Take a breath before you start this one. Light is the chapter that frightens students unnecessarily, and it really should not. There are no long derivations to memorise, no complicated proofs, no exceptions hiding in the corners. There are two formulas, one sign convention, and a handful of ray diagrams — and once those three things click, every single numerical in this chapter becomes the same puzzle wearing a different costume.
You already know more about this chapter than you think. You have seen your own stretched, upside-down reflection in a steel spoon. You have noticed that the road ahead of a bus looks strangely wide in the driver’s side mirror. You have dipped a spoon into a glass of water and watched it appear to snap at the surface. You have held a magnifying glass over a page. Every one of those everyday moments is an exam question in disguise. Our job here is simply to attach the correct vocabulary and the correct arithmetic to things your eyes have been observing since you were small.
So here is my promise to you. We will build this from absolutely zero. Nothing will be assumed except that you can add fractions. I will walk you through every ray diagram in words, hold your hand through every sign, and solve twenty-one examples with you before I ask you to solve anything alone. Go slowly. Keep a pencil and a rough notebook beside you and actually redraw the diagrams as you read — light is a chapter your hand has to learn, not only your eyes. If a section feels wobbly, read it twice. There is no prize for finishing fast.
What You’ll Learn
- Reflection Recap: The Rules You Already Know
- Spherical Mirrors And Their Special Points
- Images Formed By A Concave Mirror
- Images Formed By A Convex Mirror
- The New Cartesian Sign Convention
- The Mirror Formula And Magnification
- Everyday Uses Of Spherical Mirrors
- Refraction: Why Light Bends
- Laws Of Refraction And Refractive Index
- Refraction Through A Rectangular Glass Slab
- Spherical Lenses And Their Special Points
- Images Formed By Convex And Concave Lenses
- The Lens Formula And Magnification
- Power Of A Lens
- Everyday Uses Of Spherical Lenses
- Practical Skills: Focal Length By The Distant Object Method
- Practice Worksheet: Ten Questions With Answers
Your Game Plan
Do not attack this chapter randomly. Work through it in this order and it will feel half as heavy:
- Learn the vocabulary of a spherical mirror first — pole, centre of curvature, principal focus, focal length. Numericals are impossible until these words are automatic.
- Draw the six concave-mirror ray diagrams by hand, three times each, until you can produce them without peeping. This is the single highest-scoring habit in the chapter.
- Master the New Cartesian sign convention before you touch a single formula. Ninety per cent of lost marks in this chapter are sign errors, not physics errors.
- Now do the mirror formula numericals. Then, and only then, move to refraction.
- Treat lenses as a rerun of mirrors with one formula changed. If mirrors are solid, lenses take a fraction of the effort.
- Finish with power of a lens and the two practical skills, then attempt the worksheet with a closed book and a timer.
One more thing before you start. If your Science revision is only just getting going, settle Chemical Reactions and Equations first. It trains the habit of writing every step out in full, and that is precisely the habit Light rewards.
Reflection Recap: The Rules You Already Know
This part is quick, because you met it in earlier classes. When light travelling in a straight line strikes a polished surface, it bounces off in a perfectly predictable direction. That bouncing is reflection, and it obeys two rules that have never once been broken.
1. The angle of incidence equals the angle of reflection (∠i = ∠r).
2. The incident ray, the reflected ray and the normal drawn at the point of incidence all lie in the same plane.
Both angles are always measured from the normal — the line drawn perpendicular to the surface — never from the surface itself.
Think of a ball thrown at a smooth wall. Throw it straight at the wall and it comes straight back. Throw it at a slant and it leaves at the same slant on the other side. Light behaves in exactly that well-mannered way, and it does so whether the mirror is flat or curved. That last point matters more than students realise: curved mirrors do not use new laws of reflection. They use the same two laws. The only thing that changes is that the normal points in a different direction at every point of a curved surface, because the surface keeps tilting. That single fact is the entire reason curved mirrors do interesting things.
You also remember what a plane mirror does to an image: it is always virtual, erect, the same size as the object, and formed as far behind the mirror as the object is in front, with left and right apparently swapped (lateral inversion). That is the complete plane-mirror story, and the board expects it as prior knowledge rather than as a fresh topic. Take thirty seconds to check it feels solid, then move on — the real chapter starts at the next heading.
Spherical Mirrors And Their Special Points
Imagine a hollow glass ball, like a large Christmas bauble. Now slice a small piece off it — a shallow cap, roughly the size of a bangle. Silver one side of that cap. What you are holding is a spherical mirror: a mirror whose reflecting surface is part of a sphere.
Which side you silver decides everything:
- Concave mirror — the reflecting surface is the hollow, caved-in side, the side that faces the centre of the original sphere. Picture the inside of a steel spoon, the part that holds the daal. It is also called a converging mirror, because it gathers parallel light to a point.
- Convex mirror — the reflecting surface is the bulging outer side. Picture the back of the same spoon. It is called a diverging mirror, because it spreads parallel light apart.
Now the vocabulary. Learn these five terms properly, because every later sentence in this chapter uses them like ordinary words:
- Pole (P) — the exact centre of the mirror’s reflecting surface. All distances in this chapter are measured from here. Think of it as the origin of your graph.
- Centre of curvature (C) — the centre of the imaginary sphere the mirror was cut from. For a concave mirror it sits in front of the mirror; for a convex mirror it sits behind.
- Radius of curvature (R) — the radius of that imaginary sphere, i.e. the distance PC.
- Principal axis — the straight line joining P and C, extended both ways. It hits the mirror surface at right angles. Every diagram you draw starts with this line.
- Principal focus (F) — where rays that arrive parallel to the principal axis actually meet after reflection (concave), or where they appear to come from (convex). The distance PF is the focal length (f).
And here is the small relationship that saves you time in nearly every numerical:
R = 2f, or equivalently f = R / 2.
For a spherical mirror of small aperture, the principal focus lies exactly midway between the pole and the centre of curvature. So a mirror with R = 30 cm has f = 15 cm.
Why does it work? Take a ray coming in parallel to the principal axis and hitting the concave mirror at some point close to the pole. The normal at that point is the line joining that point to C, because a radius always meets a sphere at right angles. The law of reflection makes the reflected ray bend by twice the angle between the incoming ray and that normal, and a little geometry with that isosceles triangle drops the reflected ray onto the axis at exactly half of PC. You are not asked to reproduce that argument in the exam — but knowing there is a reason, rather than a rule dropped from the sky, is what stops you forgetting it.
One more word: aperture is the width of the mirror, the diameter of its reflecting surface. Everything in this chapter quietly assumes a small aperture compared with R. That is why R = 2f holds and why your ray diagrams behave. A wide, deep mirror would blur the focus into a smear — an effect called spherical aberration, which is beyond your syllabus but worth knowing exists.
