Why the sign convention is the whole chapter
Ask a student who finds Ray Optics hard to explain how a converging lens forms an image and they will usually do it correctly. Ask them to compute the image position for an object 30 cm from a 10 cm converging lens and the answer often comes out negative, or the magnification comes out positive when it should be negative.
That gap is not conceptual. It is procedural: the Cartesian convention has to be applied to every quantity, every time, and decided before substitution rather than during it. Treating it as bookkeeping rather than physics - and doing the bookkeeping first - converts this chapter from unreliable to mechanical.
How JEE actually asks Ray Optics
NTA publishes no chapter-wise weightage, so figures circulating online are coaching estimates from past papers; check the official NTA information bulletin for your session's syllabus and pattern.
The recurring families are mirror and lens image formation, refraction including total internal reflection and apparent depth, prisms and dispersion, and optical instruments. Combination questions - a lens followed by a mirror, or a lens partly immersed - appear regularly and are solved by applying the same formula twice, treating the first image as the object for the second element.
Key concepts, compressed
- ✓The Cartesian convention: all distances are measured from the pole or optical centre, along the incident-light direction as positive. Heights above the axis are positive.
- ✓Focal length is negative for a concave mirror and a diverging lens, positive for a convex mirror and a converging lens.
- ✓A real image forms where rays actually converge and can be projected onto a screen; a virtual image is where they only appear to come from.
- ✓Refraction bends light toward the normal on entering a denser medium and away on leaving it. The frequency is unchanged; wavelength and speed change.
- ✓Total internal reflection needs denser-to-rarer travel and an angle of incidence above the critical angle sin C = 1/μ.
- ✓For thin lenses in contact the powers add, which is why power rather than focal length is the practical quantity for combinations.
Formulas you need before attempting the questions
| Quantity | Formula | Note |
|---|---|---|
| Mirror formula | 1/v + 1/u = 1/f | plus sign |
| Lens formula | 1/v − 1/u = 1/f | minus sign |
| Mirror focal length | f = R/2 | |
| Mirror magnification | m = −v/u | |
| Lens magnification | m = v/u | |
| Snell's law | μ₁ sin i = μ₂ sin r | |
| Refractive index and speed | μ = c/v | |
| Critical angle | sin C = 1/μ | denser to rarer only |
| Apparent depth | apparent = real/μ | |
| Lens maker's formula | 1/f = (μ − 1)(1/R₁ − 1/R₂) | |
| Power of a lens | P = 1/f | f in metres, P in dioptres |
| Lenses in contact | P = P₁ + P₂ | |
| Prism deviation | δ = i + e − A | |
| Prism at minimum deviation | μ = sin((A + δm)/2) / sin(A/2) |
The five mistakes that cost the most marks
- ✓Assigning signs during the calculation rather than before it. Write u, v and f with their signs on the page first, then substitute.
- ✓Swapping the mirror and lens formulas. The mirror uses a plus and the lens a minus, and their magnification formulas differ by a sign too.
- ✓Ignoring what the sign of magnification means. Negative means inverted; a question about orientation is answered by that sign alone.
- ✓Applying total internal reflection to rarer-to-denser travel. Whatever the angle, that direction gives refraction, never TIR.
- ✓Computing power from a focal length in centimetres. Convert to metres first: 20 cm gives 5 D.
Practice set 1: reflection and mirrors
1. Describe the image formed by a plane mirror.
Virtual, erect, the same size as the object, laterally inverted, and formed as far behind the mirror as the object is in front. The lateral inversion is a depth reversal rather than a left-right one - which is why text appears reversed but the image is not upside down.
2. A concave mirror has a radius of curvature of 20 cm. What is its focal length?
10 cm, and by the Cartesian convention it is written as f = −10 cm for a concave mirror. The relation f = R/2 holds for all spherical mirrors. A convex mirror of the same radius would have f = +10 cm, and that sign difference is what makes the image virtual instead of real.
3. An object is placed 30 cm in front of a concave mirror of focal length 10 cm. Where is the image?
15 cm in front of the mirror, and real. Write the signs first: u = −30 cm, f = −10 cm. Then 1/v = 1/f − 1/u = (−1/10) − (−1/30) = −1/10 + 1/30 = −2/30, giving v = −15 cm. The negative v means the image is on the same side as the object - in front of the mirror - which is what real means for a mirror.
4. What is the magnification in that case?
−0.5, so the image is inverted and half the size of the object. m = −v/u = −(−15)/(−30) = 15/(−30) = −0.5. The negative sign is the answer to any question about orientation, and the magnitude below 1 tells you it is diminished. Both facts come free once the signs are right.
5. Where is the image when an object is placed exactly at the focus of a concave mirror?
At infinity. Substituting u = f into the mirror formula gives 1/v = 0, so v is infinite - the reflected rays emerge parallel and never converge. This is the principle of a searchlight or a car headlamp, where a source at the focus produces a parallel beam.
