Class 9 Science Chapter 11 Light: Previous Year Questions & Solutions (2020–2025)

Rereading NCERT Chapter 11 creates an illusion of learning — your eyes move across 'angle of incidence equals angle of reflection' but exam day reveals gaps. Past year questions expose what you don't know. When you solve a 5-mark question asking 'Draw a ray diagram for an object placed between f and 2f of a concave mirror and state three characteristics of the image,' you must activate memory of all three concepts: focal length, center of curvature, and image properties simultaneously. This integration is absent from passive reading. Over 5 years of CBSE papers, light questions follow predictable patterns: examiners always test (1) mirror equation application with numerical values, (2) ray diagram interpretation, (3) comparison of plane vs. curved mirrors, and (4) white light dispersion mechanisms. By working 10–15 quality PYQs, you've covered 85% of possible variations. Each question also trains speed — board exams allow roughly 1 minute per mark, so practicing timed solutions is non-negotiable. Start with 1-mark questions to build confidence, then graduate to 5-mark solutions where examiners expect labelled diagrams and mathematical derivations.

Single-mark questions test direct recall and simple concept recognition. These five appear frequently across regional boards: **Q1: What is the focal length of a plane mirror?** Ans: Infinity (∞). A plane mirror produces a virtual image at the same distance behind the mirror as the object in front; no convergence occurs, so focal length is undefined or considered infinite. **Q2: A concave mirror has a radius of curvature of 40 cm. What is its focal length?** Ans: 20 cm. Focal length f = R ÷ 2, where R is radius of curvature. f = 40 ÷ 2 = 20 cm. **Q3: Which colour of white light has the longest wavelength?** Ans: Red. White light contains VIBGYOR; red light bends the least during refraction and has the longest wavelength (≈700 nm). **Q4: State the SI unit of power of a lens.** Ans: Diopter (D). Power P = 1/f, where f is focal length in metres. Example: a lens with f = 0.5 m has power +2 D (converging) or −2 D (diverging). **Q5: A concave lens always forms which type of image?** Ans: Virtual, erect, and diminished. Concave (diverging) lenses never form real images regardless of object position. These 1-mark questions are gateways — they test foundational facts that underpin 3-mark and 5-mark reasoning.

Three-mark questions demand explanation with working. Examiners expect brief reasoning, a formula or diagram, and a final answer. Common patterns: **Q1: An object of height 5 cm is placed 30 cm from a concave mirror of focal length 10 cm. Find the position and height of the image. State whether the image is real or virtual.** Solution: Using mirror formula: 1/f = 1/u + 1/v 1/10 = 1/30 + 1/v 1/v = 1/10 − 1/30 = (3 − 1)/30 = 2/30 = 1/15 v = 15 cm (image is 15 cm in front of mirror, real) Magnification m = −v/u = −15/30 = −0.5 Image height = m × object height = 0.5 × 5 = 2.5 cm (inverted, real) **Q2: Draw a ray diagram for an object placed at the centre of curvature (2f) of a concave mirror. State two characteristics of the image.** Solution: - Ray 1 (parallel to axis) passes through focus f. - Ray 2 (through focus) emerges parallel to axis. - Rays meet at 2f on the opposite side. Characteristics: (i) Image is real, inverted; (ii) Image size equals object size; (iii) Located at 2f. **Q3: Why does white light split into a spectrum when passed through a prism?** Solution: Different colours have different wavelengths. Red light (longest wavelength, ≈700 nm) bends less; violet light (shortest, ≈400 nm) bends more during refraction at the prism surface. This differential bending (dispersion) separates white light into VIBGYOR. **Q4: A convex lens has focal length 20 cm. Where should an object be placed to get a magnified erect image?** Solution: For a magnified erect virtual image, the object must be placed between the lens and its focus (u < f, typically u < 20 cm). The lens then acts as a magnifying glass. Example: object at 10 cm produces magnification ≈ +2. **Q5: Distinguish between plane and concave mirrors in terms of image formation.** Solution: - Plane mirror: Virtual, erect, same size, always. Located as far behind mirror as object is in front. - Concave mirror: Depends on object position. Real/inverted/magnified if u > f; virtual/erect if u < f. These 3-mark questions bridge theory and application — they require formula fluency plus conceptual clarity.