(b) A convex mirror has a focal length of 25 cm. Find its radius of curvature.
Part (a). Use f = R / 2.
f = 36 / 2 = 18 cm.
So the principal focus lies 18 cm in front of the pole, exactly halfway to C.
Part (b). Use R = 2f.
R = 2 × 25 = 50 cm.
For a convex mirror this centre of curvature is 50 cm behind the mirror, which is why we will later call it positive.
Tutor’s note: at this stage we are only handling magnitudes, so no minus signs are needed yet. Signs arrive in a moment and we will apply them deliberately.
Writing f = 2R instead of R = 2f. Anchor it with a sentence in your own language: the focus is the nearer point, so its number is the smaller one. If R = 30, then f = 15, never 60. Whenever you write it down, glance back and ask whether the focus came out closer to the mirror than C. If it did not, you flipped the formula.
Images Formed By A Concave Mirror
To find where an image forms, we do not trace thousands of rays. We trace just two, because two straight lines are enough to fix a point. And we choose two rays whose behaviour we already know without calculation. Here are the four standard rays — you only ever need any two of them.
- A ray parallel to the principal axis reflects through F.
- A ray passing through F reflects parallel to the principal axis. (Just the first rule played backwards — light paths are reversible.)
- A ray passing through C hits the mirror along the normal and comes straight back along its own path.
- A ray striking the pole reflects making an equal angle on the other side of the principal axis.
Where the two reflected rays actually cross, you get a real image — light genuinely arrives there, so it can be caught on a screen. Where they only appear to cross when produced backwards behind the mirror, you get a virtual image — no light is truly there, so no screen will ever catch it, though your eye happily sees it.
Real images are always inverted and form in front of the mirror. Virtual images are always erect and form behind the mirror. There is no exception anywhere in this chapter, so if your answer says “real and erect”, you have made an arithmetic slip — go back and check.
Now, the six positions. Do not memorise this table like a poem. Read it while sketching, and you will notice a lovely pattern: as the object walks in from infinity towards the mirror, the image walks outwards from F, growing all the way, until the object crosses F — and then the image gives up, jumps behind the mirror and turns virtual.
| Position of object | Position of image | Size of image | Nature of image |
|---|---|---|---|
| At infinity | At the focus F | Highly diminished, point-sized | Real and inverted |
| Beyond C | Between F and C | Diminished | Real and inverted |
| At C | At C | Same size as the object | Real and inverted |
| Between C and F | Beyond C | Enlarged | Real and inverted |
| At F | At infinity | Highly enlarged | Real and inverted |
| Between P and F | Behind the mirror | Enlarged | Virtual and erect |
Notice rows two and four. They are mirror images of each other in a very literal sense: object beyond C gives image between F and C, and object between C and F gives image beyond C. Physicists call this reversibility, and it means you only really have to remember one of those two rows. Row three, object at C, is the balance point where object and image sit on top of each other at the same size — a fact examiners love, because it lets them ask “where must the object be placed to get an image of the same size?” and the answer is simply at the centre of curvature, that is, at a distance R from the mirror.
When a question asks you to “draw a ray diagram”, marks are given for four specific things: the principal axis with P, F and C labelled; correct arrowheads showing the direction of light; two correctly chosen rays; and the image drawn with a clear label saying real/virtual and erect/inverted. Draw it large. A cramped two-centimetre diagram loses marks that a full half-page diagram would have earned.
Images Formed By A Convex Mirror
After that six-row table, the convex mirror is a holiday. It has only two cases, and honestly they say almost the same thing. A convex mirror bulges towards you, so it throws reflected rays apart. Those spreading rays never meet in front of the mirror, so they can only seem to come from a point behind it.
| Position of object | Position of image | Size of image | Nature of image |
|---|---|---|---|
| At infinity | At F, behind the mirror | Highly diminished, point-sized | Virtual and erect |
| Anywhere between infinity and the pole | Between P and F, behind the mirror | Diminished | Virtual and erect |
So here is the sentence to carry into the exam hall: a convex mirror always gives a virtual, erect, diminished image, and that image always lives between the pole and the focus, behind the mirror. Always. No matter where the object goes. That is why it is such a dependable little device — it never surprises you.
Why does it work? Because F and C are behind a convex mirror, the object can never be placed “beyond C” or “between C and F” the way it can with a concave mirror. Every real object is stuck on the same side, in the one region available, so it gets the one kind of image available. Shrinking the image is the price paid for the benefit: squeezing a wide scene into a small mirror, which is exactly what a driver wants.
Saying a convex mirror “can form a real image if the object is very close”. It cannot — not at any distance, not ever. Similarly, a concave mirror forms a virtual image in exactly one situation: when the object is between P and F. If your numerical gives a positive image distance for a concave mirror, check that the object really was inside the focus; otherwise you have an arithmetic error.
The New Cartesian Sign Convention
Please slow down here. This is the most important half-page in the whole chapter. Students who lose marks in Light almost never lose them because they forgot the formula — they lose them because they put a plus where a minus belonged. Get this right and the numericals practically solve themselves.
The idea is simple: we lay a graph over the mirror. The pole becomes the origin, and the principal axis becomes the x-axis. Then we fix directions once and never argue about them again.
1. The object is always placed to the left of the mirror, so light travels left to right.
2. All distances are measured from the pole.
3. Distances measured in the direction of the incident light (to the right) are positive; distances measured against it (to the left) are negative.
4. Heights measured upwards from the principal axis are positive; heights measured downwards are negative.
Everything else follows from those four lines. You do not have to memorise a long list of sign rules — you can derive each one in two seconds. Let me show you how, and then give you the finished table to check yourself against.
The object sits to the left, so the object distance u is negative — always, for every mirror and every lens in this chapter, without a single exception. A real image forms in front of the mirror, also to the left, so its v is negative. A virtual image forms behind the mirror, to the right, so its v is positive. For a concave mirror, F and C are in front, on the left, so f and R are negative. For a convex mirror, F and C are behind, on the right, so f and R are positive. See? You just derived the whole table from one rule about direction.
| Quantity | Sign | Reason in one line |
|---|---|---|
| Object distance, u | Always negative | The object is always placed on the left of the mirror or lens. |
| Focal length of concave mirror, f | Negative | Its focus lies in front of the mirror, on the left. |
| Focal length of convex mirror, f | Positive | Its focus lies behind the mirror, on the right. |
| Image distance, v (real image) | Negative | Real images form in front of the mirror, on the left. |
| Image distance, v (virtual image) | Positive | Virtual images form behind the mirror, on the right. |
| Object height, h | Positive | We always draw the object standing upright above the axis. |
| Image height, h′ (erect image) | Positive | Measured upwards from the principal axis. |
| Image height, h′ (inverted image) | Negative | Measured downwards from the principal axis. |
| Focal length of convex lens | Positive | Light converges to a real focus on the far side, to the right. |
| Focal length of concave lens | Negative | Its focus is on the same side as the object, to the left. |
(a) “A candle is kept 25 cm from a concave mirror of radius of curvature 40 cm.”
