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Ray Optics and Optical Instruments — Class 12 Physics Notes & Practice

Ray Optics and Optical Instruments — Class 12 Physics Notes & Practice

Ray optics is the chapter where physics finally feels like something you can see. A mirror in the bathroom, the bend of a straw in a glass of nimbu paani, the fat lens in your grandmother’s reading glasses, the thin fibre carrying your Wi-Fi signal — all of it is in this chapter. And yet it is also the chapter where the largest number of students lose marks for the silliest reason: a sign. Not a concept, not a formula — a plus that should have been a minus.

So we are going to do something a little different here. Instead of racing through ray diagrams, we are going to build one habit and repeat it until it is boring: a four-step sign ritual that you perform identically on every single problem, whether it is a concave mirror, a prism or an astronomical telescope. Every worked example below uses the same four steps in the same order. By example twelve you will not be thinking about signs any more — your hand will do it for you.

This page is your full set of ray optics and optical instruments class 12 physics notes with solved examples — eighteen fully worked problems, a one-page formula list, a practice worksheet with answers, and a straight note on which topics have been trimmed from the 2026-27 syllabus so you do not waste a single evening on deleted material. Take it slowly. Nobody understands ray optics on the first read, and that is completely normal.

Meet Your Tutor

Ray Optics rewards a calm diagram and a disciplined sign convention more than fast formula recall. I will guide you through each mirror, lens, prism and instrument problem in the same repeatable order: draw, sign, substitute and interpret. Keep a pencil nearby and redraw the small ray sketches as you go.

What You’ll Learn

Jump straight to what you need

Your Game Plan

  1. Day 1 — Learn the ritual, not the formulas. Read the first two sections and copy the sign-convention card into your notebook by hand. Do not move on until you can state the four steps from memory.
  2. Day 2 — Mirrors. Work Examples 1 to 4 with the book closed, then check. Redo any you got wrong the same evening, not tomorrow.
  3. Day 3 — Refraction, total internal reflection, spherical surfaces. Examples 5 to 9.
  4. Day 4 — Lenses and prisms. Examples 10 to 15. This is the highest-yield day in the chapter.
  5. Day 5 — Optical instruments. Examples 16 to 18, then the formula list.
  6. Day 6 — The worksheet, timed. Forty minutes, no notes. Then mark honestly and list only the mistakes you made twice.

Study Notes

The Four-Step Sign Discipline (Your Solving Ritual)

Here is the honest truth about this chapter: the physics is not hard. The bookkeeping is. A mirror problem and a lens problem look almost identical on paper, and the only thing standing between you and full marks is whether you kept track of which side of the mirror or lens each distance was measured on. So we build a ritual. Four steps. Same four, every time.

The Ritual — do these four things in this order, always
1. DRAW the axis. A horizontal line. Put a dot on it for the pole P (mirror) or optical centre O (lens). Draw a small arrow at the top-left showing light travelling left to right. That arrow is your compass for the whole problem.
2. LABEL with signs. Write every given quantity next to its position on the axis, with its sign already attached. Object 30 cm to the left becomes u = −30 cm, not “30 cm”. Do this before you touch a formula.
3. SUBSTITUTE, signs intact. Put the signed numbers into the formula exactly as written. Do not “fix” a minus because the answer looks odd. The formula is smarter than your intuition.
4. SANITY-CHECK the sign. Read the sign of your answer back into English: “v came out negative, so for a mirror the image is in front — it is real.” If that sentence contradicts your rough diagram, something in step 2 was wrong.

Step 4 is the one everybody skips and the one that saves the most marks. It costs you eight seconds and it catches almost every arithmetic slip, because a wrong sign in step 3 nearly always produces a physically silly answer — a real image behind a convex mirror, say, which simply cannot happen.

Memory device for the two formula traps
“Mirrors ADD, lenses SUBTRACT.”
Mirror: 1/v + 1/u = 1/f   |   Lens: 1/v − 1/u = 1/f

“Mirrors carry the minus, lenses don’t.”
Mirror: m = −v/u   |   Lens: m = v/u

Notice the pattern: the mirror formula and the mirror magnification each carry exactly one extra minus sign compared with the lens versions. That is the whole difference. Say it out loud twice and it sticks.

Why it works: light reflects back off a mirror, so a real image forms on the same side as the object (negative side). Light passes through a lens, so a real image forms on the far side (positive side). The extra minus in the mirror formula is simply that reversal, written down once and for all.

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The Cartesian Sign Convention, Set Up Once and For All

CARTESIAN SIGN CONVENTION — SET-UP CARD incident light is ALWAYS drawn travelling left → right − NEGATIVE SIDE everything measured to the LEFT of x = 0 + POSITIVE SIDE everything measured to the RIGHT of x = 0 x = 0 is at P (pole of mirror) or O (optical centre of lens) principal axis + height (up) − height (down) all distances are measured FROM x = 0, along the principal axis REAL OBJECT u is ALWAYS negative because the object sits to the left of P or O SIGN OF f concave mirror −   convex mirror + convex lens +   concave lens − and f = R/2 for a mirror READING YOUR ANSWER mirror: v negative → real lens: v positive → real m negative → inverted
Copy this card into the front of your physics notebook. Every problem in this chapter starts here.

The convention itself is short enough to say in one breath: put the origin at the pole of the mirror or the optical centre of the lens, draw the incident light travelling from left to right, measure all distances from that origin, call rightward distances positive and leftward distances negative, and call heights above the principal axis positive. That is genuinely all of it.

Two consequences follow immediately and they are worth memorising as facts, because they turn up in almost every question. First, for a real object — an actual thing you can point at, sitting in front of the mirror or lens — the object distance u is always negative. Second, the focal length carries the sign of the surface: a concave mirror has a negative focal length, a convex mirror positive; a convex (converging) lens has a positive focal length, a concave (diverging) lens negative.

Common Mistake — writing u = 30 cm
The question says “an object is placed 30 cm from a concave mirror”, so students write u = 30 cm. It is not. The number 30 is a length; u is a coordinate. The object is to the left, so u = −30 cm. Attach the sign the moment you write the symbol down, in step 2 of the ritual, and you will never make this mistake in an exam.

