Class 9 Science Chapter 13 Light – Previous Year Questions (2020–2025 Solved)

Solving previous year questions achieves three things theory revision alone cannot. First, it exposes you to the exact language, diagram styles, and numerical ranges examiners use—this familiarity reduces exam-day anxiety. Second, PYQs reveal what examiners *actually* test: for Chapter 13, the focus is on applying the law of reflection (angle of incidence = angle of reflection) to mirror systems, not memorising definitions. Third, timed practice identifies your weak spots. Many Class 9 students can state the laws of reflection but stumble when asked to find the number of images in two mirrors at 60°—solving past papers exposes this gap immediately. CBSE Chapter 13 typically carries 5–8 marks in the final exam. By working through 13 solved PYQs spanning 1-mark, 3-mark, and 5-mark formats, you internalise the connection between theory and application. You'll also notice recurring sub-topics: the formula for images in two mirrors (n = 360°/θ − 1), properties of plane mirrors, convex/concave mirror sign conventions, and the path of light through a prism during dispersion. Each solved question reinforces these patterns.

1-mark questions test recall, definitions, and single-step reasoning. Here are five commonly seen versions: **Q1: What is the angle of incidence if the angle of reflection is 35°?** Answer: 35° (by the law of reflection, ∠i = ∠r) **Q2: A light ray hits a plane mirror at an angle of 60° to the normal. What is the angle between the incident and reflected rays?** Answer: 60° (the angle between incident and reflected rays = 2 × (90° − ∠i) = 2 × 30° = 60°. Or simply: if ∠i = 60°, then ∠r = 60°, and the angle between them = 180° − 60° − 60° = 60°) **Q3: How many images will be formed if an object is placed between two plane mirrors at 90° to each other?** Answer: 3 images (using n = 360°/θ − 1 = 360°/90° − 1 = 3) **Q4: Name the phenomenon responsible for the formation of a rainbow.** Answer: Dispersion (or refraction and dispersion of sunlight by water droplets) **Q5: A person suffers from myopia. Which type of lens is used to correct it?** Answer: Concave lens (−) because myopia is when the image forms in front of the retina; a concave lens diverges light rays to push the image backward onto the retina.

3-mark questions require explanation, simple calculations, or a labelled diagram. Here are five types seen frequently: **Q1: State the laws of reflection and explain with a ray diagram.** Answer: Laws of reflection: (i) The incident ray, reflected ray, and normal all lie in the same plane. (ii) The angle of incidence equals the angle of reflection (∠i = ∠r). [Diagram required: A normal line perpendicular to the mirror, incident ray approaching at angle i, reflected ray leaving at angle r, with i = r marked.] **Q2: Two plane mirrors are placed at an angle of 45° to each other. How many images of an object placed between them will be seen?** Answer: n = 360°/θ − 1 = 360°/45° − 1 = 8 − 1 = 7 images. [Show formula and substitution clearly.] **Q3: Explain why the sky appears blue during the day but red during sunset.** Answer: Blue light has a shorter wavelength and scatters more (Rayleigh scattering). During the day, blue light scatters in all directions, making the sky blue. During sunset, light travels through a thicker atmosphere. Blue light is scattered away, leaving red/orange light to reach the observer, making the sunset red. [Mention wavelength: blue ≈ 450 nm, red ≈ 700 nm.] **Q4: Draw a ray diagram showing the path of a light ray passing through a prism and undergoing dispersion.** Answer: [Diagram required: A triangular prism, incident ray entering one face, refraction at entry (ray bends towards the normal), ray travels inside the prism, emerges from the opposite face with further refraction (bends away from the normal), and finally a spectrum of colours (VIBGYOR) shown emerging. Angle of deviation marked.] **Q5: A person has hypermetropia (far-sightedness). Explain the defect and the lens used to correct it.** Answer: In hypermetropia, the eyeball is too short or the cornea is too flat. The image of nearby objects forms behind the retina. Correction: A convex lens (+) is used because it converges light rays, bringing the image forward onto the retina, allowing clear vision at normal distance.