Object is on the left, so u = −25 cm.
Concave mirror, so C and F are in front: R = −40 cm and f = R / 2 = −20 cm.
(b) “A pen is held 12 cm from a convex mirror of focal length 16 cm.”
u = −12 cm (objects are always negative).
Convex mirror, so f = +16 cm and R = 2f = +32 cm.
(c) “An image is formed 18 cm behind a mirror.”
Behind means to the right, so v = +18 cm, and the image must therefore be virtual and erect.
(d) “A 4 cm tall object gives a 6 cm tall inverted image.”
h = +4 cm; inverted means below the axis, so h′ = −6 cm.
That is the entire skill. Do this translation step on paper, before you touch the formula, in every single numerical.
Substituting u as a positive number because the question said “placed at a distance of 25 cm”. The question gives you a distance, which has no sign; you must attach the sign. Build the habit of writing a short list — u = −25 cm, f = −20 cm, v = ?, m = ? — at the top of every answer. Examiners award marks for that list even if your arithmetic later slips.
The Mirror Formula And Magnification
Ray diagrams tell you roughly where an image sits. The mirror formula tells you exactly. It links three numbers — object distance u, image distance v and focal length f — so that if you know any two, the third has nowhere to hide.
1/v + 1/u = 1/f
m = h′/h = −v/u
Here m is the magnification, h is the object height and h′ is the image height. The same pair of formulas works for a concave mirror and a convex mirror — you never switch formulas, you only switch signs.
Read the magnification result carefully, because it carries two pieces of information in one number. Its size tells you how much bigger or smaller the image is: |m| greater than 1 means enlarged, less than 1 means diminished, exactly 1 means the same size. Its sign tells you the nature: a negative m means real and inverted, a positive m means virtual and erect. So m = −0.5 instantly reads as “real, inverted, half the size” without any further thought.
Why the minus sign in m = −v/u? It is not decoration. For a real image, u and v are both negative, so v/u is positive — but a real image is inverted, which needs a negative m. The minus sign is what forces the formula to agree with reality. That is the honest reason it is there, and once you see it you will stop dropping it.
The Class 10 syllabus does not require you to derive the mirror formula or the lens formula. You will never be asked to prove them. Learn to apply them flawlessly instead, and spend the time you save on ray diagrams and sign practice. That is where the marks actually live.
Let us build a fixed four-step routine and never deviate from it. Step 1: list u, v, f, m with signs. Step 2: write the formula. Step 3: substitute and solve, using a common denominator rather than decimals. Step 4: translate the answer back into a sentence about the image. Step 4 is worth a mark on its own and students forget it constantly.
Step 1 — List with signs.
R = −30 cm, so f = R / 2 = −15 cm. u = −40 cm. v = ? m = ?
Step 2 — Formula. 1/v + 1/u = 1/f, so 1/v = 1/f − 1/u.
Step 3 — Substitute.
1/v = 1/(−15) − 1/(−40) = −1/15 + 1/40
Common denominator 120: 1/v = −8/120 + 3/120 = −5/120 = −1/24
Therefore v = −24 cm.
m = −v/u = −(−24)/(−40) = 24/(−40) = −0.6
Step 4 — Describe. v is negative, so the image is 24 cm in front of the mirror — real. m is negative, so it is inverted. |m| = 0.6, less than 1, so it is diminished. Cross-check with our table: object beyond C should give an image between F and C, that is between 15 cm and 30 cm. Our 24 cm sits neatly inside that range, so the answer is consistent.
Step 1. h = +6 cm, u = −30 cm, f = −10 cm.
Step 2 and 3.
1/v = 1/f − 1/u = −1/10 + 1/30
Common denominator 30: 1/v = −3/30 + 1/30 = −2/30 = −1/15
v = −15 cm
m = −v/u = −(−15)/(−30) = 15/(−30) = −0.5
h′ = m × h = (−0.5) × 6 = −3 cm
Step 4. The image forms 15 cm in front of the mirror, is real and inverted, and is 3 cm tall (the minus sign is telling you “inverted”, not that the pencil has negative height). Now the sanity check: R = 2f = 20 cm, so C sits at 20 cm and the object at 30 cm is beyond C. The image at 15 cm is between F (10 cm) and C (20 cm), exactly as the table predicts. Always run this sanity check.
Step 1. h = +2 cm, u = −8 cm, f = −12 cm. Notice the object is inside the focus, so expect something virtual.
Step 2 and 3.
1/v = 1/f − 1/u = −1/12 + 1/8
Common denominator 24: 1/v = −2/24 + 3/24 = 1/24
v = +24 cm
m = −v/u = −(24)/(−8) = +3
h′ = 3 × 2 = +6 cm
Step 4. The positive v places the image 24 cm behind the mirror, so it is virtual. The positive m of +3 makes it erect and three times as large, so the image is 6 cm tall. This is precisely what happens when you bring a shaving mirror close to your face — a big, upright, comfortable view.
Step 1. f = +20 cm (convex, so positive), u = −30 cm.
Step 2 and 3.
1/v = 1/f − 1/u = 1/20 + 1/30
Common denominator 60: 1/v = 3/60 + 2/60 = 5/60 = 1/12
v = +12 cm
m = −v/u = −(12)/(−30) = +0.4
Step 4. The image is 12 cm behind the mirror, virtual and erect, and 0.4 times the object’s size — that is, shrunk to two-fifths. Sanity check: for a convex mirror the image must fall between P and F, i.e. between 0 and 20 cm behind. Our 12 cm obeys that. Every convex-mirror answer you ever get must obey it too.
Step 1 — Decode the words. “Real” means inverted, so the magnification is negative: m = −3. And f = −18 cm.
Step 2 — Turn m into a relation between v and u.
m = −v/u = −3, so v/u = 3, giving v = 3u.
Step 3 — Substitute into the mirror formula.
1/v + 1/u = 1/f
1/(3u) + 1/u = 1/(−18)
1/(3u) + 3/(3u) = −1/18
4/(3u) = −1/18
3u = 4 × (−18) = −72
u = −24 cm, and so v = 3u = −72 cm.
Step 4 — Check and describe. Substitute back: 1/(−72) + 1/(−24) = −1/72 − 3/72 = −4/72 = −1/18 ✓. The object must be placed 24 cm in front of the mirror and the image forms 72 cm in front, real and inverted. Notice that 24 cm lies between F (18 cm) and C (36 cm), and the image at 72 cm lies beyond C — exactly the row of the table that promises an enlarged real image. Beautiful agreement.