Why it works: the convention is not a law of nature, it is an agreement — the same agreement you already use in coordinate geometry, where points left of the origin have negative x. Because everyone agrees, one single formula can describe concave and convex mirrors at once, and one single formula can describe converging and diverging lenses at once. Without the convention you would need four formulas and four sets of rules. The signs are doing real work for you.

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Reflection at Spherical Mirrors: Mirror Formula and Magnification

A spherical mirror is just a slice cut out of a shiny sphere. If the reflecting side is the hollow inside, you have a concave mirror; if it is the bulging outside, a convex mirror. Think of a steel katori: the inside curve is concave, the outside curve is convex. Everything else is vocabulary attached to that one picture.

Diagram: Ray construction for a concave mirror — object beyond C, at C, between C and F, and inside F, with pole, centre of curvature and principal focus labelled — to be added by illustrator

Four words to fix in your head, all measured along the principal axis. The pole P is the centre of the mirror’s surface — this is your origin. The centre of curvature C is the centre of the sphere the mirror was cut from. The distance PC is the radius of curvature R. The principal focus F is the point where rays travelling parallel to the axis converge after reflection (concave), or appear to diverge from (convex), and it sits exactly halfway: f = R/2.

If you studied Light: Reflection and Refraction in Class 10 Science, all of this will feel familiar — the difference in Class 12 is that we now insist on signs and treat the formula, not the ray diagram, as the primary tool.

Key Rule — the mirror equations
Mirror formula:   1/v + 1/u = 1/f   and   f = R/2
Magnification:   m = −v/u = h′/h
Here u is the object distance, v the image distance, h the object height and h′ the image height — all signed, all measured from P.

These are derived using the paraxial approximation: we assume every ray stays close to the principal axis and makes a small angle with it, so that the small arc of the mirror behaves like a section of a circle whose sagitta we can ignore. That assumption is why the formula is exact only near the axis, and why real mirrors show spherical aberration at the edges.

Object position (concave mirror)Image positionNature and size
At infinityAt FReal, inverted, a point
Beyond CBetween F and CReal, inverted, diminished
At CAt CReal, inverted, same size
Between C and FBeyond CReal, inverted, magnified
At FAt infinityReal, inverted, hugely magnified
Between F and PBehind the mirrorVirtual, erect, magnified
Anywhere (convex mirror)Between P and F, behindVirtual, erect, diminished
Example 1 — Concave mirror, object beyond the centre of curvature
An object stands 40 cm in front of a concave mirror of focal length 15 cm. Find the image distance and the magnification.

1. Draw. Axis, pole P at the origin, light travelling left to right, object on the left.
2. Label. u = −40 cm (object is to the left). f = −15 cm (concave).
3. Substitute. 1/v = 1/f − 1/u = (−1/15) − (−1/40) = −8/120 + 3/120 = −5/120 = −1/24.
So v = −24 cm. Then m = −v/u = −(−24)/(−40) = −0.6.
4. Sanity-check. v is negative, so the image is 24 cm in front of the mirror — real. It lies between F (15 cm) and C (30 cm), exactly as the table predicts. m is negative and smaller than 1 in size, so the image is inverted and diminished to 0.6 of the object’s height. Everything agrees.
Example 2 — The same mirror, object moved inside the focus
The same concave mirror (f = 15 cm), but now the object is only 10 cm away. What changes?

2. Label. u = −10 cm, f = −15 cm.
3. Substitute. 1/v = −1/15 + 1/10 = −2/30 + 3/30 = 1/30, so v = +30 cm.
m = −v/u = −(30)/(−10) = +3.
4. Sanity-check. v came out positive — the image is 30 cm behind the mirror, so it is virtual. m is positive and equal to 3, so the image is erect and three times as tall. This is exactly what happens when you bring your face close to a shaving mirror: a big, upright reflection. The formula told us that without a single ray drawn.
Example 3 — Convex mirror (the one on every scooter)
A convex mirror has radius of curvature 40 cm. A car is 30 cm from it. Locate the image and find m.

2. Label. Convex, so C is behind the mirror: R = +40 cm and f = R/2 = +20 cm. u = −30 cm.
3. Substitute. 1/v = 1/20 + 1/30 = 3/60 + 2/60 = 5/60 = 1/12, so v = +12 cm.
m = −12/(−30) = +0.4.
4. Sanity-check. Positive v → virtual, behind the mirror; positive m < 1 → erect and diminished. And notice v = 12 cm lies between P and F (20 cm), just as the last row of the table promises. That shrinking is precisely why convex mirrors are used as rear-view mirrors — a wide field of view squeezed into a small glass.
Example 4 — Working backwards from the magnification
A concave mirror of focal length 12 cm forms a real image three times the size of the object. Where is the object, and how tall is the image if the object is 2 cm tall?

2. Label. f = −12 cm. A real image from a concave mirror is inverted, so m = −3 (not +3 — this is the step people rush).
3. Substitute. From m = −v/u = −3 we get v = 3u. Put that into the mirror formula:
1/(3u) + 1/u = 1/f ⇒ 4/(3u) = 1/f ⇒ u = 4f/3 = 4(−12)/3 = −16 cm, and v = 3(−16) = −48 cm.
Image height h′ = mh = (−3)(2) = −6 cm, i.e. 6 cm tall and inverted.
4. Sanity-check. Substitute back: 1/(−48) + 1/(−16) = −1/48 − 3/48 = −4/48 = −1/12 = 1/f. ✓ Both u and v are negative, so object and image are both in front — correct for a real image. The object at 16 cm sits between F (12) and C (24), and the image at 48 cm is beyond C: row four of the table again.
Common Mistake — forgetting that a real image means m is negative
In Example 4, taking m = +3 gives u = −8 cm, an object inside the focus — which produces a virtual image, contradicting the question. Whenever a problem says “real image, three times magnified”, write m = −3. Whenever it says “virtual, three times magnified”, write m = +3. Decide the sign of m in step 2, before any algebra.

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Reading the v–u Graph: What the Formula Actually Looks Like

Before we leave formulas behind, it is worth seeing one. If you rearrange the lens formula 1/v − 1/u = 1/f you get v = uf/(u + f). That is a hyperbola, and plotting it explains, in one picture, several things students otherwise memorise as unrelated facts. The graph below is for a converging lens of focal length +10 cm, and every plotted point was computed from that formula.

