} = \frac{1}{0.125 m} = + 8 D$
(c) Only for an object at (near) infinity does the image form at the focus, so the lens-screen distance equals the focal length.

Answer: Focal length is about 12.5 cm; power is +8 D; the object must be distant so that the image forms at the focus.

Light – Reflection and Refraction: previous year questions

Asked in 21 of the 21 board papers we hold (2023 to 2026), on average 7.6 marks per paper; 72 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Light – Reflection and Refraction

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

2026, set 31/1/1, Q30 · 1 mark · MCQ

A convex lens of focal length 15 cm, is forming a real image. If the size of image is same as the size of object, then position of object and position of image will be, respectively:

(A) $- 15$ cm and $- 15$ cm from lens
(B) $- 15$ cm and $+ 15$ cm from lens
(C) $- 30$ cm and $+ 30$ cm from lens
(D) $- 30$ cm and $- 30$ cm from lens

Try it first, then show our solutionSolution

A convex lens forms a real image of the same size when the object is at $2 F_{1}$ . The image is then at $2 F_{2}$ .
$2 f = 2 \times 15 = 30$ cm. By the sign convention, $u = - 30$ cm and $v = + 30$ cm. Check: $\frac{1}{30} - \frac{1}{- 30} = \frac{1}{15}$ . Correct option: (C).
1 mark

Answer: −30 cm and +30 cm, option (C)

Where the marks are

Object at 2F₁ and image at 2F₂ with correct signs (1 mark)

See it in the full 2026 board paper, set 31/1/1

2026, set 31/1/1, Q33 · 2 marks · Short answerCompetency-based

The given figure shows image formation by a lens. Analyse the figure and answer the following questions:

(a)
What is the type of lens used for image formation in the given ray diagram?
0 marks
(b)
If the real image is formed at a distance of 30 cm from the lens and the size of image is twice the size of the object, then where was the object placed?
0 marks

Figure in the paper: Ray diagram: object AB stands vertically on the principal axis to the left of a lens drawn as a tall rectangle. Image A'B' is formed further left, on the same side as the object, with B' above the axis. Two rays from B are shown: one parallel to the axis refracting through the lens, and one passing through the optical centre; both are drawn as dashed lines extended back to B'. See it in the official paper.

Try it first, then show our solutionSolution

(a) The image is enlarged and erect and forms on the same side as the object, with the refracted rays traced back to meet at B'. Only a converging lens does this, so the lens is a convex (converging) lens.
1 mark
(b) The image is real and twice the size of the object, so it is inverted: $m = - 2$ . Also $v = + 30$ cm. For a lens, $m = \frac{v}{u}$ .
½ mark
$u = \frac{v}{m} = \frac{30}{- 2} = - 15$ cm. The object was placed 15 cm in front of the lens (on the left side).
½ mark

Answer: (a) Convex lens; (b) u = −15 cm, i.e. 15 cm in front of the lens

Where the marks are

Identifies convex/converging lens (1 mark)
Uses m = v/u with m = −2 for a real image (½ mark)
u = −15 cm (½ mark)

See it in the full 2026 board paper, set 31/1/1

2026, set 31/1/1, Q38 · 4 marks · Case-basedCompetency-based

Read the following passage and answer the questions given:
Lenses can form different types of images depending upon their focal length and position of object. A convex lens can create real, inverted or virtual, erect images, while a concave lens forms only virtual and diminished images. The focal length determines the power of lens. Convex lenses have positive focal length while concave lenses have negative focal length by convention. When lenses are placed together, their combined power is determined by the sum of their individual powers. Ray diagrams help to visualize how light converges or diverges through lens to form an image.