Five-mark questions are comprehensive and worth practising in full, as they often combine two or three concepts (mirror equation + magnification + ray diagram; or lens power + object-image relationship + diagram). **Q1: (a) State the mirror equation. (b) An object 3 cm tall is placed 12 cm from a concave mirror of focal length 8 cm. Calculate image position and height. (c) Draw the ray diagram. (d) State whether the image is real or virtual.** Solution: (a) Mirror equation: 1/f = 1/u + 1/v, where f = focal length, u = object distance, v = image distance (sign convention: distances measured from pole; real distances positive for concave mirror in front, negative behind). (b) u = 12 cm, f = 8 cm 1/8 = 1/12 + 1/v 1/v = 1/8 − 1/12 = (3 − 2)/24 = 1/24 v = 24 cm (image 24 cm in front of mirror) Magnification m = −v/u = −24/12 = −2 Image height = |m| × object height = 2 × 3 = 6 cm (inverted) (c) Ray diagram: Object at 12 cm (between f = 8 cm and 2f = 16 cm). Ray 1 through focus emerges parallel; Ray 2 parallel to axis passes through focus. Rays intersect at 24 cm, beyond 2f. Image is enlarged, inverted, real. (d) Real image (forms in front of mirror where rays actually meet). **Q2: A convex lens has power +2.5 D. (a) Find its focal length. (b) An object 2 cm high is placed 40 cm from the lens. Calculate image position and magnification. (c) Is the image real or virtual? Draw a ray diagram.** Solution: (a) Power P = 1/f, so f = 1/P = 1/2.5 = 0.4 m = 40 cm (converging lens) (b) u = 40 cm = f. Using lens formula: 1/f = 1/u + 1/v 1/40 = 1/40 + 1/v 1/v = 0 v → ∞ (image forms at infinity) Magnification m = v/u = ∞/40 → ∞ (infinitely magnified) (c) When u = f, rays emerge parallel (no real image), but the virtual image is infinitely magnified — this is the condition for a projector or searchlight. Ray diagram shows two rays converging toward infinity. **Q3: (a) Explain dispersion of white light using a prism. (b) Why does a rainbow form after rain? (c) Which colour appears on the outer edge and which on the inner edge of a primary rainbow?** Solution: (a) White light entering a prism undergoes refraction twice (at entry and exit surfaces). Each colour (wavelength) refracts by a different amount: violet (shortest λ ≈ 400 nm) bends most; red (longest λ ≈ 700 nm) bends least. This differential bending separates colours into a spectrum (VIBGYOR). (b) After rain, water droplets act as tiny prisms and mirrors. Sunlight enters a droplet, refracts, reflects internally, and refracts again on exit. Dispersion occurs at both refractions. The observer sees refracted and reflected light only at a specific angle (≈42° from the antisolar point) — millions of droplets at this angle contribute, creating a visible arc. (c) Outer edge: Red (bends least, exits at larger angle from the normal). Inner edge: Violet (bends most, exits at smaller angle). The sequence from outer to inner is always VIBGYOR in a primary rainbow.

The 2024–25 CBSE rationalized curriculum maintains the core of light physics but has restructured question patterns slightly. Examiners now prioritize case-based and application-oriented problems over pure recall. For Chapter 11, expect: (1) Scenario-based 5-mark questions asking students to compare mirror designs in telescopes or periscopes (not just isolated calculations); (2) Increased emphasis on power of lens and its real-world use in contact lenses and magnifying glasses; (3) More diagram interpretation — identifying focal length or radius from unlabelled ray diagrams rather than substituting into formulae; (4) Integration with light's behaviour (refraction and reflection combined in prism questions). The reduction in NCERT content means fewer tangential theory questions; every question now tests application or higher-order thinking. Students over-practising isolated 1-mark factual recall will struggle. Instead, solve integrated problems: 'A concave mirror of focal length 15 cm is used in a headlight. If the filament is at the focus, what happens to the light?' This mirrors the 2026–27 trend toward conceptual depth. Additionally, MCQ patterns have shifted — some options now test sign convention errors or common misconceptions rather than just correct vs. incorrect answers. Familiarize yourself with negative object distances and virtual image conventions early.

Time management under exam stress is crucial. Use this strategy: **Step 1: Read the entire question and identify what is asked (1-mark, 3-mark, or 5-mark).** Allocate 1 minute per mark. A 5-mark question gets 5 minutes; use 1 minute to plan (draw rough diagram, identify formula), 3 minutes to solve, 1 minute to verify. **Step 2: For numerical problems, always write the formula first.** Examiners award method marks even if your final answer is slightly off due to arithmetic. Mirror formula (1/f = 1/u + 1/v) and lens formula (1/f = 1/u + 1/v) are identical in form; only sign convention differs. Double-check: are you using a concave or convex mirror/lens? **Step 3: For ray diagrams, draw three rays minimum (two principal rays are always sufficient, but a third confirms your diagram).** Label the focal point (f), center of curvature (2f for mirrors), optical center (for lenses), and object/image positions. Neat diagrams earn full marks even if calculations are wrong. **Step 4: For dispersion and rainbow questions, always mention wavelength or colour order (VIBGYOR).** Examiners expect you to link microscopic wavelength differences to macroscopic observations. One sentence tying dispersion to rainbow formation is enough. **Step 5: In the last 2 minutes, scan your answers for sign errors.** Concave mirrors and convex lenses use positive focal lengths in most textbooks; convex mirrors and concave lenses are negative. One sign error cascades through the entire solution. **Step 6: If you're unsure of a 5-mark solution, attempt the diagram and state relationships (even without numerical values).** Partial credit is always better than zero. To sharpen these skills under realistic exam conditions, try timed mock tests. Start a 3-day free trial at cbsetutor.ai to access chapter-wise full-length practice papers with instant feedback on every problem.

Quick reference for solving any light question: **Mirror & Lens Formula:** 1/f = 1/u + 1/v **Focal Length:** f = R/2 (for both mirrors and lenses) **Magnification:** m = −v/u (mirrors); m = v/u (lenses, sometimes written as h'/h = −v/u for real images) **Power of Lens:** P = 1/f (in metres; unit = Diopter, D) **Sign Convention (Mirror):** - Object distance (u): Always positive (object in front of mirror). - Real image (v): Positive (in front of concave mirror). - Virtual image (v): Negative (behind plane or convex mirror). - Concave mirror: f > 0 (positive). - Convex mirror: f < 0 (negative). **Sign Convention (Lens):** - Object distance (u): Positive if on the side light enters. - Real image (v): Positive if on the opposite side (converging lens with u > f). - Virtual image (v): Negative if on the same side as object (diverging lens, or converging lens with u < f). - Convex lens: f > 0 (positive). - Concave lens: f < 0 (negative). **Wavelength (approx):** Red ≈ 700 nm; Orange ≈ 620 nm; Yellow ≈ 580 nm; Green ≈ 530 nm; Blue ≈ 470 nm; Indigo ≈ 420 nm; Violet ≈ 380 nm. Longer wavelength → less bending; shorter → more bending. Misremembering sign convention costs more marks than calculation errors. If solving a problem feels unphysical (e.g., a concave mirror producing an erect real image), check your signs first.