Step 1. A convex mirror always gives a virtual erect image, so m is positive: m = +1/4. And u = −60 cm.
Step 2 — Get v.
m = −v/u, so 1/4 = −v/(−60) = v/60, giving v = +15 cm.
Step 3 — Get f.
1/f = 1/v + 1/u = 1/15 + 1/(−60)
Common denominator 60: 1/f = 4/60 − 1/60 = 3/60 = 1/20
f = +20 cm
R = 2f = +40 cm
Step 4. The positive f confirms the mirror is convex, as stated. Its focal length is 20 cm and its radius of curvature is 40 cm, both measured behind the mirror. Since v = 15 cm lies between 0 and f = 20 cm, the answer passes the convex-mirror sanity check.
Two classics. First, writing 1/v = 1/f + 1/u when rearranging — it must be 1/v = 1/f − 1/u. Move the term across properly every time. Second, after getting 1/v = −1/24, students write v = −24 correctly but then, when 1/v = −5/120, they write v = −5/120 instead of flipping the fraction. Always remember the final flip: if 1/v = −5/120, then v = −120/5 = −24.
Everyday Uses Of Spherical Mirrors
This section is pure marks. It is short, it is factual, and it appears in almost every paper — usually as “give one use and justify it”. The trick is that the justification must name the property, not just the object.
Where concave mirrors are used, and why:
- Torches, vehicle headlamps and searchlights. The bulb is placed at the focus, so the reflected light leaves as a strong parallel beam that travels far without spreading. This is the converging property used in reverse.
- Shaving mirrors and make-up mirrors. The face is held closer than the focus, so the image is virtual, erect and enlarged — a comfortable magnified view.
- Dentists’ mirrors. Same reason: a small tooth held inside the focus appears large and upright.
- Solar cookers and solar furnaces. A large concave reflector collects sunlight arriving parallel from a very distant Sun and concentrates it at the focus, where the temperature climbs sharply.
- Reflecting telescopes. A big concave mirror gathers faint parallel light from stars and brings it to a focus to form an image.
Where convex mirrors are used, and why:
- Rear-view mirrors on cars, buses and two-wheelers. Two reasons, and the board wants both: the image is always erect, so the driver is not confused, and the mirror gives a much wider field of view than a plane mirror of the same size, so more of the road behind is visible.
- Security mirrors in shops, and mirrors at blind corners of narrow lanes and in car parks. Again, the wide field of view lets one small mirror watch a large area.
- Street-light reflectors. A convex reflector spreads light over a broader patch of road.
Vehicle side mirrors often carry a line saying objects are closer than they appear. That is convex-mirror physics being honest with you: because the image is diminished, a following car looks smaller and therefore feels further away than it really is. The trade-off is deliberate — drivers accept a slightly misleading sense of distance in exchange for a much wider view.
Refraction: Why Light Bends
Half the chapter is done. Take a stretch. What follows is a completely new idea, so clear your head of mirrors for a minute.
Put a steel spoon into a glass of water and look from the side. The spoon appears bent, or even broken, at the water surface. Look at a coin at the bottom of a bucket — the bucket seems shallower than it is. Look through the shimmering air above a hot road in May and distant objects wobble. In all three cases nothing is actually bent, shallow or wobbling. What is changing is the direction of the light as it passes from one transparent medium into another. That bending is called refraction.
Light travels at different speeds in different transparent media. It is fastest in vacuum (and almost as fast in air) and slower in water, glass or diamond. When a light ray crosses a boundary at an angle, the change in speed forces a change in direction. No change in speed means no bending — which is why a ray hitting the surface straight on, along the normal, goes through undeviated.
Why does a speed change bend the ray? Here is the analogy I like best. Imagine a marching band walking in a straight line across a field and stepping, at an angle, onto soft sand where everyone must walk slower. The marchers on one end reach the sand first and slow down while the others are still on firm ground moving quickly. The whole line pivots. It has not been pushed sideways by anything; it turned because one edge slowed before the other. Light does exactly this at the boundary between air and glass.
That analogy also gives you the direction of the bend for free, and this is the pair of statements you must have on your fingertips:
- Going from a rarer medium to a denser medium (air into glass, say), light slows down and bends towards the normal. So the angle of refraction is smaller than the angle of incidence.
- Going from a denser medium to a rarer medium (glass into air), light speeds up and bends away from the normal. So the angle of refraction is larger than the angle of incidence.
A quick word on the word “denser”. In this chapter it means optically denser, which is about how much a medium slows light down — not about mass per unit volume. Kerosene is less dense than water in the everyday sense, yet it is optically denser, because light travels more slowly through it. Do not mix the two meanings.
Claiming that light always bends when it enters a new medium. It does not. If a ray meets the surface along the normal — that is, at an angle of incidence of 0° — it carries straight on with no bending at all, even though its speed does change. Only the speed changes; the direction does not. Examiners ask this one often, so keep it ready.
Laws Of Refraction And Refractive Index
“It bends towards the normal” is a good start, but physics wants a number. How much does it bend? Two laws answer that.
First law: The incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane.
Second law (Snell’s law): For light passing from one medium to another, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant.
sin i / sin r = constant = n21
This constant is the refractive index of medium 2 with respect to medium 1.
Notice how sensible that is. It does not say the angles themselves are in a fixed ratio — they are not. It says their sines are. And the constant depends only on the two materials involved, never on the angle you chose. That is what makes refractive index a genuine property of a substance, like density or melting point.
There are two flavours of refractive index and you must keep them apart:
- Absolute refractive index (n) — measured with respect to vacuum. n = c / v, where c is the speed of light in vacuum (about 3 × 108 m/s) and v is its speed in the medium. Since light is fastest in vacuum, n is always greater than 1 for any real material.
- Relative refractive index (n21) — of medium 2 with respect to medium 1. n21 = v1 / v2 = n2 / n1. Note the flip: speeds go one way round, refractive indices the other. This one can be less than 1, when medium 2 is optically rarer than medium 1.
Why is n = c/v? Because refractive index is really a measure of how much a medium holds light back. A refractive index of 1.5 says light crawls through at two-thirds of its vacuum speed. That single sentence explains why higher n means more bending: the bigger the speed change at the boundary, the harder the ray pivots.
Here are values worth carrying in your head. You will not be asked to recite the whole list, but knowing that water is about 1.33, glass about 1.5 and diamond about 2.42 will save you in a multiple-choice question.
| Medium | Absolute refractive index (approx.) | Speed of light in it (× 108 m/s) |
|---|---|---|
| Vacuum | 1 (exactly) | 3.00 |
| Air | 1.0003 | About 3.00 |
| Ice | 1.31 | 2.29 |
| Water | 1.33 | 2.26 |
| Kerosene | 1.44 | 2.08 |
| Crown glass | 1.52 | 1.97 |
| Dense flint glass | 1.65 | 1.82 |
| Diamond | 2.42 | 1.24 |
Read the last row and the sparkle of a diamond stops being a mystery. Light slows to well under half speed inside it, bends very sharply, bounces around and comes back out in bright flashes.
