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Light – Reflection and Refraction – NCERT Exercises

🎓 Class 10 Science CBSE Theory Ch 9 — Light – Reflection and Refraction ⏱ ~21 min
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This MCQ module is based on: Light – Reflection and Refraction – NCERT Exercises

[myaischool_lt_science_assessment grade_level="class_10" science_domain="physics" difficulty="intermediate"]

Chapter 9 — Summary of Light: Reflection and Refraction

Key Ideas at a Glance

  • Laws of reflection: \(\angle i = \angle r\); incident ray, reflected ray and normal all lie in one plane. Valid for plane and spherical mirrors.
  • Plane mirror image — virtual, erect, same size, laterally inverted, as far behind as the object is in front.
  • Spherical mirrors: concave (converging) and convex (diverging). Focal length \(f = R/2\).
  • Six positions of object before a concave mirror ⇒ six image types (infinity, beyond C, at C, between C and F, at F, between P and F). Convex mirror always gives virtual, erect, diminished image.
  • Sign convention — origin at pole/optical centre, incoming light positive; object distance \(u\) is negative; real image distance for mirror is negative, for lens positive.
  • Mirror formula: \( \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}\). Magnification \(m = -v/u = h'/h\).
  • Refraction — bending of light on changing medium. Snell's law: \(n_1\sin i = n_2\sin r\).
  • Refractive index \(n = c/v\); the higher the value, the denser (optically).
  • Glass slab: emergent ray is parallel to incident ray with lateral shift.
  • Spherical lens: convex (converging, \(P > 0\)) and concave (diverging, \(P < 0\)).
  • Lens formula: \(\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}\); magnification \(m = v/u\).
  • Power: \(P = 1/f(\text{m})\); unit dioptre (D). Combination: \(P = P_1 + P_2\).

Key Terms

ReflectionBouncing back of light from a surface.
RefractionBending of light as it crosses media.
NormalPerpendicular to a surface at the point of incidence.
Plane mirrorFlat reflecting surface.
Concave mirrorReflecting surface is the inside of a sphere; converging.
Convex mirrorReflecting surface is the outside; diverging.
Pole (P)Centre of the mirror's surface.
Centre of curvature (C)Centre of the sphere of which the mirror is part.
Radius of curvature (R)Distance PC.
Principal focus (F)Where parallel axial rays meet (or appear to meet) after reflection/refraction.
Focal length (f)Distance from P (or O) to F; \(f = R/2\) for mirror.
ApertureEffective diameter of mirror/lens.
Real imageFormed by actual meeting of rays; can be caught on screen.
Virtual imageFormed by apparent meeting of rays; cannot be caught on screen.
Magnification (m)Ratio of image height to object height.
Refractive index (n)\(c/v\); measure of optical density.
Snell's law\(n_1 \sin i = n_2 \sin r\).
Lateral shiftSideways displacement of emergent ray through a slab.
Convex lensConverging; thicker in middle.
Concave lensDiverging; thinner in middle.
Optical centre (O)Geometric centre of a lens.
Power (P)Reciprocal of focal length in metres.
Dioptre (D)SI unit of power; 1 D = 1 m⁻¹.
Lateral inversionLeft–right reversal by a plane mirror.

NCERT Exercises — Full Solutions

Q1.
Which one of the following materials cannot be used to make a lens?
  • (a) Water
  • (b) Glass
  • (c) Plastic
  • (d) Clay
Answer: (d) Clay. Lenses must be transparent so that light can refract through them. Water, glass and plastic are all transparent; clay is opaque and therefore cannot be used to make a lens.
Q2.
The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should the position of the object be?
  • (a) Between the principal focus and centre of curvature
  • (b) At the centre of curvature
  • (c) Beyond the centre of curvature
  • (d) Between the pole of the mirror and its principal focus
Answer: (d) Between the pole and the principal focus. Only in this position does a concave mirror produce a virtual, erect and enlarged image (this is the shaving-mirror case).
Q3.
Where should an object be placed in front of a convex lens to get a real image of the size of the object?
  • (a) At the principal focus of the lens
  • (b) At twice the focal length
  • (c) At infinity
  • (d) Between the optical centre of the lens and its principal focus
Answer: (b) At twice the focal length (i.e. at 2F1). Then the image forms at 2F2 on the opposite side, same size as the object, real and inverted.
Q4.
A spherical mirror and a thin spherical lens have each a focal length of –15 cm. The mirror and the lens are likely to be:
  • (a) both concave
  • (b) both convex
  • (c) the mirror is concave and the lens is convex
  • (d) the mirror is convex, but the lens is concave
Answer: (a) Both concave. By the New Cartesian sign convention, the focal length of a concave mirror and a concave lens are both negative.
Q5.
No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be:
  • (a) plane
  • (b) concave
  • (c) convex
  • (d) either plane or convex
Answer: (d) Either plane or convex. A plane mirror always gives an erect image (same size). A convex mirror also always gives an erect (virtual, diminished) image. A concave mirror would give an inverted (real) image for most positions.
Q6.
Which of the following lenses would you prefer to use while reading small letters found in a dictionary?
  • (a) A convex lens of focal length 50 cm
  • (b) A concave lens of focal length 50 cm
  • (c) A convex lens of focal length 5 cm
  • (d) A concave lens of focal length 5 cm
Answer: (c) A convex lens of focal length 5 cm. A magnifying glass is a convex lens used with the object inside its focal length. Shorter focal length (5 cm) gives more magnification than 50 cm. A concave lens always diminishes the image and cannot be used as a magnifier.
Q7.
We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.