6. What kind of image does a convex mirror always form?
Virtual, erect and diminished, regardless of where the object is placed. Because f is positive and u is negative, v always comes out positive - behind the mirror. This is why convex mirrors are used as rear-view and shop-security mirrors: they give a wide field of view at the cost of size.
7. A plane mirror is rotated by 10° while the incident ray stays fixed. Through what angle does the reflected ray turn?
20°. Rotating the mirror by θ rotates the normal by θ, which changes both the angle of incidence and the angle of reflection by θ - so the reflected ray turns by 2θ. This doubling is what makes the optical lever a sensitive measuring instrument, and it is asked directly often enough to be worth remembering.
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Practice set 2: refraction
8. Light enters glass of refractive index 1.5 from air at an angle of incidence of 30°. What is the angle of refraction?
About 19.5°. By Snell's law, sin r = sin i/μ = sin30°/1.5 = 0.5/1.5 = 0.333, so r = arcsin(0.333) ≈ 19.5°. The ray bends toward the normal because it is entering a denser medium, which is a useful check: if your answer exceeded 30°, the calculation has been inverted.
9. What is the critical angle for a glass-air interface with μ = 1.5?
About 41.8°. sin C = 1/μ = 1/1.5 = 0.667, so C = arcsin(0.667) ≈ 41.8°. Beyond this angle, light travelling from glass to air is totally internally reflected. Note that a higher refractive index gives a smaller critical angle, which is why diamond (μ ≈ 2.42, C ≈ 24°) sparkles so strongly.
10. What two conditions are required for total internal reflection?
Light must travel from a denser medium to a rarer one, and the angle of incidence must exceed the critical angle. Both are necessary. Light going from air into glass will never undergo TIR no matter how large the angle - it simply refracts, and options suggesting otherwise are testing this exact point.
11. A coin lies at the bottom of a tank of water 8 cm deep. At what depth does it appear? (μ = 4/3)
6 cm. Apparent depth = real depth/μ = 8/(4/3) = 8 × 3/4 = 6 cm. The pool looks shallower than it is, and the effect scales with refractive index. This is a standard question and the arithmetic is only safe if you remember to divide, not multiply, by μ.
12. What is the speed of light in a medium of refractive index 1.5?
2 × 10⁸ m/s. Since μ = c/v, we have v = c/μ = 3 × 10⁸/1.5 = 2 × 10⁸ m/s. Refractive index is fundamentally a ratio of speeds, and defining it that way makes Snell's law a consequence rather than a separate rule.
13. When light passes from air into glass, what happens to its frequency, wavelength and speed?
Frequency is unchanged; speed and wavelength both decrease by the factor μ. The frequency is set by the source and cannot change at a boundary - if it did, wave crests would have to accumulate there. Since v = fλ and v falls while f is fixed, λ must fall proportionally. Answering that frequency changes is the standard error.
14. How does an optical fibre transmit light around bends?
By total internal reflection at the core-cladding boundary. The core has a higher refractive index than the cladding, so light striking the interface above the critical angle is reflected entirely with no refracted loss. Repeated thousands of times, this guides the signal around curves with very little attenuation.
Practice set 3: lenses
15. An object is placed 30 cm from a converging lens of focal length 10 cm. Where is the image?
15 cm on the far side of the lens, and real. Signs first: u = −30 cm, f = +10 cm. Then 1/v = 1/f + 1/u = 1/10 + (−1/30) = 3/30 − 1/30 = 2/30, giving v = +15 cm. A positive v means the image is on the opposite side from the object, which for a lens means real.
16. What is the magnification in that case?
−0.5: inverted and half size. For a lens, m = v/u = 15/(−30) = −0.5. Note the contrast with question 4: the mirror uses m = −v/u and the lens uses m = v/u, yet both give −0.5 here because the sign of v differs between the two cases. Using the wrong formula would have given +0.5 and the wrong orientation.
17. What is the power of a converging lens of focal length 20 cm?
+5 D. Power is the reciprocal of the focal length in metres: P = 1/0.2 = 5 D. Forgetting to convert from centimetres gives 0.05 D, which is the classic error. The positive sign indicates a converging lens.
18. What is the power of a diverging lens of focal length 25 cm?
−4 D. A diverging lens has a negative focal length, f = −0.25 m, so P = 1/(−0.25) = −4 D. The sign is part of the answer, not decoration - prescriptions for short-sightedness are written as negative powers precisely because they require diverging lenses.
19. Two thin lenses of powers +5 D and −2 D are placed in contact. What is the combined power and focal length?
+3 D, giving a focal length of 1/3 m ≈ 33.3 cm. Powers add directly for thin lenses in contact: P = P₁ + P₂ = 5 − 2 = 3 D. This is exactly why power is the more convenient quantity - adding focal lengths would be wrong, and the reciprocal combination is far messier.