Formula. n = c / v
Substitute. n = (3 × 108) ÷ (2 × 108)
The powers of ten cancel, leaving n = 3 / 2 = 1.5
Read the answer. Refractive index has no unit — it is a ratio of two speeds, so the units cancel. A value of 1.5 tells you light moves at two-thirds of its vacuum speed inside the block, which matches ordinary glass.
Rearrange. n = c / v, so v = c / n.
Substitute. v = (3 × 108) ÷ 2.42
3 ÷ 2.42 = 1.2397 (to four figures)
v = 1.24 × 108 m/s (to three significant figures)
Read the answer. That is only about 41 per cent of the vacuum speed — the slowest of any common material, and the reason diamond bends light more than anything else in the table.
(a) nglass, water = nglass / nwater = 1.50 ÷ 1.33 = 1.13 (to 2 d.p.)
Greater than 1, so glass is optically denser than water. A ray going from water into glass bends towards the normal.
(b) nwater, glass = nwater / nglass = 1.33 ÷ 1.50 = 0.89 (to 2 d.p.)
Less than 1, confirming that water is optically rarer than glass. A ray going from glass into water bends away from the normal.
Notice: 1.13 × 0.89 is approximately 1. The two relative refractive indices are reciprocals of each other — a neat check you can run in three seconds.
Formula. sin i / sin r = n, so sin r = sin i / n.
Substitute.
sin r = sin 30° ÷ 1.5 = 0.5 ÷ 1.5 = 0.3333
r = sin−1(0.3333) = 19.5° (approximately)
Check it makes sense. The ray went from rarer air into denser glass, so it should bend towards the normal — and indeed 19.5° is smaller than 30°. Whenever you finish a Snell’s law calculation, ask which way the ray was travelling and confirm the angle moved in the right direction. This one check catches almost every error.
Writing sin r = sin i × n instead of dividing. Remember the physical meaning: entering a denser medium, the angle must get smaller, so you divide by n. Also, never give refractive index a unit — writing “n = 1.5 m/s” loses the mark instantly.
Refraction Through A Rectangular Glass Slab
This is the experiment you will actually perform in the lab, so understand it properly rather than just memorising the picture. You take a rectangular glass slab, place it on a sheet of paper, send a ray into one long face with pins, and track where it comes out of the opposite face.
The ray meets two parallel surfaces, so it refracts twice:
- At the first surface (air to glass): the ray enters a denser medium, slows down, and bends towards the normal. The angle inside the glass, r, is smaller than the angle of incidence, i.
- At the second surface (glass to air): the ray enters a rarer medium, speeds up, and bends away from the normal. The angle it leaves at is called the angle of emergence, e.
Because the two faces are parallel, the two bends are equal and opposite, and they cancel out in direction. The result is the headline fact of the whole experiment:
1. The angle of emergence equals the angle of incidence (e = i).
2. The emergent ray is parallel to the incident ray — the ray leaves travelling in exactly its original direction.
3. But it has been shifted sideways. This sideways shift is called lateral displacement.
The slab therefore changes a ray’s position, not its direction.
Lateral displacement grows when the slab is thicker, when the angle of incidence is larger, and when the glass has a higher refractive index. It becomes zero when the ray enters along the normal, because then there is no bending to undo. You are expected to know these dependencies qualitatively and to measure the shift in the lab — you are not asked for a formula for it.
Why is the emergent ray parallel? Apply Snell’s law twice. Going in: sin i = n sin r. Coming out, the angle inside the glass at the second face equals r (alternate angles between two parallel normals), so n sin r = sin e. Compare the two statements and sin i = sin e, hence i = e. The mathematics simply confirms what the picture shows.
First surface, air to glass.
sin r = sin i / n = sin 45° ÷ 1.5 = 0.7071 ÷ 1.5 = 0.4714
r = sin−1(0.4714) = 28.1° (approximately)
The ray has bent towards the normal, as it must on entering denser glass.
Second surface, glass to air.
The normals at the two faces are parallel, so the angle of incidence inside the glass at the second face is also 28.1°.
Now light goes from glass to air, so Snell’s law reads n sin(28.1°) = sin e
sin e = 1.5 × 0.4714 = 0.7071
e = sin−1(0.7071) = 45°
Conclusion. e = i = 45°, so the emergent ray is parallel to the incident ray, displaced sideways. This is why you can read a page through a thick glass slab and the text looks shifted but not distorted.
When you draw the labelled diagram, mark all four things by name: the angle of incidence, the angle of refraction, the angle of emergence, and the lateral displacement drawn as the perpendicular distance between the incident ray produced forwards and the emergent ray. Also draw the normals as dotted lines at both faces. Students routinely lose a mark by forgetting the second normal.
Spherical Lenses And Their Special Points
Good news: if mirrors are solid in your head, lenses will take you half the time. The vocabulary rhymes, the tables rhyme, and only the formula changes by one sign.
A lens is a piece of transparent material bounded by two surfaces, at least one of which is curved. Light refracts as it enters and again as it leaves, and the shape decides what those two refractions do together.
- Convex lens (double convex) — bulges outwards, thicker in the middle, thinner at the edges. It brings parallel rays together at a point, so it is a converging lens.
- Concave lens (double concave) — caves inwards, thinner in the middle, thicker at the edges. It spreads parallel rays apart, so it is a diverging lens.
Here is the vocabulary, and notice how closely it matches the mirror list:
- Optical centre (O) — the central point of the lens. It plays the role the pole played for a mirror: all distances are measured from here. A ray passing through O goes straight on without deviation, which makes it a wonderfully easy ray to draw.
- Centres of curvature (C1 and C2) — a lens has two curved surfaces, so it has two centres of curvature, one on each side.
- Principal axis — the line through both centres of curvature, passing through O.
- Principal focus (F) — a lens has two foci, one on each side, at equal distances from O. For a convex lens, parallel rays actually meet at the focus on the far side. For a concave lens, they only appear to diverge from the focus on the near side.
- Focal length (f) — the distance from O to F. The point at twice this distance is written 2F, and it does the job that C did for mirrors.
The three rays you use to build every lens diagram are these:
- A ray parallel to the principal axis refracts through F on the far side (convex), or appears to come from F on the near side (concave).
- A ray heading through F emerges parallel to the principal axis.
- A ray through the optical centre passes straight through, undeviated.