Convex lens, f = +10 cm  —  v = uf / (u + f) COMPUTED POINTS (marked ●) u = −40 cm → v = +13.33 cm u = −30 cm → v = +15.00 cm u = −20 cm → v = +20.00 cm u = −15 cm → v = +30.00 cm u = −  5 cm → v = −10.00 cm u = −f = −10 cm object sits at F → image runs off to infinity (the asymptote) HOW TO READ IT blue branch (v > 0): real image, far side of lens red branch (v < 0): virtual image, same side the branches never meet — that gap is u = −f −40−30−20−100 −50−250+20+40+60 object distance u (cm) — always negative for a real object image distance v (cm)
Every marked coordinate was computed from v = uf/(u + f) with f = +10 cm, not sketched by eye.

Three things jump out of that picture. One: as the object goes far away (u → −∞) the blue curve flattens towards v = +10 cm — the image collapses onto the focal point, which is exactly why distant objects image at F. Two: at u = −20 cm (that is, u = −2f) we get v = +20 cm, the famous same-size case. Three: the curve tears apart at u = −10 cm. Approach the focus from the left and v rockets to +∞; step just inside the focus and v reappears at −∞ on the virtual branch. That is not a glitch — it is the mathematical fingerprint of an object at the focus producing a beam of parallel rays and no image at all.

Exam Tip — the graph is a marks question in itself
Examiners like to ask for a sketch of 1/v against 1/u rather than v against u, because that one is a straight line: 1/v = 1/u + 1/f has slope 1 and intercept 1/f. If a graph question feels messy, try plotting reciprocals — the hyperbola straightens out and the focal length is just the intercept.

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Refraction of Light, Snell’s Law and Apparent Depth

Drop a spoon into a glass of water and it looks broken at the surface. Nothing happened to the spoon; something happened to the light. Light travels at different speeds in different materials, and when it crosses a boundary at an angle it changes direction. That bending is refraction.

The refractive index of a medium is n = c/v, the speed of light in vacuum divided by its speed in that medium. Since nothing outruns light in vacuum, n is always at least 1: about 1.33 for water, 1.5 for ordinary glass, 2.42 for diamond. If you have already met light as an electromagnetic disturbance in Electromagnetic Waves, this is the same story from the geometry side.

Key Rule — Snell’s law
n1 sin i = n2 sin r,   equivalently   sin i/sin r = n2/n1 = 1n2
Angles are measured from the normal, never from the surface. The incident ray, refracted ray and normal all lie in one plane.
Going into a denser medium (n2 > n1): the ray bends towards the normal. Coming out into a rarer medium: it bends away.
Example 5 — Snell’s law, air into glass
Light in air strikes a flat glass slab (n = 1.5) at 30° to the normal. Find the angle of refraction.

Substitute. 1 × sin 30° = 1.5 × sin r ⇒ sin r = 0.5/1.5 = 1/3 = 0.3333.
r = sin−1(0.3333) = 19.47° ≈ 19.5°.
Sanity-check. r < i, so the ray bent towards the normal — correct, because glass is optically denser than air. Had the answer come out larger than 30°, we would know a reciprocal had been flipped.

A neat consequence of refraction at a flat surface is apparent depth. Look straight down into water and the bottom seems closer than it is, because rays from the bottom bend away from the normal as they leave the water and your eye traces them back to a shallower point. For near-normal viewing, apparent depth = real depth / n, and the apparent upward shift is t(1 − 1/n) for a slab of thickness t.

Example 6 — The coin at the bottom of the bucket
A coin lies at the bottom of a bucket holding water (n = 4/3) to a depth of 12 cm. Viewed from directly above, how deep does the coin appear, and by how much has it apparently risen?

Substitute. Apparent depth = 12 ÷ (4/3) = 12 × 3/4 = 9 cm.
Apparent shift = 12 − 9 = 3 cm, which matches t(1 − 1/n) = 12(1 − 0.75) = 3 cm. ✓
Sanity-check. The apparent depth must be less than the real depth (things look shallower, never deeper), and 9 < 12. This is exactly why a swimming pool always turns out deeper than it looked — and why you should never judge water depth by eye.

Why it works: refraction is not light “choosing” to bend. A wavefront entering a slower medium at an angle has one edge slowed before the other, so the whole front pivots — like a marching column swinging round when the soldiers on one side take shorter steps. The amount of pivot is fixed entirely by the speed ratio, which is what n measures.

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Total Internal Reflection and Optical Fibres

Diagram: Ray paths at a denser-to-rarer boundary showing partial refraction below the critical angle, grazing emergence at the critical angle, and total internal reflection beyond it — plus a zig-zag ray inside an optical fibre core — to be added by illustrator

Send light the other way — from glass out into air — and the ray bends away from the normal. Increase the angle of incidence inside the glass and the refracted ray leans further and further over, until at one particular angle it grazes along the surface at 90°. Push past that angle and there is no refracted ray at all: every bit of light is reflected back into the glass. That angle is the critical angle C, and the effect is total internal reflection.

Key Rule — the two conditions and the formula
Total internal reflection happens only when both are true:
(i) light travels from the denser medium to the rarer medium, and
(ii) the angle of incidence exceeds the critical angle.

Putting r = 90° in Snell’s law: sin C = n2/n1, and for a medium of index n against air, sin C = 1/n.
Example 7 — Critical angles for glass, water and diamond
Find the critical angle at the surface with air for glass (n = 1.5), water (n = 4/3) and diamond (n = 2.42).

Glass: sin C = 1/1.5 = 0.6667 ⇒ C = 41.81°
Water: sin C = 3/4 = 0.7500 ⇒ C = 48.59°
Diamond: sin C = 1/2.42 = 0.4132 ⇒ C = 24.41°

Sanity-check. The denser the medium, the smaller the critical angle — and 24.41° < 41.81° < 48.59° follows exactly that order. Diamond’s tiny critical angle means light that enters a cut stone bounces around inside many times before escaping. That is the whole secret of a diamond’s sparkle: geometry, not magic.
Example 8 — Inside an optical fibre
An optical fibre has a core of refractive index 1.50 and cladding of refractive index 1.45. (a) Find the critical angle at the core–cladding boundary. (b) Find the largest angle at which light can enter the flat end face from air and still be trapped (the acceptance angle).

