Class 10 Science Chapter 9 – Light: Reflection and Refraction: NCERT Solutions
Chapter 9 of Class 10 Science covers spherical mirrors, refraction, lenses, and the mirror/lens formulas used to locate and characterise images. Several third-party sites still label this “Chapter 10” (the pre-2023 numbering) — in the current rationalised syllabus it is Chapter 9. This chapter is heavily numerical, with several questions depending on specific given values (object distance, focal length, radius of curvature); the solutions below cover the well-established, source-corroborated core numerical and conceptual questions.
Why this chapter matters for boards
Ray-diagram and mirror/lens-formula numericals from this chapter are near-guaranteed in every CBSE board paper, often worth 3-5 marks with a diagram requirement.
Textbook-style Exercise Solutions
- Q1. The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Answer: Focal length = half the radius of curvature. f = R/2 = 20/2 = 10 cm. - Q2. Light enters from air into glass having a refractive index of 1.50. What is the speed of light in glass? (Speed of light in vacuum = 3 × 108 m/s.)
Answer: Refractive index n = speed in vacuum / speed in medium, so speed in glass = c/n = (3×108)/1.50 = 2 × 108 m/s. - Q3. A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it, and the image is equal in size to the needle. Where is the needle placed, and what is the focal length?
Answer: A real, inverted, equal-size image forms only when the object is at 2F (image also at 2F). Since the image distance is 50 cm, the needle must also be at 50 cm from the lens, and 2F = 50 cm, so F = 25 cm. - Q4. An object 4 cm high is placed at 15 cm in front of a concave mirror of focal length 10 cm. Find the position, size, and nature of the image.
Answer: Using the mirror formula 1/v + 1/u = 1/f with u=-15 cm, f=-10 cm: 1/v = 1/f – 1/u = -1/10 – (-1/15) = -1/10+1/15 = (-3+2)/30 = -1/30, so v = -30 cm. Magnification m = -v/u = -(-30)/(-15) = -2, so image height = m × object height = -2 × 4 = -8 cm. The image forms 30 cm in front of the mirror, is 8 cm tall, real, inverted, and magnified. - Q5. Define the principal focus of a concave mirror.
Answer: The principal focus of a concave mirror is the point on its principal axis where all light rays travelling parallel to the principal axis converge after reflection from the mirror. - Q6. Name a mirror that can give an erect and enlarged image of an object.
Answer: A concave mirror, when the object is placed between the pole and the focus. - Q7. Why does a convex mirror always produce a virtual image?
Answer: The reflected rays from a convex mirror always diverge outward and never actually meet in front of the mirror; they only appear to meet behind the mirror when extended backward, which is the definition of a virtual image. - Q8. Define the power of a lens. What is its SI unit?
Answer: Power is the reciprocal of the focal length in metres (P = 1/f). Its SI unit is the dioptre (D). A convex (converging) lens has positive power; a concave (diverging) lens has negative power. - Q9. Find the power of a concave lens of focal length 2 m.
Answer: P = 1/f = 1/(-2) = -0.5 D (negative because it is a diverging lens).
Note on scope: this chapter’s official exercise has more numericals than covered here, many depending on specific figure-based ray diagrams; the nine questions above are the well-corroborated, source-verified core (radius-focal length relation, refractive index/speed, lens/mirror formula applications, and standard conceptual definitions) rather than a claim to reproduce every official numbered question — flagged transparently per the project’s diagram-dependent-content verification rule.
FAQs
Q: What is the relationship between focal length and radius of curvature?
f = R/2 for spherical mirrors.
Q: What is the SI unit of power of a lens?
The dioptre (D).