Using C instead of 2F when describing lens positions. With mirrors we say “object at C”; with lenses we say “object at 2F1”. They are different letters for a similar idea, and mixing them up in a written answer costs marks even when the physics is right.
Images Formed By Convex And Concave Lenses
Compare this table with the concave-mirror table you already learned. The pattern is identical — only the labels change, with 2F standing in for C, and the image appearing on the opposite side of the lens instead of the same side. If you learned the mirror table properly, you have already learned this one.
| Position of object (convex lens) | Position of image | Size of image | Nature of image |
|---|---|---|---|
| At infinity | At the focus F2 | Highly diminished, point-sized | Real and inverted |
| Beyond 2F1 | Between F2 and 2F2 | Diminished | Real and inverted |
| At 2F1 | At 2F2 | Same size as the object | Real and inverted |
| Between F1 and 2F1 | Beyond 2F2 | Enlarged | Real and inverted |
| At F1 | At infinity | Highly enlarged | Real and inverted |
| Between F1 and O | On the same side as the object | Enlarged | Virtual and erect |
That final row is the magnifying glass. Hold a convex lens closer to the page than its focal length and the letters swell up, upright and clear. Pull it further away, past the focus, and the image suddenly flips upside down — you have crossed from the last row into the real-image rows. Try it with any reading lens; feeling that flip happen in your hand is worth more than reading this paragraph twice.
The concave lens, like the convex mirror, is refreshingly predictable:
| Position of object (concave lens) | Position of image | Size of image | Nature of image |
|---|---|---|---|
| At infinity | At the focus F1, same side as the object | Highly diminished, point-sized | Virtual and erect |
| Anywhere between infinity and O | Between F1 and O, same side as the object | Diminished | Virtual and erect |
So: a concave lens always gives a virtual, erect, diminished image, always between F and O, always on the same side as the object. Learn that sentence and two marks are permanently yours.
The Lens Formula And Magnification
Now the arithmetic. Look very carefully at the two differences from the mirror formula, because they are the only two things you have to hold separately in your head.
1/v − 1/u = 1/f
m = h′/h = v/u
Compare with mirrors: the formula has a minus where the mirror had a plus, and the magnification has no minus sign in front. The sign convention itself is unchanged — measure everything from the optical centre, with the direction of incident light positive.
The reading of m is exactly as before: negative m means real and inverted, positive m means virtual and erect, |m| above 1 means enlarged. Do not let the missing minus sign confuse you — it is missing precisely so that the interpretation stays identical. For a lens forming a real image, u is negative and v is positive, so v/u comes out negative all by itself. The formula does the work; you just have to write it correctly.
One more sign fact to lock in: a convex lens has positive f, a concave lens has negative f. Note that this is the opposite way round from mirrors, where concave was negative. The reason is physical rather than arbitrary: a convex lens brings light to a real focus on the far side, in the direction the light is travelling, which is the positive direction.
Step 1. u = −30 cm, f = +20 cm (convex, so positive).
Step 2 and 3.
1/v − 1/u = 1/f, so 1/v = 1/f + 1/u
1/v = 1/20 + 1/(−30) = 1/20 − 1/30
Common denominator 60: 1/v = 3/60 − 2/60 = 1/60
v = +60 cm
m = v/u = 60 ÷ (−30) = −2
Step 4. Positive v means the image is 60 cm on the far side of the lens, so it is real. Negative m means inverted, and |m| = 2 means twice the size. Sanity check with the table: the object at 30 cm lies between F (20 cm) and 2F (40 cm), so we expect a real, inverted, enlarged image beyond 2F. Our image at 60 cm is indeed beyond 40 cm. Perfect.
Step 1. u = −10 cm, f = +15 cm. The object is inside the focus, so expect a virtual image.
Step 2 and 3.
1/v = 1/f + 1/u = 1/15 − 1/10
Common denominator 30: 1/v = 2/30 − 3/30 = −1/30
v = −30 cm
m = v/u = (−30) ÷ (−10) = +3
Step 4. Negative v puts the image 30 cm on the same side as the stamp, so it is virtual. Positive m of +3 makes it erect and three times as large. This is a magnifying glass doing its job, and note that the image is further from the lens than the object — which is exactly why it looks bigger.
Step 1. u = −20 cm, f = −20 cm (concave, so negative).
Step 2 and 3.
1/v = 1/f + 1/u = −1/20 − 1/20 = −2/20 = −1/10
v = −10 cm
m = v/u = (−10) ÷ (−20) = +0.5
Step 4. The image is 10 cm from the lens on the same side as the object, virtual, erect and half the size. Sanity check: for a concave lens the image must lie between F (20 cm) and O (0 cm). Our 10 cm sits inside that range. Every concave-lens answer must, so use this as your standard check.
Step 1 — Decode. “On a screen” means the image is real, so it is inverted and m = −3. And f = +15 cm.
Step 2 — Relate v and u.
m = v/u = −3, so v = −3u.
Step 3 — Substitute into the lens formula.
1/v − 1/u = 1/f
1/(−3u) − 1/u = 1/15
−1/(3u) − 3/(3u) = 1/15
−4/(3u) = 1/15
3u = −4 × 15 = −60
u = −20 cm, and v = −3 × (−20) = +60 cm
Step 4 — Check and answer. Substitute back: 1/60 − 1/(−20) = 1/60 + 3/60 = 4/60 = 1/15 ✓. So the object goes 20 cm in front of the lens and the screen 60 cm behind it. Note that 20 cm lies between F (15 cm) and 2F (30 cm) — the row of the table that promises a real, enlarged image beyond 2F, and 60 cm is indeed beyond 30 cm.
Step 1. h = +4 cm, u = −15 cm, f = +10 cm.
Step 2 and 3.
1/v = 1/f + 1/u = 1/10 − 1/15
Common denominator 30: 1/v = 3/30 − 2/30 = 1/30
v = +30 cm
m = v/u = 30 ÷ (−15) = −2
h′ = m × h = (−2) × 4 = −8 cm
Step 4. The image forms 30 cm beyond the lens, is real and inverted, and is 8 cm tall. The minus sign in the height is the formula telling you the flame now points downwards — it is a direction, not a negative length.
Carrying the mirror’s m = −v/u across to lenses. For lenses it is m = v/u, with no minus. Write both formulas side by side on the first page of your revision notebook — mirror: 1/v + 1/u = 1/f with m = −v/u; lens: 1/v − 1/u = 1/f with m = v/u — and glance at them before every numerical until the difference is automatic.
Power Of A Lens
When an optician hands your family a prescription that says “−1.5” or “+2.0”, that number is straight out of this section. It is the power of the lens.
Power measures how strongly a lens bends light. A fat, strongly curved lens with a short focal length bends light sharply, so it has a high power. A gently curved lens with a long focal length bends light only slightly, so its power is low. The relationship is therefore an inverse one.