(a) sin C = ncladding/ncore = 1.45/1.50 = 0.96667 ⇒ C = 75.16°.
(b) A ray entering the end face at angle i refracts to r inside, and strikes the side wall at (90° − r). Requiring 90° − r ≥ C gives the numerical aperture
NA = sin imax = √(ncore2 − ncladding2) = √(2.2500 − 2.1025) = √0.1475 = 0.3841
so imax = sin−1(0.3841) = 22.59°.
Sanity-check. The critical angle is large (75°) because core and cladding are so close in index — which is deliberate: it keeps the trapped rays nearly parallel to the fibre axis so that pulses do not smear out over kilometres. Note also that C is measured from the normal to the side wall, which is why the useful quantity is 90° − r, not r. Mixing those two up is the single most common error in fibre problems.
Exam Tip — where total internal reflection shows up
Three applications are asked again and again: optical fibres (endoscopes, broadband cables), totally reflecting prisms in binoculars and periscopes (a 45°–90°–45° glass prism turns light through 90° or 180° with no silvering, because 45° > 41.81°), and mirage on a hot road, where hot air near the surface acts as the rarer medium. Learn one clean sentence of explanation for each.

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Refraction at a Spherical Surface

Diagram: Refraction of a paraxial ray at a single convex spherical surface separating two media, with pole, centre of curvature, object point and image point labelled — to be added by illustrator

This is the bridge between refraction and lenses, and it is the section students most often skip — a mistake, because the lens maker’s formula is built out of it. The setup is one curved boundary between two media: light starts in medium 1 (index n1), meets a spherical surface of radius R, and continues in medium 2 (index n2).

Key Rule — refraction at a single spherical surface
n2/v − n1/u = (n2 − n1)/R
Same sign convention as always. R is positive if the centre of curvature lies to the right of the surface (a surface bulging towards the incoming light has R positive), negative if to the left.
Example 9 — Image formed inside a glass paperweight
A long glass rod (n = 1.5) has one end ground into a convex surface of radius 20 cm. A small object sits on the axis in air, 100 cm from that end. Where is the image?

2. Label. n1 = 1 (air, where the light starts), n2 = 1.5 (glass), u = −100 cm, R = +20 cm (the centre of curvature is inside the glass, to the right).
3. Substitute. 1.5/v − 1/(−100) = (1.5 − 1)/20 = 0.5/20 = 0.025
1.5/v = 0.025 − 0.010 = 0.015 ⇒ v = 1.5/0.015 = +100 cm.
4. Sanity-check. v is positive, so the image lies 100 cm to the right of the surface — i.e. inside the glass, which is where light actually goes after refracting. Real image, correctly on the transmission side. If your answer had come out negative you would know you had put the wrong medium as n1: n1 is always the medium the light is coming from.

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Thin Lenses: Lens Formula, Lens Maker’s Formula and Power

Diagram: Ray constructions for a convex lens (object beyond 2F, between F and 2F, and inside F) and for a concave lens, with optical centre, both focal points and 2F marked — to be added by illustrator

A lens is simply two refracting surfaces back to back. Apply the spherical-surface formula at the first surface, feed its image in as the object for the second surface, and let the glass thickness go to zero — out drops everything you know about lenses. That is the entire derivation in one sentence, and it is worth carrying in your head because it tells you why the lens maker’s formula has two radii in it.

Key Rule — the three lens equations
Thin lens formula:   1/v − 1/u = 1/f
Magnification:   m = v/u = h′/h  (no minus sign — that is the mirror’s)
Lens maker’s formula:   1/f = (n2/n1 − 1)(1/R1 − 1/R2), which for a lens in air becomes 1/f = (n − 1)(1/R1 − 1/R2)
Power:   P = 1/f in dioptres when f is in metres, or P = 100/f when f is in centimetres.
Example 10 — Convex lens, object between F and 2F
A converging lens has focal length 20 cm. An object 5 cm tall is placed 30 cm in front of it. Find the image distance, magnification and image height.

2. Label. u = −30 cm, f = +20 cm (convex).
3. Substitute. 1/v = 1/f + 1/u = 1/20 − 1/30 = 3/60 − 2/60 = 1/60 ⇒ v = +60 cm.
m = v/u = 60/(−30) = −2.   h′ = mh = (−2)(5) = −10 cm.
4. Sanity-check. Positive v → the image is 60 cm beyond the lens, on the far side — real. Negative m → inverted; |m| = 2 → twice as tall, so 10 cm and upside down. The object at 30 cm lies between f = 20 and 2f = 40, and the rule says the image should then be beyond 2f — and 60 > 40. ✓
Example 11 — Concave lens, same object distance
Now swap in a diverging lens of focal length 20 cm, object still 30 cm away.

2. Label. u = −30 cm, f = −20 cm (concave — this single minus sign is the whole question).
3. Substitute. 1/v = −1/20 − 1/30 = −3/60 − 2/60 = −5/60 = −1/12 ⇒ v = −12 cm.
m = (−12)/(−30) = +0.4.
4. Sanity-check. Negative v → the image is on the same side as the object, 12 cm from the lens — virtual. Positive m < 1 → erect and diminished. A diverging lens can only ever do this, whatever the object distance, which is a useful thing to state in one line if a question asks you to justify your answer.
Example 12 — Lens maker’s formula, in air and then under water
An equiconvex lens is made of glass with n = 1.5 and each face has radius of curvature 20 cm. (a) Find its focal length and power in air. (b) Find its focal length when the whole lens is immersed in water (nw = 4/3).

