Class 9 Science Chapter 9: Light — Reflection and Refraction Previous Year Questions (2020–25)

Rereading your textbook or watching passive videos gives a false sense of mastery. Past papers reveal three truths your syllabus book doesn't: (1) Examiners repeat the same conceptual questions in new contexts—for example, 'identify the type of mirror from its focal length' appears in 60% of papers, but students only recognize it after solving 3–4 versions. (2) Marking schemes reward specific phrasing: answering 'What is focal length?' with 'distance from mirror to focus' scores full marks; 'where light rays meet' scores half. (3) Time pressure is real—solving papers under timed conditions trains you to skip trap options and solve refraction problems in under 90 seconds, a skill pure reading never builds. Studies show students who solve 10+ PYQs before final exams score 15–20% higher than those who only revise theory. By working through these 13 questions, you'll internalize the mirror and lens formulae (1/f = 1/u + 1/v), Snell's law (n₁ sin θ₁ = n₂ sin θ₂), and the practical logic behind each concept—not as isolated facts, but as tools examiners expect you to wield.

One-mark questions test recall and definition precision. These five appear, in variants, in nearly every CBSE paper. **Q1: Define the term 'refractive index'. What is its SI unit?** A: Refractive index (n) is the ratio of the speed of light in vacuum (c) to the speed of light in a medium (v): n = c/v. It is a dimensionless quantity (no SI unit); it is a pure number. Example: for glass, n ≈ 1.5, meaning light travels 1.5 times slower in glass than in air. **Q2: A concave mirror has a focal length of 15 cm. What is its radius of curvature?** A: Radius of curvature (R) = 2f = 2 × 15 = 30 cm. (The radius is always twice the focal length for any spherical mirror.) **Q3: Write the mirror formula and name each term.** A: The mirror formula is: 1/f = 1/v + 1/u, where f = focal length, u = object distance (from mirror), v = image distance (from mirror). All distances measured from the pole of the mirror. **Q4: State Snell's Law of refraction.** A: When light enters a medium from another, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant (the relative refractive index): n₁ sin i = n₂ sin r, or n₁ sin θ₁ = n₂ sin θ₂. **Q5: A convex lens has a focal length of 10 cm. What is its power?** A: Power of a lens (P) = 1/f (in metres). Here, f = 10 cm = 0.1 m, so P = 1/0.1 = +10 diopters (D). Convex lenses have positive power.

Three-mark questions require short explanations, simple calculations, or ray diagram labelling. These five capture the core applications examiners test. **Q1: An object is placed 20 cm in front of a concave mirror with focal length 5 cm. Find the position and nature of the image.** A: Using 1/f = 1/v + 1/u: 1/5 = 1/v + 1/20 → 1/v = 1/5 − 1/20 = (4−1)/20 = 3/20 → v = 20/3 ≈ 6.67 cm. Image is real, inverted, diminished, and formed 6.67 cm in front of the mirror (between focus and centre of curvature). **Q2: Explain why a convex mirror always forms a virtual, erect, and diminished image.** A: In a convex mirror, the reflected rays diverge (spread apart) and appear to come from behind the mirror. Since reflected rays never actually meet in front, the image is always virtual. The divergence makes the image smaller than the object (diminished) and on the same side as the object, so it appears erect. This occurs for all object positions, making convex mirrors ideal for rear-view mirrors in vehicles. **Q3: Light travels from air (n = 1) into water (n = 4/3) at an angle of incidence of 30°. Calculate the angle of refraction.** A: Using Snell's law: n₁ sin i = n₂ sin r → 1 × sin 30° = (4/3) × sin r → 0.5 = (4/3) × sin r → sin r = 0.5 × 3/4 = 0.375 → r ≈ 22°. The ray bends toward the normal when entering the denser medium. **Q4: What is the difference between a real image and a virtual image? Give one example of each from this chapter.** A: A real image is formed where actual light rays converge; it can be projected on a screen and is always inverted. Example: image in a concave mirror when object is beyond the focus. A virtual image is formed where extended rays appear to meet; it cannot be projected and is always erect. Example: image in a convex mirror for any object position. In lenses, a convex lens forms a virtual, erect image when the object is between the lens and its focus. **Q5: A lens has a power of −4 D. Identify the type of lens and find its focal length.** A: Since power is negative, the lens is concave (diverging). Power P = 1/f (in metres), so f = 1/P = 1/(−4) = −0.25 m = −25 cm. The negative focal length confirms it is a concave lens used for correcting myopia (short-sightedness).