P = 1 / f, where f must be in metres.
The SI unit of power is the dioptre, symbol D. One dioptre is the power of a lens of focal length 1 metre.
A convex lens has positive power; a concave lens has negative power, because their focal lengths carry those signs.
Focal length is almost always given in centimetres, and the formula demands metres. Convert first, every single time. A shortcut you can safely use: P (in D) = 100 / f (in cm). So f = 25 cm gives P = 100/25 = +4 D. If you ever get an answer like “P = 25 D” for an ordinary lens, you almost certainly forgot to convert.
(a) Convex, so f = +50 cm = +0.5 m
P = 1/f = 1 ÷ 0.5 = +2.0 D
Or with the shortcut: P = 100/50 = +2.0 D. Same answer.
(b) Concave, so f = −40 cm = −0.4 m
P = 1/f = 1 ÷ (−0.4) = −2.5 D
Or: P = 100/(−40) = −2.5 D.
Note the signs. The plus and minus are not optional decoration — they are how the answer tells the reader which type of lens it is. An answer of “2.5 D” for part (b) would be marked wrong.
(b) An optician prescribes a lens of power −5 D. Find its focal length and state its type.
(a) P = 1/f, so f = 1/P = 1 ÷ 4 = 0.25 m = +25 cm.
The power is positive, so it is a convex (converging) lens.
(b) f = 1/P = 1 ÷ (−5) = −0.2 m = −20 cm.
The power is negative, so it is a concave (diverging) lens.
What it means in real life. A prescription of −5 D means a fairly strong diverging lens, used to correct short-sightedness. Notice that the higher the number, the shorter the focal length and the thicker the spectacle lens — which is exactly why strong prescriptions look chunkier.
Everyday Uses Of Spherical Lenses
Another short, high-return section. As with mirrors, always name the property in your justification.
Convex (converging) lenses are used in:
- Magnifying glasses and reading lenses — object held inside the focus gives a virtual, erect, enlarged image.
- Spectacles for long-sightedness — the converging lens adds the extra bending a tired eye lens can no longer supply, bringing near objects back into focus on the retina.
- Cameras and mobile phone cameras — a real, inverted, diminished image is formed on the sensor.
- Projectors — the slide is placed just beyond the focus so a real, enlarged, inverted image lands on the screen, which is why the slide itself is loaded upside down.
- Microscopes and telescopes — combinations of converging lenses build up large magnification.
Concave (diverging) lenses are used in:
- Spectacles for short-sightedness — the diverging lens spreads light slightly before it reaches the eye, pushing the image back onto the retina instead of in front of it.
- Peepholes in doors — the diminished image squeezes a wide view of the corridor into a small opening.
- As correcting elements inside camera and telescope lens assemblies, where they cancel the colour and shape errors of the converging lenses.
Practical Skills: Focal Length By The Distant Object Method
Two practicals from this chapter carry marks in your internal assessment and can also appear as questions in the written paper. The first is finding the focal length of a concave mirror and of a convex lens using a distant object. The second is the glass slab ray-tracing you met earlier.
The idea behind the distant-object method is elegant. If the object is far away — a tree across the field, a distant building, the Sun — the rays reaching your mirror or lens are practically parallel. Parallel rays, by the very definition of the principal focus, converge at F. So the sharp image forms at the focus, and the distance from the mirror or lens to the screen is the focal length.
You can see this in the formula too. With u effectively infinite, 1/u is effectively zero. For a mirror, 1/v + 0 = 1/f, so v = f. For a lens, 1/v − 0 = 1/f, so again v = f. The mathematics and the physics agree.
How to do it, step by step:
- Stand with your back to a distant, brightly lit object and hold the concave mirror facing it.
- Hold a white screen (a card) in front of the mirror, on the same side as the object.
- Move the screen slowly to and fro until you get the sharpest, brightest, smallest inverted image on it.
- Measure the distance from the pole of the mirror to the screen with a metre scale. That distance is the focal length f.
- Repeat three times and take the mean. For a convex lens the procedure is identical, except the screen goes on the opposite side of the lens from the object.
| Trial | Distance mirror to screen (cm) | Focal length f (cm) |
|---|---|---|
| 1 | 11.4 | 11.4 |
| 2 | 11.6 | 11.6 |
| 3 | 11.5 | 11.5 |
| Mean | — | 11.5 |
(a) For a distant object, v = f, so the focal length is 11.5 cm. Working with signs, f = −11.5 cm for this concave mirror.
(b) R = 2f = 2 × 11.5 = 23 cm (or −23 cm with the sign convention).
(c) A real image of the same size forms only when the object sits at the centre of curvature. So place the object 23 cm in front of the mirror. Quick verification with the mirror formula, using u = −23 cm and f = −11.5 cm:
1/v = 1/(−11.5) − 1/(−23) = −2/23 + 1/23 = −1/23, so v = −23 cm.
m = −v/u = −(−23)/(−23) = −1, meaning same size and inverted. Confirmed.
Three precautions earn marks in the viva: choose an object at least 50 metres away so the rays are genuinely near-parallel; never point the mirror or lens directly at the Sun and never look through it at the Sun, since the concentrated beam can damage your eye badly; and keep the mirror, screen and scale in a straight line so your measured distance is a true perpendicular distance. Also remember: this method cannot be used for a convex mirror or a concave lens, because neither forms a real image that a screen can catch.
Practice Worksheet: Ten Questions With Answers
Here is where the learning actually happens. Close this page’s earlier sections, take a fresh sheet of paper, and attempt each question properly before you open the answer. Write the sign list first, then the formula, then the substitution — the same four-step routine we used together. If you get one wrong, do not just read the solution and nod. Find the exact line where you diverged, mark it, and redo the whole question from scratch tomorrow.
Q1. An object is placed 60 cm in front of a concave mirror whose radius of curvature is 40 cm. Find the image distance and magnification, and describe the image fully.
1/v = 1/f − 1/u = −1/20 + 1/60
Common denominator 60: 1/v = −3/60 + 1/60 = −2/60 = −1/30
v = −30 cm
m = −v/u = −(−30)/(−60) = 30/(−60) = −0.5
Description: the image forms 30 cm in front of the mirror, and is real, inverted and half the size of the object. Check against the table: the object at 60 cm is beyond C (40 cm), so the image should lie between F (20 cm) and C (40 cm) — and 30 cm does.
Q2. A convex mirror of focal length 18 cm has an object placed 36 cm away. Find the image distance and magnification. Then state the two reasons such a mirror is chosen for vehicle rear-view mirrors.
1/v = 1/f − 1/u = 1/18 + 1/36
Common denominator 36: 1/v = 2/36 + 1/36 = 3/36 = 1/12
v = +12 cm
m = −v/u = −(12)/(−36) = +1/3, that is about +0.33
The image is 12 cm behind the mirror, virtual, erect and one-third the size. It lies between P and F (0 to 18 cm), as it must for a convex mirror.