2. Label. Light hits the first surface, whose centre of curvature is to the right: R1 = +20 cm. The second surface curves the other way, its centre to the left: R2 = −20 cm.
3(a). Substitute. 1/f = (1.5 − 1)(1/20 − 1/(−20)) = 0.5 × (0.05 + 0.05) = 0.5 × 0.10 = 0.05
⇒ f = +20 cm, and P = 100/20 = +5 D.
3(b). In water the ratio becomes n2/n1 = 1.5 ÷ (4/3) = 1.125, so
1/f′ = (1.125 − 1)(0.10) = 0.125 × 0.10 = 0.0125 ⇒ f′ = +80 cm, P′ = +1.25 D.
4. Sanity-check. The focal length grew four-fold, so the lens became much weaker in water. That is exactly right: the lens bends light because of the contrast between its index and the surroundings, and water shrinks that contrast. It is also why your vision is blurry underwater but sharp behind goggles — the goggles restore an air gap in front of your eye.
Common Mistake — using the mirror formula’s plus sign for a lens
Writing 1/v + 1/u = 1/f for a lens will get you a plausible-looking wrong number every single time, and there is no sanity-check that catches it reliably. Before you substitute, say the memory device to yourself: mirrors ADD, lenses SUBTRACT. Two seconds, and it protects a five-mark question.

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Combination of Thin Lenses in Contact

Put two thin lenses flat against each other and they behave as a single lens. The image made by the first becomes the object for the second, and when you write that out the algebra collapses to something beautifully simple: focal lengths combine as reciprocals, and powers simply add.

Key Rule — lenses in contact
1/F = 1/f1 + 1/f2 + …   and equivalently   P = P1 + P2 + …
The total magnification is the product: m = m1 × m2 × …
Work in powers whenever you can — adding is far safer than adding fractions.
Example 13 — A converging and a diverging lens in contact
A convex lens of focal length 25 cm is placed in contact with a concave lens of focal length 50 cm. Find the power and focal length of the combination, and say whether it converges or diverges.

2. Label. f1 = +25 cm, f2 = −50 cm.
3. Substitute. P1 = 100/25 = +4 D,   P2 = 100/(−50) = −2 D.
P = 4 + (−2) = +2 D ⇒ F = 100/2 = +50 cm.
4. Sanity-check. P is positive, so the pair acts as a converging lens — and it should, because the convex lens is the stronger of the two (4 D beats 2 D). Check the fraction route too: 1/F = 1/25 − 1/50 = 2/50 − 1/50 = 1/50, giving F = 50 cm. ✓ Same answer by both roads.

Why it works: power measures how sharply a lens bends light per unit distance, and bending is additive as long as the light has no room to travel between the two lenses. The moment you separate them by a distance d, the simple sum fails and you must use the two-lens sequence step by step — which is exactly what a compound microscope and a telescope are.

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Refraction of Light Through a Prism

Diagram: A ray passing through a triangular prism, showing angles i₁, r₁, r₂, i₂, the refracting angle A and the angle of deviation D, plus the D-versus-i curve with its minimum — to be added by illustrator

A prism is a wedge of glass with two flat refracting faces meeting at the refracting angle A. A ray refracts on the way in, travels through the glass, and refracts again on the way out — and because both bends push it the same way, it emerges turned through an overall angle of deviation D.

Key Rule — the prism relations
A = r1 + r2  (the two internal angles always add to the prism angle)
D = i1 + i2 − A
At minimum deviation the path is symmetric: i1 = i2 = i and r1 = r2 = A/2, so i = (A + Dm)/2, giving the prism formula
n = sin[(A + Dm)/2] ÷ sin(A/2)
Example 14 — Finding n from the minimum deviation
A prism of refracting angle 60° gives a minimum deviation of 30°. Find the refractive index of its glass, and the angle of incidence at which this happens.

3. Substitute. n = sin[(60° + 30°)/2] / sin(60°/2) = sin 45° / sin 30° = 0.70711 / 0.5 = 1.4142, i.e. n = √2.
Angle of incidence: i = (A + Dm)/2 = 90°/2 = 45°, and inside the prism r1 = r2 = A/2 = 30°.
4. Sanity-check. Verify with Snell’s law at the first face: sin 45°/sin 30° = 0.70711/0.5 = 1.4142 = n. ✓ And r1 + r2 = 30 + 30 = 60 = A. ✓ Also D = i1 + i2 − A = 45 + 45 − 60 = 30°. ✓ Three independent checks all agree, which is how you should finish a prism question.
Example 15 — The same question run backwards
A 60° prism is made of glass with n = 1.5. What is its minimum deviation?

3. Substitute. sin[(60° + Dm)/2] = n sin(A/2) = 1.5 × sin 30° = 1.5 × 0.5 = 0.75
(60° + Dm)/2 = sin−1(0.75) = 48.59° ⇒ 60° + Dm = 97.18° ⇒ Dm = 37.18°.
4. Sanity-check. Glass with n = 1.5 is denser than the √2 ≈ 1.414 glass of Example 14, so it should deviate light more — and 37.18° > 30°. ✓ The incidence angle here would be (60 + 37.18)/2 = 48.59°.
Exam Tip — check your current syllabus for prism dispersion
Refraction of light through a prism — the relations above — is squarely in the 2026-27 course. Extended treatment of dispersion by a prism and of the angular spread of a spectrum sits at the boundary of what is currently examinable, and the scattering topics that used to sit alongside it have been removed. Do the prism relations properly; treat detailed dispersion questions as optional enrichment and confirm against your own school’s current syllabus copy before spending an evening on them.

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The Simple Microscope (Magnifying Glass)

Every optical instrument in this chapter is answering the same question: how do I make something take up a bigger angle at my eye? That is what magnification really means — not making the object bigger, but making its image spread over a wider angle on your retina. The reference distance is the least distance of distinct vision, D = 25 cm, the closest a normal eye can focus comfortably.

A simple microscope is one converging lens of short focal length held close to the eye, with the object just inside its focus. It produces an enlarged, erect, virtual image.

Key Rule — magnifying power of a simple microscope
Image at the near point (relaxed viewing is sacrificed for maximum magnification):  M = 1 + D/f
Image at infinity (normal adjustment, eye fully relaxed):  M = D/f
The near-point value is always exactly 1 larger. Every instrument in this chapter has these same two settings — learn the pair, not eight separate formulas.
Example 16 — A 5 cm magnifying glass, both settings
A magnifying glass has focal length 5 cm. Find its magnifying power (a) with the image at the near point and (b) with the image at infinity. In case (a), where exactly must the object be held? Take D = 25 cm.