Five-mark questions demand multi-step calculations, derivations, or detailed ray diagrams with explanations. These three are the gold standard in CBSE papers. **Q1: An object 4 cm tall is placed 25 cm in front of a concave mirror with radius of curvature 20 cm. (a) Find the position and size of the image. (b) Draw the ray diagram. (c) State the nature of the image.** A: (a) First, find focal length: f = R/2 = 20/2 = 10 cm. Using mirror formula: 1/f = 1/v + 1/u → 1/10 = 1/v + 1/25 → 1/v = 1/10 − 1/25 = (5−2)/50 = 3/50 → v = 50/3 ≈ 16.67 cm. Magnification m = −v/u = −(50/3)/25 = −2/3. Image height = m × object height = (−2/3) × 4 = −2.67 cm (negative indicates inverted). (b) Ray diagram: Draw mirror with pole, focus at 10 cm, centre at 20 cm. Object at 25 cm. Two rays: one through centre of curvature (reflects back), one parallel to axis (reflects through focus). They meet at v = 16.67 cm. (c) Image is real, inverted, diminished (2.67 cm vs 4 cm), formed between focus and centre of curvature. **Q2: A ray of light passes from a medium of refractive index n₁ = 1.5 to another medium of refractive index n₂ = 1.0 (air). If the angle of incidence is 30°, find: (a) the angle of refraction, (b) the critical angle for this pair of media, and (c) explain total internal reflection.** A: (a) Snell's law: n₁ sin i = n₂ sin r → 1.5 × sin 30° = 1 × sin r → 1.5 × 0.5 = sin r → sin r = 0.75 → r ≈ 48.6°. The ray bends away from the normal (away from denser medium). (b) Critical angle θc satisfies: n₁ sin θc = n₂ × 1 → 1.5 × sin θc = 1 → sin θc = 2/3 ≈ 0.667 → θc ≈ 41.8°. (c) Total internal reflection occurs when the angle of incidence exceeds the critical angle. All light reflects back into the denser medium; no refraction occurs. This is the principle behind optical fibres and diamond's sparkle. **Q3: A convex lens of focal length 20 cm forms a virtual, erect image at a distance of 10 cm from the lens. (a) Find the object distance. (b) Find the magnification. (c) If the object is 3 cm tall, calculate the image height. (d) Explain why the image is virtual.** A: (a) For a virtual image, v = −10 cm (negative by sign convention). Using lens formula: 1/f = 1/v + 1/u → 1/20 = 1/(−10) + 1/u → 1/u = 1/20 + 1/10 = (1+2)/20 = 3/20 → u = 20/3 ≈ 6.67 cm. Object is placed 6.67 cm from the lens. (b) Magnification m = −v/u = −(−10)/(20/3) = 10 × 3/20 = +1.5. Positive magnification indicates erect image. (c) Image height = m × object height = 1.5 × 3 = 4.5 cm (erect and enlarged). (d) The image is virtual because the object is placed between the lens and its focus. Refracted rays diverge and appear to come from behind the lens. No actual light rays converge on a screen.

The rationalized CBSE 2024–25 syllabus maintains the core mirror and lens content, but assessment patterns show subtle evolution. First, the board has reduced emphasis on purely theoretical definitions (like 'state Snell's Law') in favour of contextual application questions (e.g., 'why does a swimming pool appear shallower than it actually is?'). This shift rewards conceptual clarity over memorization. Second, multi-step problems combining mirrors and lenses are appearing more frequently—for instance, a 5-mark question asking students to find the magnification when a concave mirror is used as a shaving mirror and comparing it to a magnifying lens. Third, case-study style questions linking optics to real-world devices (periscopes, telescopes, cameras) are becoming standard. Students should practice explaining 'how it works' in 60–80 words. Fourth, numerical accuracy is stricter: answers like v = 16.67 cm are expected, not approximations like v ≈ 17 cm, especially in term exams. Finally, ray diagrams are now routinely marked for labelling (principal axis, optical centre, focus, centre of curvature), not just shape correctness. Practising these 13 PYQs directly aligns you with 2025 and 2026 board patterns; the conceptual depth required here will transfer seamlessly to upcoming variations.

Manage time and accuracy with this exam-room strategy. **Start with 1-mark questions (3 minutes).** Read quickly; if it's a definition, write in one sentence. If it's 'find the focal length given radius,' use R = 2f and move on. Don't overthink. **Next, attempt 3-mark questions (12 minutes for three questions).** Identify which formula applies (mirror, lens, or Snell's law). Write it clearly. Substitute values step-by-step and box your final answer. If a ray diagram is asked, spend 90 seconds drawing and labelling (axis, focus, centre, object, image). **Reserve 15 minutes for 5-mark questions.** Read the full question first to spot what's being asked: position? magnification? both? Build a solution roadmap in 30 seconds, then execute. Show each algebraic step—examiners award partial marks for correct method even if your final number is slightly off (e.g., calculation error). **Prioritize accuracy over speed.** A correct 3-mark answer is worth more than a rushed, half-wrong 5-mark response. **In the last 2 minutes, proofread** your mirror formulae (check sign conventions for u, v, f) and Snell's law (is the refracted ray bent correctly?). **Reflex tip:** If stuck on a numerical problem, state the formula and attempt at least one substitution—you'll earn method marks. Avoid leaving blanks in optics; a partial calculation is always safer than silence. This systematic approach, reinforced by solving these 13 PYQs multiple times, typically raises optics scores by 8–15% on the final exam.

This resource is designed for active, spaced learning, not passive reading. **Day 1–2:** Solve the five 1-mark questions without looking at answers. Time yourself to 1 minute per question. Check answers; if you miss more than one, review the corresponding NCERT definitions. **Day 3–4:** Work through 3-mark questions. For each, write the solution on paper (don't just read). Compare your working to the model answer. Identify where you took shortcuts or misapplied a formula. **Day 5–6:** Tackle 5-mark questions in exam conditions—25 minutes total, no reference materials. Mark yourself strictly using the given solutions. Note any steps you skipped. **Week 2:** Revisit questions you scored < 3/5 on. Redo them from scratch, then compare. Repeat this cycle until you score consistently 90%+. **Spaced repetition:** Return to this guide 1 week before mid-term and 3 weeks before finals. On each visit, you'll notice you're faster and more confident—that's how learning compounds. If you find yourself struggling with conceptual leaps (e.g., why magnification is negative for real images, or why convex mirrors never form real images), live tutoring with cbsetutor.ai can accelerate clarity through interactive problem-solving sessions where doubts are resolved in real-time.