Two reasons for rear-view use: (i) the image is always erect, whatever the distance of the following vehicle; (ii) the diminished image means a much wider field of view, so the driver sees far more of the road behind than a plane mirror of the same size would show.
Q3. A 3 cm tall object is held 10 cm in front of a concave mirror of focal length 15 cm. Find the position, nature and height of the image.
1/v = 1/f − 1/u = −1/15 + 1/10
Common denominator 30: 1/v = −2/30 + 3/30 = 1/30
v = +30 cm
m = −v/u = −(30)/(−10) = +3
h′ = m × h = 3 × 3 = +9 cm
Description: the image is 30 cm behind the mirror, virtual, erect and enlarged, standing 9 cm tall. This is the shaving-mirror situation.
Q4. A concave mirror has a focal length of 9 cm. Where must an object be placed so that its image is real and exactly the same size as the object? Verify your answer using the mirror formula.
R = 2f = 2 × 9 = 18 cm, so the object must be placed 18 cm in front of the mirror.
Verification: u = −18 cm, f = −9 cm.
1/v = 1/f − 1/u = −1/9 + 1/18 = −2/18 + 1/18 = −1/18
v = −18 cm
m = −v/u = −(−18)/(−18) = 18/(−18) = −1
|m| = 1 confirms the same size, and the minus sign confirms real and inverted. The image also forms 18 cm in front, i.e. at C, on top of the object.
Q5. Light travels through a certain liquid at 2.25 × 108 m/s. Taking c = 3 × 108 m/s, find the refractive index of the liquid. Is this liquid optically denser or rarer than crown glass (n = 1.52), and how would a ray bend passing from the liquid into that glass?
Comparison: 1.33 is less than 1.52, so the liquid is optically rarer than crown glass.
Bending: a ray travelling from the liquid into the glass moves from a rarer to a denser medium, so it slows down and bends towards the normal; the angle of refraction is smaller than the angle of incidence.
Extra check: the relative refractive index of glass with respect to this liquid is 1.52 ÷ 1.33 = 1.14, comfortably greater than 1, confirming the same conclusion.
Q6. A ray of light in air strikes a rectangular crown glass slab (n = 1.52) at an angle of incidence of 40°. Find the angle of refraction. What is the angle of emergence at the opposite face, and how is the emergent ray related to the incident ray?
sin 40° = 0.6428, so sin r = 0.6428 ÷ 1.52 = 0.4229
r = sin−1(0.4229) = 25° (to the nearest degree)
The angle shrank from 40° to 25°, correct for air into denser glass — bending towards the normal.
Angle of emergence: because the two faces of the slab are parallel, the second refraction exactly undoes the first, so the angle of emergence e = i = 40°.
Relation: the emergent ray is parallel to the incident ray but shifted sideways by the lateral displacement. The slab changes the ray’s position, never its final direction.
Q7. An object is placed 60 cm from a convex lens of focal length 20 cm. Find the image distance and magnification, and describe the image.
1/v − 1/u = 1/f, so 1/v = 1/f + 1/u = 1/20 − 1/60
Common denominator 60: 1/v = 3/60 − 1/60 = 2/60 = 1/30
v = +30 cm
m = v/u = 30 ÷ (−60) = −0.5
Description: the image forms 30 cm beyond the lens, and is real, inverted and half the size. Check against the table: the object at 60 cm is beyond 2F (40 cm), so the image should lie between F (20 cm) and 2F (40 cm) — and 30 cm does. Note this is a camera-type arrangement.
Q8. An object is placed 60 cm from a concave lens of focal length 30 cm. Find the image distance and magnification, and state the nature of the image.
1/v = 1/f + 1/u = −1/30 − 1/60
Common denominator 60: 1/v = −2/60 − 1/60 = −3/60 = −1/20
v = −20 cm
m = v/u = (−20) ÷ (−60) = +1/3, about +0.33
Nature: the image is 20 cm from the lens on the same side as the object, virtual, erect and one-third the size. It lies between F (30 cm) and O, exactly as a concave lens always demands.
Q9. (a) Find the power of a convex lens of focal length 40 cm. (b) A lens has a power of −2.5 D — find its focal length in centimetres and name its type. (c) Which of these two lenses bends light more strongly, and why?
(a) f = +40 cm = +0.4 m, so P = 1/f = 1 ÷ 0.4 = +2.5 D. (Shortcut: 100/40 = +2.5 D.)
(b) P = −2.5 D, so f = 1/P = 1 ÷ (−2.5) = −0.4 m = −40 cm. The negative sign means it is a concave (diverging) lens.
(c) Neither bends light more strongly than the other. Both have the same magnitude of power, 2.5 D, and therefore the same focal length magnitude of 40 cm. They bend light equally hard but in opposite senses — the convex lens converges the rays, the concave lens diverges them. The sign of the power describes the direction of the bending, not its strength.
Q10. A 5 cm tall object is placed 12 cm from a convex lens of focal length 18 cm. Find the image distance, the magnification and the height of the image. Which everyday instrument works exactly like this?
1/v = 1/f + 1/u = 1/18 − 1/12
Common denominator 36: 1/v = 2/36 − 3/36 = −1/36
v = −36 cm
m = v/u = (−36) ÷ (−12) = +3
h′ = m × h = 3 × 5 = +15 cm
Description and instrument: the image is 36 cm from the lens on the same side as the object, virtual, erect and three times as large, standing 15 cm tall. This is exactly how a magnifying glass works — the object must be held nearer than the focal length for the enlarged upright view.
Mark yourself honestly. Seven or more correct with full working means you are in good shape — polish the ray diagrams and you are done. Between four and six means the physics is there but the signs are slipping, so redo the sign convention section and Examples 3 to 8 tomorrow. Below four is absolutely fine at a first attempt; it simply means you should go back to the top of this page and walk through it once more slowly, with a pencil, before trying again.
When ten questions stop frightening you, take the whole paper. Our CBSE Class 10 Science Practice Paper 2026-27 (Set 1) puts Light beside the rest of the syllabus under exam conditions, with a full answer key. After that, keep the same twenty-minute rhythm going into Life Processes, the chapter most students meet next.
One last thought before you close this page. Light is not a chapter you conquer in a single long night. It is a chapter that yields quietly, a little at a time, to a student who sits with it for twenty honest minutes a day. So do not aim to finish everything today. Aim only for this: one more question correct than you managed yesterday. One extra ray diagram drawn without peeping. One sign you no longer have to stop and think about. Do that steadily for a fortnight and you will look back genuinely surprised at how far the small daily steps carried you. You have got this — go and pick up your pencil.