(a)
A convex lens of focal length 20 cm is used to form an image. If an object is placed at 40 cm from the lens, what will be the position and nature of image?
1 mark
(b)
Illustrate the formation of image with the help of ray diagram, when the object is placed between the optical centre and principal focus of concave lens.
1 mark
(c)(i)
A lens combination consists of a convex lens of focal length 30 cm and a concave lens of focal length 15 cm placed together. Find the equivalent focal length and power of this lens combination.
2 marks
OR
(c)(ii)
Two lenses are placed in contact. One is a concave lens with focal length 2 m and the other is a convex lens with focal length 1.5 m. What type of lens will the combination behave as (convex or concave)? Give reason.
2 marks

Try it first, then show our solutionSolution

(a) Given $f = + 20$ cm and $u = - 40$ cm. Using $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ : $\frac{1}{v} = \frac{1}{20} + \frac{1}{- 40} = \frac{2 - 1}{40} = \frac{1}{40}$ , so $v = + 40$ cm. The image is 40 cm on the other side of the lens, at $2 F_{2}$ .
½ mark
(a) Nature: the image is real and inverted, and the same size as the object. This is because the object is at $2 F_{1}$ .
½ mark
(b) Ray diagram: draw a concave lens with $F_{1}$ and $F_{2}$ marked, and object AB between O and $F_{1}$ . Ray 1 from B goes parallel to the axis. After refraction it diverges, and its dashed back-extension passes through $F_{1}$ . Ray 2 from B passes undeviated through O. The extension and ray 2 meet at B′. The image A′B′ lies between O and $F_{1}$ on the same side as the object, and is virtual, erect and diminished.
1 mark
(c)(i) $f_{1} = + 30$ cm $= + 0.3$ m and $f_{2} = - 15$ cm $= - 0.15$ m. Using $P = \frac{1}{f}$ (f in metres): $P_{1} = \frac{1}{0.3} = + 3.33$ D.
½ mark
(c)(i) $P_{2} = \frac{1}{- 0.15} = - 6.67$ D.
½ mark
(c)(i) Equivalent power: $P = P_{1} + P_{2} = 3.33 - 6.67 = - 3.33$ D.
½ mark
(c)(i) Equivalent focal length: $f = \frac{1}{P} = \frac{1}{- 3.33} = - 0.3$ m $= - 30$ cm. The combination acts as a diverging lens.
½ mark
(c)(ii) $f_{1} = - 2$ m (concave) gives $P_{1} = - \frac{1}{2} = - 0.5$ D. $f_{2} = + 1.5$ m (convex) gives $P_{2} = + \frac{1}{1.5} = + 0.67$ D. Then $P = P_{1} + P_{2} = - \frac{1}{2} + \frac{2}{3} = + \frac{1}{6}$ D, so $f = + 6$ m.
1 mark
(c)(ii) The combined power (and focal length) is positive, so the combination behaves as a convex (converging) lens. Reason: the convex lens has the smaller focal length, so its power is greater and outweighs the concave lens.
1 mark

Answer: (a) 40 cm behind the lens (at 2F₂), real and inverted, same size; (b) virtual, erect, diminished image between O and F₁ on the object side; (c)(i) f = −30 cm, P = −3.33 D; (c)(ii) behaves as a convex lens (P = +1/6 D, f = +6 m)

Where the marks are

(a) Position: 40 cm on other side / at 2F (½ mark)
(a) Nature: real and inverted (½ mark)
(b) Correct ray diagram for concave lens with object between O and F (1 mark)
(c)(i) Uses $P = 1/ f$ with powers of each lens (½ + ½ mark)
(c)(i) Combined power $- 3.33$ D (½ mark)
(c)(i) Equivalent focal length $- 30$ cm (½ mark)
(c)(ii) States combination behaves as convex lens (1 mark)
(c)(ii) Reason: net power $+ 1/6$ D / focal length +6 m positive (1 mark)

See it in the full 2026 board paper, set 31/1/1

Every Light – Reflection and Refraction question, by year

For 2027

Sample paper 2026-27: Q30 (1 mark, solved), Q32 (1 mark, solved), Q33 (2 marks, solved), Q35 (3 marks, solved)

2026

Set 31/1/1: Q30 (1 mark, solved), Q33 (2 marks, solved), Q38 (4 marks, solved)

Set 31/2/1: Q30 (1 mark, solved), Q33 (2 marks, solved), Q35 (3 marks, solved)