20. What does the lens maker's formula tell you?
That focal length depends on the refractive index of the lens material relative to its surroundings and on the two surface radii: 1/f = (μ − 1)(1/R₁ − 1/R₂). It explains why the same lens has different focal lengths in different media, and why a lens made of material with μ equal to its surroundings has no focusing power at all.
21. What happens to the focal length of a glass lens when it is immersed in water?
It increases, so the lens becomes weaker. In the lens maker's formula the relevant quantity is the relative refractive index of glass with respect to the surrounding medium, which drops from about 1.5 in air to about 1.13 in water. A smaller (μ − 1) means a larger f. If the surrounding medium were denser than the lens, a converging lens would become diverging.
22. An object is placed between a converging lens and its focus. Describe the image.
Virtual, erect and magnified, on the same side as the object. With |u| less than f, the refracted rays diverge and only appear to come from a point behind the object. This is the magnifying-glass configuration, and it is the only arrangement in which a converging lens produces an erect image.
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Practice set 4: prisms and dispersion
23. What is the relation between the angles for a ray passing through a prism?
δ = i + e − A, where i and e are the angles of incidence and emergence and A is the prism angle. There is also the geometric relation r₁ + r₂ = A for the two internal refraction angles. Together these two relations solve essentially every standard prism question.
24. What characterises the condition of minimum deviation in a prism?
The angle of incidence equals the angle of emergence, i = e, and the ray passes symmetrically through the prism parallel to its base. At that point μ = sin((A + δm)/2)/sin(A/2), which is the standard laboratory method for measuring refractive index. Deviation is minimum there, so small changes in incidence barely alter it - which is what makes the measurement reliable.
25. Why does a prism disperse white light into colours?
Because refractive index depends on wavelength: μ is larger for shorter wavelengths. Violet therefore deviates most and red least, spreading the colours into a spectrum. A rectangular glass slab also disperses light internally, but the two parallel faces recombine the colours, which is why no spectrum emerges.
26. What is dispersive power, and what does it depend on?
It is the ratio of angular dispersion to mean deviation, ω = (μᵥ − μᵣ)/(μ − 1). It depends only on the material of the prism, not on its angle - which is why two prisms of different materials can be combined to produce dispersion without deviation (a direct-vision prism) or deviation without dispersion (an achromatic combination).
Practice set 5: optical instruments and the eye
27. What is the magnifying power of a simple microscope when the image is formed at the near point?
M = 1 + D/f, where D is the near-point distance, conventionally 25 cm. If the image is instead formed at infinity for relaxed viewing, the magnification is D/f - one less. Questions specify which viewing condition applies, and using the wrong one is a common slip.
28. How does a compound microscope achieve high magnification?
In two stages. The objective forms a real, inverted, magnified image just inside the focus of the eyepiece, and the eyepiece then acts as a simple microscope on that image. The total magnification is the product of the two, which is why a compound instrument far exceeds what a single lens can achieve.
29. What is the magnifying power of an astronomical telescope in normal adjustment?
M = f(objective)/f(eyepiece), with the tube length equal to the sum of the two focal lengths. High magnification therefore needs a long-focus objective and a short-focus eyepiece. Note the contrast with the microscope: a telescope wants a large objective focal length, a microscope a small one.
30. Which lens corrects myopia, and why?
A diverging (concave) lens, of negative power. In myopia the eye converges light too strongly and forms images in front of the retina, so a diverging lens is placed in front to reduce the total converging power and push the image back onto the retina. Hypermetropia is the opposite case and is corrected with a converging lens of positive power.
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How to study this chapter efficiently
- ✓Write u, v and f with their signs on the page before substituting into any formula. This one habit fixes most of the chapter's errors.
- ✓Keep the mirror and lens formulas visibly separate in your notes, along with their differing magnification formulas.
- ✓Sketch a rough ray diagram even when the question does not ask for one. It tells you immediately whether the image should be real or virtual, and catches sign errors before you finish.
- ✓Convert focal lengths to metres the moment a question mentions power.
- ✓For multi-element systems, solve sequentially: the image formed by the first element becomes the object for the second, carrying its sign.
- ✓Learn the two prism relations - δ = i + e − A and r₁ + r₂ = A - as a pair. Between them they handle almost every prism question.
Turn this into active practice
Sign errors are systematic rather than random, which means they are also fixable. A student who consistently gets the lens magnification sign wrong will keep doing so until a question forces them to notice - and reading a solved example never does, because the signs are already there.
The JEE Ray Optics quiz on QUFF generates fresh questions across mirrors, refraction, lenses, prisms and instruments, marks them instantly and explains each answer. Do a timed set, and for every wrong answer check whether the physics or only the sign was wrong. If it is the sign, that is good news - it is the fastest thing in this chapter to repair.
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