5-mark questions test deep understanding, multi-step problem-solving, and combined concepts. Here are three detailed solutions: **Q1: Explain the structure and function of the human eye. Draw a labelled diagram and describe how the eye focuses light onto the retina.** Solution: The human eye is a complex optical organ consisting of: (1) Cornea – transparent front of the eye, primary light-bending structure; (2) Aqueous humour – transparent fluid maintaining eye shape; (3) Iris – controls pupil size to regulate light entry; (4) Lens – adjustable convex lens, focuses light on the retina via accommodation; (5) Vitreous humour – gel filling the eyeball; (6) Retina – light-sensitive layer containing photoreceptor cells (rods and cones); (7) Optic nerve – transmits image signals to the brain. Focusing mechanism: When an object is close, ciliary muscles contract, increasing lens thickness (focal length decreases). When an object is far, muscles relax, lens flattens (focal length increases). This accommodation allows sharp images to form on the retina across a range of distances (near point ≈ 25 cm, far point = ∞ for normal eye). [Diagram: Cross-section of eye with all parts labelled, light rays entering and converging on retina.] **Q2: Two plane mirrors are inclined at an angle of 60° to each other. An object is placed on the bisector of the angle between the mirrors. How many images will be formed? Draw a diagram showing the positions of all images.** Solution: Formula: n = 360°/θ − 1 = 360°/60° − 1 = 6 − 1 = 5 images. When the object is on the bisector, all 5 images are visible (if 360°/θ is an even number, and the object is on the bisector, all images appear; if it's odd and the object is on the bisector, all images still appear in this case). [Diagram: Two mirrors at 60° angle, object O on the bisector, five image positions marked as I₁, I₂, I₃, I₄, I₅ arranged symmetrically around O.] Explanation: Each mirror produces an image; these images act as objects for the other mirror, creating multiple reflections until all rays have undergone sufficient reflections and exit the mirror system. **Q3: A ray of light enters a triangular glass prism and emerges after refraction. Explain the path of light and the principle of dispersion. Why are different colours dispersed at different angles?** Solution: Path of light in a prism: (1) Incident ray strikes the first refracting surface at an angle of incidence i₁. (2) By Snell's law (n₁ sin i₁ = n₂ sin r₁), the ray refracts towards the normal (since glass has n > 1) and travels inside the prism at angle of refraction r₁. (3) At the second surface, the ray hits at an angle i₂ (inside glass). (4) It refracts away from the normal (exiting to air) and emerges at angle e₂. Deviation (δ) is the angle between the incident and emergent rays. Principle of dispersion: Different colours have different wavelengths (red ≈ 700 nm, violet ≈ 400 nm). The refractive index of glass varies with wavelength—shorter wavelengths (violet) are refracted more than longer wavelengths (red). Thus, violet light deviates more, red deviates less, and they separate into a spectrum (VIBGYOR). [Diagram: Prism with white light entering, path inside prism, and spectrum emerging from the second surface.] Example: For a 60° prism, if white light is incident at 45°, violet emerges at one angle and red at a slightly smaller angle, creating a spectrum of 1–2° angular separation.

The 2024–25 rationalized CBSE Class 9 Science syllabus has refined Chapter 13. Key shifts from earlier years include: (1) Increased emphasis on ray diagrams—examiners now expect detailed, to-scale diagrams in 3-mark and 5-mark answers; (2) Reduced focus on mirror formula derivations, but heightened demand for numerical applications of the image formula (n = 360°/θ − 1); (3) Greater integration of real-world contexts—vision defects now appear in 5-mark questions pairing anatomy with optical principles, rather than as isolated 1-mark definitions; (4) Minimal change in reflection laws themselves, but greater complexity in multi-mirror setups (e.g., three mirrors, or mirrors not perpendicular). Dispersion questions have shifted from 'name the phenomenon' (1-mark) to explaining *why* colours separate at different angles using wavelength and refractive index (3–5 marks). The new pattern expects students to use the relationship: sin i / sin r = n, where n = 1.5 for typical glass. For vision defects, the 2026–27 focus includes presbyopia (age-related) and astigmatism (less common at Class 9 level, but appearing in sample papers). Recommend: practise labelled diagrams regularly, memorise the image formula and its conditions, and link optical defects directly to corneal/lens shape using ray diagrams. Mock papers from 2024–25 onward reflect this shift.

1. **Read all questions before starting (2 min).** Identify 1-mark questions (usually 4–5 total for this chapter). These are quickest; answer them first to build confidence and secure easy marks. Typical examples: law of reflection angle, image count in mirrors, lens type for a defect. 2. **Prioritise 3-mark questions (7–8 min per question).** These appear most frequently in board papers. Always include a ray diagram, even if the question doesn't explicitly ask for one—diagrams earn 1–2 bonus marks. For 'state the law' questions, write exactly two sentences (law + condition). For calculations (mirror images), show the formula and substitution step-by-step. 3. **Reserve time for 5-mark questions (12–15 min per question).** Begin with the one you find easiest (often eye structure or a two-mirror problem). Write headings (e.g., 'Structure of the eye,' 'Dispersion process'). Use bullet points for long explanations. Always include a labelled diagram. If stuck, write what you know (e.g., 'cornea bends light,' 'lens accommodates') even if incomplete—partial marks are awarded. 4. **Avoid common mistakes.** Do not confuse angle of incidence with angle of deviation. Do not use n = 360°/θ (without subtracting 1) when calculating image count—this is the most frequent error. Do not write 'refraction = bending' without specifying Snell's law or the reason. 5. **Check your work (2 min).** Verify that your image count formula result makes sense (usually 1–7 images for angles 30°–360°). Confirm that your diagram is clear and labelled with all essential parts. If describing light paths, ensure the word 'normal' appears at least once.

**Reflection and mirrors:** Law of reflection: ∠i = ∠r. Image count in two plane mirrors: n = 360°/θ − 1 (where θ is the angle between mirrors, in degrees). Magnification in plane mirror = 1 (image same size as object). **Refraction:** Snell's law: n₁ sin i = n₂ sin r, where n is the refractive index. Critical angle: sin θc = 1/n (for total internal reflection). **Dispersion:** Different wavelengths refract at different angles. Red light (≈ 700 nm) refracts least; violet light (≈ 400 nm) refracts most. Refractive index increases with decreasing wavelength: nviolet > nred. **Human eye:** Near point (minimum distance for clear vision) ≈ 25 cm for a normal eye. Far point (maximum distance) = ∞. Power of accommodation allows lens thickness to vary. Defects: Myopia (image in front of retina) → corrected with concave lens (−). Hypermetropia (image behind retina) → corrected with convex lens (+). Presbyopia (lens loses elasticity with age) → bifocals or progressive lenses. All formulas and definitions align with the 2024–25 CBSE Class 9 Science textbook (NCERT).