Set 31/3/1: Q33 (2 marks, solved), Q38 (4 marks, solved)

Set 31/4/1: Q35 (3 marks, solved), Q38 (4 marks, solved)

Set 31/5/1: Q30 (1 mark, solved), Q33 (2 marks, solved), Q35 (3 marks, solved), Q36 (3 marks, solved)

Sample paper 2025-26: Q30 (1 mark, solved), Q32 (1 mark, solved), Q33 (2 marks, solved), Q38 (4 marks, solved)

2025

Set 31/1/1: Q13 (1 mark, solved), Q24 (2 marks, solved), Q31 (3 marks, solved), Q36(a) (5 marks, solved), Q36(b) (5 marks, solved)

Set 31/2/1: Q11 (1 mark, solved), Q12 (1 mark, solved), Q13 (1 mark, solved), Q25(a) (2 marks, solved), Q25(b) (2 marks, solved), Q31 (3 marks, solved)

Set 31/3/1: Q11 (1 mark, solved), Q36(a) (5 marks, solved), Q36(b) (5 marks, solved)

Set 31/4/1: Q10 (1 mark, solved), Q11 (1 mark, solved), Q25(A) (2 marks, solved), Q31 (3 marks, solved), Q39 (4 marks, solved)

Set 31/5/1: Q13 (1 mark, solved), Q24 (2 marks, solved), Q31 (3 marks, solved), Q36(A) (5 marks, solved), Q36(B) (5 marks, solved)

Set 31/6/1: Q13 (1 mark, solved), Q24 (2 marks, solved), Q31 (3 marks, solved)

Sample paper 2024-25: Q13 (1 mark, solved), Q19 (1 mark, solved), Q24 (2 marks, solved), Q39 (4 marks, solved)

2024

Set 31/1/1: Q12 (1 mark, solved), Q24 (2 marks, solved), Q39 (4 marks, solved)

Set 31/2/1: Q24(a) (2 marks, solved), Q24(b) (2 marks, solved), Q36 (5 marks, solved)

Set 31/3/1: Q36(a) (5 marks, solved), Q36(b) (5 marks, solved)

Set 31/4/1: Q11 (1 mark, solved), Q36(a) (5 marks, solved), Q36(b) (5 marks, solved)

Set 31/5/1: Q11 (1 mark, solved), Q25 (2 marks, solved), Q39 (4 marks, solved)

Sample paper 2023-24: Q13 (1 mark, solved), Q24 (2 marks, solved), Q31 (3 marks, solved), Q36(a) (5 marks, solved), Q36(b) (5 marks, solved)

2023

Set 31/1/1: Q30 (3 marks, solved), Q39 (4 marks, solved)

Set 31/2/1: Q13 (1 mark, solved), Q31(a) (3 marks, solved), Q31(b) (3 marks, solved), Q36 (5 marks, solved)

Set 31/4/1: Q30(A) (3 marks, solved), Q30(B) (3 marks, solved), Q31 (3 marks, solved), Q39 (4 marks, solved)

Set 31/5/1: Q30(a) (3 marks, solved), Q30(b) (3 marks, solved), Q39 (4 marks)

Set 31/6/1: Q7 (1 mark, solved), Q25(a) (2 marks, solved), Q25(b) (2 marks, solved), Q39 (4 marks, solved)