(a) M = 1 + D/f = 1 + 25/5 = 6.
Object position: the virtual image is at the near point, so v = −25 cm and f = +5 cm.
1/u = 1/v − 1/f = −1/25 − 1/5 = −0.04 − 0.20 = −0.24 ⇒ u = −4.17 cm.
(b) M = D/f = 25/5 = 5, with the object exactly at the focus, u = −5 cm.
4. Sanity-check. In case (a) the object sits 4.17 cm away, just inside the 5 cm focus — which is the only place a single convex lens gives an erect virtual image. Moving the object from 4.17 cm out to 5 cm drops the magnification from 6 to 5 but lets your eye relax completely. That trade-off is the entire physics of “normal adjustment”.

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The Compound Microscope and Its Magnifying Power

Diagram: Compound microscope ray layout — objective forming a real inverted intermediate image inside the focus of the eyepiece, which forms the final magnified virtual image; tube length and both focal lengths labelled — to be added by illustrator

One lens can only do so much. A compound microscope magnifies twice: a short-focus objective makes a real, inverted, enlarged image inside the tube, and a longer-focus eyepiece then acts as a simple microscope on that image. Two magnifications multiply, which is how you reach hundreds of times rather than five or six.

Key Rule — compound microscope
M = mo × me  where  mo = vo/uo
Image at near point:  me = 1 + D/fe  →  M = (vo/uo)(1 + D/fe)
Image at infinity:  me = D/fe  →  M = (vo/uo)(D/fe), and tube length L = vo + fe
Both fo and fe are small, with fo < fe. The final image is inverted with respect to the object.
Example 17 — A full compound microscope calculation
A compound microscope has an objective of focal length 1.0 cm and an eyepiece of focal length 5.0 cm. A specimen is placed 1.2 cm from the objective. Find (a) the objective’s magnification, (b) the total magnifying power with the final image at the near point and the separation of the lenses, and (c) the same two quantities in normal adjustment. D = 25 cm.

(a) Objective. uo = −1.2 cm, fo = +1.0 cm.
1/vo = 1/1.0 − 1/1.2 = 1 − 0.83333 = 0.16667 ⇒ vo = +6.0 cm.
mo = vo/uo = 6.0/(−1.2) = −5.
(b) Near point. me = 1 + 25/5 = 6, so M = (−5)(6) = −30, i.e. 30× and inverted.
For the eyepiece with ve = −25 cm: 1/ue = −1/25 − 1/5 = −0.24 ⇒ ue = −4.17 cm.
Separation L = vo + |ue| = 6.0 + 4.17 = 10.17 cm.
(c) Normal adjustment. me = 25/5 = 5, so M = (−5)(5) = −25, and L = vo + fe = 6.0 + 5.0 = 11.0 cm.
4. Sanity-check. The specimen at 1.2 cm sits just outside the objective’s 1.0 cm focus — exactly where it must be to give a real, strongly enlarged intermediate image. Near-point viewing gives more magnification (30 vs 25) but a shorter tube (10.17 vs 11.0 cm), which is the same trade-off as the magnifying glass. And the near-point magnification is larger by the factor 6/5, precisely the ratio of the two eyepiece formulas. ✓
Common Mistake — quoting the tube length as vo + fe for near-point viewing
L = vo + fe is only true in normal adjustment, where the intermediate image sits exactly at the eyepiece’s focus. For near-point viewing the intermediate image must sit slightly inside the focus, so you must first find ue from the lens formula — here 4.17 cm, not 5 cm. Read the question for the words “relaxed eye”, “image at infinity” or “normal adjustment” before you pick a route.

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The Astronomical Telescope: Refracting and Reflecting

Diagram: Astronomical telescope in normal adjustment — parallel rays from a distant object, objective forming a real image at its focus which coincides with the eyepiece focus, final image at infinity; alongside a Cassegrain reflecting telescope with concave primary mirror and secondary mirror — to be added by illustrator

A telescope faces the opposite problem to a microscope. The object is not small — it is enormous — but it is so far away that it subtends a tiny angle. So the objective is given a long focal length and a large aperture: long focal length for angular magnification, large aperture to gather enough light.

Key Rule — astronomical telescope
Normal adjustment (final image at infinity):  M = fo/fe  and  L = fo + fe
Final image at the near point:  M = (fo/fe)(1 + fe/D)
Here fo is large and fe small — the exact opposite of the microscope, where both are small.
Example 18 — Telescope, both adjustments
An astronomical telescope has an objective of focal length 150 cm and an eyepiece of focal length 6 cm. Find the magnifying power and tube length (a) in normal adjustment and (b) with the final image at the near point (D = 25 cm).

(a) Magnitude |M| = fo/fe = 150/6 = 25; with sign, M = −25 because the final image is inverted.   L = 150 + 6 = 156 cm.
(b) M = (150/6)(1 + 6/25) = 25 × 1.24 = 31.
Tube length: for the eyepiece, ve = −25 cm, fe = +6 cm, so
1/ue = −1/25 − 1/6 = −(6 + 25)/150 = −31/150 ⇒ ue = −150/31 = −4.84 cm.
L = fo + |ue| = 150 + 4.84 = 154.84 cm.
4. Sanity-check. Near-point viewing again gives more magnification (31 vs 25) and a shorter tube (154.84 vs 156 cm) — the same pattern as the microscope, for the same reason. Note 31/25 = 1.24 = 1 + fe/D, so the two answers are consistent by construction. ✓

A reflecting telescope replaces the objective lens with a large concave mirror. Two advantages follow immediately, and both are standard exam answers. First, a mirror shows no chromatic aberration at all, because reflection does not depend on wavelength the way refraction does. Second, a big mirror can be supported across its whole back surface, whereas a big lens can only be held at its rim and sags under its own weight — which is why every giant telescope built in the last century is a reflector. In the Cassegrain design the light bounces off a concave primary, then off a small convex secondary, and out through a hole in the primary to the eyepiece.

FeatureCompound microscopeAstronomical telescope
Object distanceJust beyond fo, a few mmEffectively infinite
Objective focal lengthVery smallVery large
Objective apertureSmallAs large as possible
M (normal adjustment)(vo/uo)(D/fe)|M| = fo/fe; signed M = −fo/fe (inverted)
Tube length (normal)vo + fefo + fe
PurposeEnlarge a tiny nearby objectWiden the angle of a distant one

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Real or Virtual? The Decision Flow

Step 4 of the ritual asks you to read the sign of your answer back into English. Here is that step drawn out as a flow. Learn this one picture and you will never again write “real image behind the mirror” in an exam.