Practice questions on Light – Reflection and Refraction

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
The power of a lens of focal length −25 cm is:
(a) +4 D
(b) −4 D
(c) −0.25 D
(d) −25 D
Hint
Convert centimetres to metres before taking the reciprocal.
Answer
(b) −4 D. $P = \frac{1}{- 0.25 m} = - 4 D$ ; the lens is concave.
Q2 · 2 marks · Very short answerBoard level
Define the principal focus of a convex lens and of a concave lens. Why does a lens have two principal foci? (2 marks)
Hint
Think of rays parallel to the principal axis, then send them in from the other side.
Answer
Convex lens: the point on the principal axis where rays parallel to the axis meet after refraction. Concave lens: the point on the axis from which such rays appear to diverge after refraction. Parallel light can enter from either side, giving a focus on each side ( $F_{1}$ and $F_{2}$ ).
Q3 · 1 mark · Assertion-reasonBoard level
Assertion (A): Kerosene is optically denser than water, even though its mass density is less than that of water.
Reason (R): Optical density is decided by refractive index, and kerosene has a higher refractive index than water.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Compare 1.44 and 1.33 from Table 9.3.
Answer
(a). Kerosene's refractive index (1.44) is more than water's (1.33), so light is slower in kerosene and it is optically denser; mass density does not decide this.
Q4 · 3 marks · Short answerBoard level
An object 8 cm tall is placed 60 cm in front of a convex mirror of focal length 20 cm. Find the position, size and nature of the image. (3 marks)
Hint
For a convex mirror, f is positive; use $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ .
Answer
$u = - 60$ , $f = + 20$ : $\frac{1}{v} = \frac{1}{20} + \frac{1}{60} = \frac{4}{60}$ , so $v = + 15 cm$ (behind the mirror).
$m = - \frac{v}{u} = + 0.25$ ; $h^{'} = + 2 cm$ .
Virtual, erect, diminished, 2 cm tall.
Q5 · 5 marks · Long answerHarder board level
A concave mirror has a focal length of 10 cm. Find the two positions of an object for which the image is twice the size of the object. For each, state the image position and nature, and say which one can be used as a shaving mirror. (5 marks)
Hint
Try m = −2 (real) and m = +2 (virtual), and use $v = - m u$ in the mirror formula.
Answer
Real: $m = - 2$ , $v = 2 u$ : $\frac{1}{2 u} + \frac{1}{u} = \frac{3}{2 u} = - \frac{1}{10}$ , so $u = - 15 cm$ , $v = - 30 cm$ : real, inverted, enlarged, in front.
Virtual: $m = + 2$ , $v = - 2 u$ : $- \frac{1}{2 u} + \frac{1}{u} = \frac{1}{2 u} = - \frac{1}{10}$ , so $u = - 5 cm$ , $v = + 10 cm$ : virtual, erect, enlarged, behind the mirror.
The second (face within 10 cm, between P and F) is the shaving-mirror case.
Q6 · 3 marks · Short answerBoard level
The absolute refractive index of glass is 1.5 and that of a liquid X is 1.2. (a) Find the speed of light in each. (b) Find the refractive index of glass with respect to X. (c) A ray goes obliquely from X into glass. Does it bend towards or away from the normal? ( $c = 3 \times 1 0^{8} m s^{- 1}$ ) (3 marks)
Hint
Use $v = \frac{c}{n}$ , then $n_{21} = \frac{v _{1}}{v _{2}}$ .
Answer
(a) Glass: $\frac{3 \times 1 0 ^{8}}{1.5} = 2 \times 1 0^{8} m s^{- 1}$ ; X: $\frac{3 \times 1 0 ^{8}}{1.2} = 2.5 \times 1 0^{8} m s^{- 1}$ .
(b) $n_{g X} = \frac{2.5 \times 1 0 ^{8}}{2 \times 1 0 ^{8}} = 1.25$ .
(c) Towards the normal: glass is optically denser, so light slows down.
Q7 · 3 marks · Short answerBoard level
An object is placed 30 cm from a lens. The image is erect and one-quarter the size of the object. Identify the lens and find the image distance, the focal length and the power. (3 marks)
Hint
An erect diminished image from a lens at any position points to one type only. Use $m = \frac{v}{u}$ .
Answer
Concave lens (only it gives an erect, diminished image).
$m = + \frac{1}{4} = \frac{v}{- 30}$ , so $v = - 7.5 cm$ .