You have solved for v now do step 4 — read the sign MIRROR 1/v + 1/u = 1/f  ·  m = −v/u LENS 1/v − 1/u = 1/f  ·  m = v/u v is NEGATIVE REAL image in front of mirror inverted, m < 0 v is POSITIVE VIRTUAL image behind the mirror erect, m > 0 v is POSITIVE REAL image far side of lens inverted, m < 0 v is NEGATIVE VIRTUAL image same side as object erect, m > 0 Then check the size: |m| > 1 magnified, |m| < 1 diminished, |m| = 1 same size If any of this contradicts your rough sketch, go back and recheck the signs you wrote in step 2
Green = real image, red = virtual image. Notice the mirror and lens columns are exact mirror images of each other.

Look at the symmetry in that diagram. For a mirror, negative v means real; for a lens, positive v means real. They are opposite, and they are opposite for one physical reason: reflected light goes back the way it came, refracted light carries on forwards. If you can explain that sentence, you have understood the sign convention rather than memorised it.

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Ray Optics Class 12 Physics Formula List (One-Page Revision)

Print this, stick it inside your cupboard door, and read it once every morning for the week before the exam. This is the complete ray optics and optical instruments class 12 physics formula list for the current syllabus — nothing extra, nothing missing.

SituationFormulaWatch out for
Spherical mirror1/v + 1/u = 1/f,  f = R/2Concave f negative, convex positive
Mirror magnificationm = −v/u = h′/hThe minus sign is part of the formula
Refractive indexn = c/vmediumAlways ≥ 1
Snell’s lawn1 sin i = n2 sin rAngles from the normal, not the surface
Apparent depthapparent = real/n; shift = t(1 − 1/n)Near-normal viewing only
Critical anglesin C = n2/n1  (= 1/n against air)Denser → rarer only
Optical fibreNA = sin imax = √(ncore2 − nclad2)Wall angle is 90° − r
Spherical surfacen2/v − n1/u = (n2 − n1)/Rn1 is where light comes from
Thin lens1/v − 1/u = 1/f,  m = v/uMinus in the formula, none in m
Lens maker’s formula1/f = (n − 1)(1/R1 − 1/R2)Use nlens/nmedium if not in air
Power of a lensP = 1/f(m) = 100/f(cm)Dioptres need metres
Lenses in contact1/F = 1/f1 + 1/f2;  P = P1 + P2Only when they touch
PrismA = r1 + r2;  D = i1 + i2 − ABoth hold for every ray, not just at Dm
Prism formulan = sin[(A + Dm)/2] / sin(A/2)Minimum deviation only
Simple microscopeM = 1 + D/f  (near pt); M = D/f (∞)D = 25 cm
Compound microscopeM = (vo/uo)(1 + D/fe) or (vo/uo)(D/fe)L = vo + fe only at ∞
TelescopeM = fo/fe;  L = fo + feNear point: multiply by (1 + fe/D)

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Check the Current 2026-27 Scope

The current official CBSE 2026-27 Physics outline lists mirrors, refraction, total internal reflection and optical fibres, spherical surfaces, lenses, prisms, microscopes and astronomical telescopes for this chapter. The topics below do not appear in that Class 12 Ray Optics outline. Treat them as enrichment only unless your school gives separate instructions.

Not listed in the current official Class 12 outline
• Scattering of light — the blue of the sky, the reddish sun at sunrise and sunset
• The human eye, image formation on the retina, and accommodation
• Correction of eye defects — myopia and hypermetropia, and the lens powers needed
• Resolving power of a microscope and of a telescope

Several of these ideas are useful enrichment, and the same careful sign-convention habit also matters in the Class 10 Electricity notes and practice guide. For Class 12 board preparation, prioritise the topics that the current CBSE outline explicitly lists.

This page prioritises the topics explicitly named in the official outline: mirrors and the mirror formula; refraction, total internal reflection and optical fibres; spherical surfaces; thin lenses, lens maker’s formula, magnification, power and lenses in contact; refraction through a prism; microscopes; and reflecting and refracting astronomical telescopes with magnifying power. For anything beyond that list, use your school’s current instructions as the deciding reference.

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Ray Optics Class 12 Physics Important Questions — How They Are Set

Across recent papers, questions from this chapter fall into a small number of recognisable shapes. Knowing the shapes is worth more than knowing extra facts, because it tells you what the first line of your answer should be.

  • Plug-and-read (1–2 marks). A mirror or lens with given u and f; find v, m, or the nature of the image. Pure ritual. Never lose these.
  • Work backwards (2–3 marks). Magnification and focal length given, find the object distance — like Example 4. The trap is always the sign of m.
  • Two-element chains (3 marks). A lens followed by a mirror, or a lens in contact with another. Image of the first becomes object of the second; carry the sign through, do not restart.
  • Medium-change questions (2–3 marks). “What happens to the focal length if the lens is put in water?” Answer with the lens maker’s formula and the ratio nlens/nmedium, as in Example 12.
  • Derivations (3–5 marks). Mirror formula, refraction at a spherical surface, lens maker’s formula, prism formula, magnifying power of a compound microscope or telescope. Learn the sequence of steps, and always state the paraxial assumption.
  • Reasoning one-liners (1–2 marks). Why convex mirrors are used as rear-view mirrors; why a diamond sparkles; why big telescopes use mirrors; why a lens is weaker in water. Each needs one clean sentence, and you now have all four above.
  • Graph and assertion-reason. The 1/v versus 1/u straight line, or the v–u hyperbola from the section above.
Exam Tip — show the signed substitution line
In a numerical, write the line “u = −30 cm, f = +20 cm” on its own before the formula. Examiners award method marks for it, and if your arithmetic slips later you still keep those marks. It takes one line and it is the cheapest insurance in the paper.

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Practice Worksheet

Twelve questions, all original, all solvable with the ritual. Give yourself forty minutes with a calculator and no notes. Write the signed substitution line for every numerical — that habit is what you are really practising here.

Q1. An object is placed 15 cm in front of a concave mirror of focal length 10 cm. Find the image distance and magnification, and describe the image.

Show Answer
u = −15 cm, f = −10 cm.  1/v = −1/10 + 1/15 = −3/30 + 2/30 = −1/30 ⇒ v = −30 cm. m = −(−30)/(−15) = −2. The image is 30 cm in front of the mirror: real, inverted, twice as large. (Object between F and C → image beyond C. ✓)

Q2. A convex mirror has a radius of curvature of 30 cm. An object stands 10 cm from it. Locate the image and give its magnification.

Show Answer
R = +30 cm ⇒ f = +15 cm; u = −10 cm.  1/v = 1/15 + 1/10 = 2/30 + 3/30 = 5/30 = 1/6 ⇒ v = +6 cm. m = −6/(−10) = +0.6. Virtual, erect, diminished, 6 cm behind the mirror — between P and F, as a convex mirror always gives.

Q3. A transparent block has refractive index 1.6. Find its critical angle with air, and state whether a ray striking the inside of a face at 35° will emerge.

Show Answer
sin C = 1/1.6 = 0.6250 ⇒ C = 38.68°. Since 35° < 38.68°, the angle is below the critical angle, so the ray will emerge (refracted away from the normal), with partial reflection back inside.

Q4. A tank holds water (n = 4/3) to a real depth of 2.0 m. Viewed from directly above, what is the apparent depth, and by how much does the bottom appear raised?

Show Answer
Apparent depth = 2.0 ÷ (4/3) = 2.0 × 0.75 = 1.5 m. Apparent rise = 2.0 − 1.5 = 0.5 m. Check with t(1 − 1/n) = 2.0 × 0.25 = 0.5 m. ✓

Q5. An object is placed 10 cm from a converging lens of focal length 15 cm. Find v and m, and say what kind of image forms.

Show Answer
u = −10 cm, f = +15 cm.  1/v = 1/15 − 1/10 = 2/30 − 3/30 = −1/30 ⇒ v = −30 cm. m = (−30)/(−10) = +3. Virtual, erect, three times enlarged, on the same side as the object — the object is inside the focus, so the lens is acting as a magnifying glass.

Q6. A plano-convex lens is made of glass with n = 1.5. Its curved face has radius 15 cm and the flat face is turned away from the incoming light. Find its focal length and power in air.

Show Answer
R1 = +15 cm, R2 = ∞ so 1/R2 = 0.  1/f = (1.5 − 1)(1/15 − 0) = 0.5/15 = 1/30 ⇒ f = +30 cm. P = 100/30 = +3.33 D. Converging, as the positive sign confirms.

Q7. A lens of power +6 D is placed in contact with one of power −4 D. Find the power and focal length of the combination and state its nature.

Show Answer
P = P1 + P2 = +6 + (−4) = +2 D. f = 100/2 = +50 cm (= 0.5 m). Positive power → the combination is converging.

Q8. A prism of refracting angle 60° is made of glass of refractive index 1.6. Calculate the angle of minimum deviation and the angle of incidence at which it occurs.

Show Answer
sin[(A + Dm)/2] = n sin(A/2) = 1.6 × sin 30° = 1.6 × 0.5 = 0.800.  (A + Dm)/2 = sin−1(0.800) = 53.13° ⇒ A + Dm = 106.26° ⇒ Dm = 46.26°. Angle of incidence i = (A + Dm)/2 = 53.13°. (Compare with 37.18° for n = 1.5 — denser glass, greater deviation. ✓)

Q9. An astronomical telescope has an objective of focal length 100 cm and an eyepiece of focal length 4 cm. Find its magnifying power and tube length in normal adjustment.

Show Answer
|M| = fo/fe = 100/4 = 25; signed M = −25 because the final image is inverted.  L = fo + fe = 100 + 4 = 104 cm. The final image is at infinity, so the eye is fully relaxed.

Q10. A compound microscope has fo = 2.0 cm and fe = 6.25 cm. The object is 2.5 cm from the objective and the final image is at the near point. Find vo, mo and the total magnifying power. Take D = 25 cm.

Show Answer
uo = −2.5 cm, fo = +2.0 cm.  1/vo = 1/2.0 − 1/2.5 = 0.500 − 0.400 = 0.100 ⇒ vo = +10.0 cm. mo = 10.0/(−2.5) = −4. me = 1 + 25/6.25 = 1 + 4 = 5. Total M = (−4)(5) = −20, i.e. 20× and inverted.

Q11. A glass rod (n = 1.5) has a convex end of radius 10 cm. A point object in air lies on the axis, 30 cm from that end. Find the image position.

Show Answer
n1 = 1, n2 = 1.5, u = −30 cm, R = +10 cm.  1.5/v − 1/(−30) = 0.5/10 = 0.0500, so 1.5/v = 0.0500 − 0.03333 = 0.016667 ⇒ v = +90 cm. The image is real, 90 cm inside the glass.

Q12. A concave mirror of focal length 20 cm produces a virtual image four times the size of the object. Where is the object placed, and where is the image?

Show Answer
Virtual and enlarged ⇒ m = +4 (positive, because a virtual image in a mirror is erect). From m = −v/u: v = −4u. Substituting into 1/v + 1/u = 1/f: −1/(4u) + 1/u = 3/(4u) = 1/f ⇒ u = 3f/4 = 3(−20)/4 = −15 cm, and v = −4(−15) = +60 cm. Check: 1/60 + 1/(−15) = 1/60 − 4/60 = −3/60 = −1/20 = 1/f. ✓ The object sits 15 cm from the mirror, inside the 20 cm focus, exactly as a virtual image requires.

Once you have marked yourself, do not simply note the score. Note the step that failed. Almost every wrong answer in this chapter traces back to step 2 — a sign written down without thinking — and once you see that pattern in your own work, it stops happening.

When you are comfortable here, the natural next stop in your Class 12 Physics revision is Current Electricity, which shares the same discipline of signs and directions in a completely different setting. When you are ready to see what happens where the ray model finally breaks down, Wave Optics picks up exactly where this chapter stops.

One more than yesterday. You do not need to master ray optics tonight. You need one more correct sign than you got right yesterday, and then one more the day after. Six days of that and this chapter will feel like arithmetic. Kaizen — small, steady, unglamorous improvement — is what actually moves a physics score, and it is completely within your reach.

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