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = - \frac{4}{30} + \frac{1}{30} = - \frac{3}{30}$ , so $f = - 10 cm$ ; $P = - 10 D$ .
Q8 · 5 marks · Long answerHarder board level
(a) State the sign convention for focal lengths of lenses. (b) Compare the images formed by a convex lens and a concave lens when an object is placed (i) far beyond 2F and (ii) between F and O. (c) A concave lens has power −4 D. An object is placed 25 cm from it. Find the image position and magnification. (5 marks)
Hint
Find f in cm from the power first.
Answer
(a) Distances are measured from the optical centre; f is positive for a convex lens and negative for a concave lens.
(b) (i) Convex: real, inverted, diminished, between $F_{2}$ and $2 F_{2}$ ; concave: virtual, erect, diminished, between $F_{1}$ and O. (ii) Convex: virtual, erect, enlarged, same side; concave: virtual, erect, diminished, between $F_{1}$ and O.
(c) $f = \frac{1}{- 4} = - 0.25 m = - 25 cm$ ; $\frac{1}{v} = - \frac{1}{25} - \frac{1}{25}$ , so $v = - 12.5 cm$ ; $m = \frac{- 12.5}{- 25} = + 0.5$ .
Q9 · 3 marks · Short answerHOTS / competency
A student wants a convex lens that gives a real image of the same size as an object, with the object and the screen exactly 50 cm apart. What focal length and power must the lens have? Where should the lens be placed? (3 marks)
Hint
Same size means object at $2 F_{1}$ and image at $2 F_{2}$ . What is the total distance then?
Answer
Same-size real image: object at $2 F_{1}$ , image at $2 F_{2}$ , so object-to-screen distance $= 4 f = 50 cm$ .
$f = + 12.5 cm$ ; $P = \frac{1}{0.125} = + 8 D$ .
Lens exactly midway, 25 cm from both object and screen.
Q10 · 3 marks · Short answerHOTS / competency
A student solves: 'An object is 5 cm from a convex lens of focal length 10 cm.' She writes $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{10} + \frac{1}{5}$ and gets $v = + 3.3 cm$ , a real image. Find her error, solve correctly, and describe the image. (3 marks)
Hint
Check which formula belongs to lenses and which to mirrors.
Answer
She used the mirror formula. For a lens, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ , so $\frac{1}{v} = \frac{1}{10} - \frac{1}{5} = - \frac{1}{10}$ , $v = - 10 cm$ .
$m = \frac{v}{u} = \frac{- 10}{- 5} = + 2$ .
Virtual, erect, twice enlarged, on the same side as the object (object between $F_{1}$ and O).
Q11 · 3 marks · Short answerHarder board level
An object is placed 20 cm in front of a spherical mirror and, separately, 20 cm in front of a thin lens. The mirror gives an erect image twice the size of the object. The lens gives an erect image half the size of the object. Using the sign convention, find the focal length of each, with its sign, and name the mirror and the lens. (3 marks)
Hint
Erect means m is positive. Use $v = - m u$ for the mirror and $v = m u$ for the lens.
Answer
Mirror: $m = + 2$ , so $v = - m u = - 2 \times (- 20) = + 40$ cm; $\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{40} - \frac{1}{20} = - \frac{1}{40}$ , so $f = - 40$ cm: a concave mirror (object within F).
Lens: $m = + \frac{1}{2}$ , so $v = m u = - 10$ cm; $\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = - \frac{1}{10} + \frac{1}{20} = - \frac{1}{20}$ , so $f = - 20$ cm: a concave lens.
Both focal lengths are negative, so both are concave.
Q12 · 3 marks · Short answerHarder board level
An object is moved from 60 cm to 30 cm in front of a concave mirror of focal length 20 cm. Find the image position and magnification in each case. In which direction does the image move, by how much, and how does its size change? (3 marks)
Hint
Find v and m for both positions with $f = - 20$ cm, then compare.
Answer
At 60 cm: $\frac{1}{v} = - \frac{1}{20} + \frac{1}{60} = - \frac{2}{60}$ , so $v = - 30$ cm; $m = - \frac{- 30}{- 60} = - 0.5$ .
At 30 cm: $\frac{1}{v} = - \frac{1}{20} + \frac{1}{30} = - \frac{1}{60}$ , so $v = - 60$ cm; $m = - \frac{- 60}{- 30} = - 2$ .
The image moves 30 cm further from the mirror (from 30 cm to 60 cm in front) and grows from half to twice the object's size; it stays real and inverted.
Q13 · 3 marks · Short answerHarder board level
A convex lens of focal length 12 cm forms a virtual image three times the size of an object. Find the object distance and the image distance. Where is the object relative to the focus? (3 marks)
Hint
A virtual image from a lens is erect, so m = +3 and v = 3u.
Answer
$m = + 3 = \frac{v}{u}$ , so $v = 3 u$ .
$\frac{1}{3 u} - \frac{1}{u} = - \frac{2}{3 u} = \frac{1}{12}$ , so $u = - 8$ cm and $v = - 24$ cm.
The object is 8 cm from the lens, between $F_{1}$ and O; the image is 24 cm from the lens on the same side as the object.
Q14 · 5 marks · Long answerHarder board level
(a) Describe, as a labelled ray diagram in words, the path of a ray of light falling obliquely on a rectangular glass slab, marking the angles of incidence, refraction and emergence.
(b) Why is the emergent ray parallel to the incident ray but not along the same line?
(c) The refractive index of the glass is 1.5. Find the speed of light in it. ( $c = 3 \times 1 0^{8} m s^{- 1}$ )
(d) If the slab were replaced by one of the same shape with a refractive index of 1.25, would the ray bend more or less at the first face? Give a reason. (5 marks)
Hint
Measure every angle from the normal; use $v = \frac{c}{n}$ .
Answer
(a) Draw and label: slab outline ABCD; incident ray EO meeting AB at O; normal NN′ at O; refracted ray OO′ bent towards the normal; normal MM′ at O′ on CD; emergent ray O′H bent away from the normal; i between EO and NN′, r between OO′ and NN′, e between O′H and MM′; EO extended as a dotted line to show the sideways shift. Common slip: measuring angles from the surface. (2)
(b) The bending at AB (air to glass) and at CD (glass to air) is equal and opposite because the faces are parallel, so the emergent ray is parallel to the incident ray; but the ray has travelled along a slanting path inside the glass, so it comes out shifted sideways. (1)
(c) $v = \frac{c}{n} = \frac{3 \times 1 0 ^{8}}{1.5} = 2 \times 1 0^{8} m s^{- 1}$ . (1)
(d) Less. A smaller refractive index means light slows down less on entering, so its change of direction is smaller. (1)
Q15 · 4 marks · Case-basedHOTS / competency
Read the passage and answer the questions.
Suppose a science club tests three clear materials, P, Q and R, and finds their absolute refractive indices to be 1.25, 1.5 and 2.0. One member says, 'R must be the heaviest material, because it is the densest.' Another member sends a narrow ray obliquely from P into R, and then from R out into air.
(a) In which material does light travel slowest? Find its speed there. (1)
(b) Find the refractive index of R with respect to Q. (1)
(c) Describe the bending of the ray at the P-R boundary and at the R-air boundary. (1)
(d) Is the first member's reasoning correct? Justify your answer. (1)
( $c = 3 \times 1 0^{8} m s^{- 1}$ ) (4 marks)
Hint
Use $v = \frac{c}{n}$ and $n_{21} = \frac{v _{1}}{v _{2}}$ ; remember optical density is not mass density.
Answer
(a) R, which has the largest refractive index: $v = \frac{3 \times 1 0 ^{8}}{2.0} = 1.5 \times 1 0^{8} m s^{- 1}$ .
(b) In Q, $v = \frac{3 \times 1 0 ^{8}}{1.5} = 2 \times 1 0^{8} m s^{- 1}$ , so $n_{R Q} = \frac{v _{Q}}{v _{R}} = \frac{2 \times 1 0 ^{8}}{1.5 \times 1 0 ^{8}} = \frac{4}{3} \approx 1.33$ (no unit).
(c) P to R: optically rarer to denser, light slows and bends towards the normal. R to air: denser to rarer, light speeds up and bends away from the normal.
(d) No. R is optically the densest because it has the largest refractive index, but optical density is not mass density, so this tells us nothing about how heavy R is; for example, kerosene is optically denser than water though its mass density is less.

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.What is the pole of a spherical mirror?Show answer
The centre of its reflecting surface, a point on the mirror itself, labelled P.
2.How are R and f related for a small-aperture spherical mirror?Show answer
R = 2f, so the focus lies midway between P and C.
3.Where must an object be for a concave mirror to give a virtual, erect, enlarged image?Show answer
Between the pole P and the principal focus F.
4.What kind of image does a convex mirror always form?Show answer
Virtual, erect and diminished, behind the mirror between P and F.
5.Why does a ray through C of a concave mirror come back along its path?Show answer
It strikes the mirror along the normal, so the angle of incidence is zero.
6.Define the absolute refractive index of a medium.Show answer
The ratio of the speed of light in vacuum (or air) to the speed of light in the medium, n = c/v.
7.What happens to a ray passing through the optical centre of a lens?Show answer
It goes straight through without deviation.
8.What is 1 dioptre?Show answer
The power of a lens whose focal length is 1 metre; 1 D = 1 per metre.

Best order to study it

Revise the laws of reflection, then learn the spherical mirror terms (P, C, R, F, f, principal axis) and R = 2f until you can label a diagram from memory.
Learn the concave mirror image table and the four ray rules together by drawing each case, then the two convex mirror cases and the uses of both mirrors.
Learn the New Cartesian Sign Convention, then practise mirror formula and magnification numericals, always writing a signed data line first.
Study refraction: the bending rule, the glass slab activity, the two laws, Snell's law and refractive index with speed of light, including reading the refractive index table.
Learn lenses: terms, three ray rules and the convex and concave lens tables, then the lens formula, lens magnification and power, and finish with mixed numericals and practical-based questions.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you label P, C, F, R and the principal axis on a concave and a convex mirror, and say where C lies for each?
Can you state all six concave mirror cases (object position, image position, size, nature) without looking?
Can you name two uses each of concave and convex mirrors, with the reason for each?
Can you write the sign of u, v and f for a concave mirror, convex mirror, convex lens and concave lens?
Can you write the mirror formula, lens formula and both magnification formulas, and say where the minus signs go?
Can you explain why the emergent ray from a glass slab is parallel to the incident ray but shifted?
Can you state both laws of refraction and Snell's law?
Can you find refractive index from speeds, and say which of two media is optically denser?
Can you convert a power in dioptres to a focal length in cm, with sign and lens type, and back again?

Light – Reflection and Refraction: frequently asked questions

Do I need to derive the mirror formula or the lens formula for the board exam?

No. The 2026-27 syllabus says the derivation is not required for either formula. You must know both formulas, apply the sign convention correctly and solve numericals with them.

Do I have to memorise the refractive index table?

No. NCERT says you need not memorise the data. You should be able to read the table, pick the optically densest or rarest medium, and compare speeds of light in different media.

Why is there a minus sign in mirror magnification but not in lens magnification?

For a mirror, a real image forms in front, on the same side as the object, so v and u are both negative; the minus sign makes m negative for real images. For a lens, a real image forms on the other side, so v is positive and u negative, and v divided by u is already negative. Both rules give negative m for real, inverted images.

Is the formula for the power of lenses in contact part of the syllabus?

The idea that powers of lenses in contact add up appears in a More to Know box in NCERT, and CBSE says information in text boxes is not assessed in the board exam. It is simple to understand, but focus your preparation on the main text: power, focal length and the sign of power.

How do I remember which rays to use in a ray diagram?

Choose the two rays whose paths you know for sure. For mirrors, the ray parallel to the axis and the ray through or towards C are usually easiest; for lenses, the ray parallel to the axis and the ray through the optical centre. Always add arrowheads and use dotted lines for virtual rays and images.

Based on: CBSE Science (086) syllabus 2026-27; NCERT Science, Textbook for Class X (Reprint 2026-27)