To get an erect (virtual) image from a concave mirror, the object must lie between the pole and the principal focus. With \(f = 15\) cm, the object distance \(u\) must be between 0 and 15 cm (exclusive).

Nature: virtual, erect and enlarged (larger than the object).

Ray diagram: A parallel-to-axis ray reflects through F; a ray heading toward C is reflected back; the two reflected rays diverge. Their backward extensions behind the mirror meet at the enlarged, erect image — exactly Fig 9.5d in Part 1.

Q8.
Name the type of mirror used in the following situations: (a) Headlights of a car. (b) Side/rear-view mirror of a vehicle. (c) Solar furnace. Support your answer with reason.
  • (a) Concave mirror — the bulb is placed at the focus so that after reflection the rays emerge parallel, producing a powerful directed beam.
  • (b) Convex mirror — always gives an erect, diminished image with a wide field of view, so the driver sees more of the road behind.
  • (c) Concave mirror (large) — parallel rays from the Sun converge at the principal focus, concentrating heat energy to very high temperatures.
Q9.
One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.

Yes, a complete image is still formed. Every point of the object sends light in all directions; some of these rays fall on the uncovered half of the lens and, after refraction, converge at the corresponding image point. Thus each point of the object is still imaged, so the full image is produced.

However, the image becomes dimmer — because only half of the light reaches it — since the other half is blocked by the black paper. Experimentally the image on the screen is complete but less bright.

Q10.
An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.

Given: \(h = 5\) cm, \(u = -25\) cm, \(f = +10\) cm.

\[ \dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} = \dfrac{1}{10} + \dfrac{1}{-25} = \dfrac{5-2}{50} = \dfrac{3}{50} \]

\(v = +50/3 \approx +16.67\) cm (on the opposite side, real image).

\(m = v/u = (50/3)/(-25) = -2/3\).

\(h' = m h = (-2/3)(5) = -10/3 \approx -3.33\) cm (inverted).

Nature: real, inverted, diminished (about 3.33 cm tall), at 16.67 cm on the other side. (Object is beyond 2F1 since \(2f = 20\) cm < 25 cm.)

Q11.
A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.

A concave lens forms a virtual image on the same side as the object. Given: \(f = -15\) cm, \(v = -10\) cm.

\[ \dfrac{1}{u} = \dfrac{1}{v} - \dfrac{1}{f} = \dfrac{1}{-10} - \dfrac{1}{-15} = -\dfrac{1}{10} + \dfrac{1}{15} = \dfrac{-3+2}{30} = -\dfrac{1}{30} \]

\(u = -30\) cm. The object is placed 30 cm in front of the lens.

Q12.
An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.

For a convex mirror \(f = +15\) cm; \(u = -10\) cm.

\[ \dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = \dfrac{1}{15} - \dfrac{1}{-10} = \dfrac{1}{15} + \dfrac{1}{10} = \dfrac{2+3}{30} = \dfrac{5}{30} = \dfrac{1}{6} \]

\(v = +6\) cm (behind the mirror).

\(m = -v/u = -6/(-10) = +0.6\).

Nature: virtual, erect, diminished (0.6 × object), 6 cm behind the mirror.

Q13.
The magnification produced by a plane mirror is +1. What does this mean?

Magnification \(m = h'/h = +1\) means:

  • The image has the same size as the object (|m| = 1).
  • The positive sign shows the image is erect (not inverted) and virtual.

So a plane mirror gives a virtual, erect image exactly the same size as the object.

Q14.
An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.

\(f = R/2 = +15\) cm (convex); \(u = -20\) cm; \(h = 5\) cm.

\[ \dfrac{1}{v} = \dfrac{1}{15} - \dfrac{1}{-20} = \dfrac{1}{15} + \dfrac{1}{20} = \dfrac{4+3}{60} = \dfrac{7}{60} \]

\(v = 60/7 \approx +8.57\) cm (virtual, behind the mirror).

\(m = -v/u = -(8.57)/(-20) = +0.43\).

\(h' = m h = 0.43 \times 5 \approx +2.15\) cm.

Nature: virtual, erect, diminished; about 2.15 cm tall, 8.57 cm behind the mirror.

Q15.
An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed so that a sharp focused image can be obtained? Find the size and nature of the image.

\(f = -18\) cm; \(u = -27\) cm; \(h = 7\) cm.

\[ \dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = \dfrac{1}{-18} - \dfrac{1}{-27} = -\dfrac{1}{18} + \dfrac{1}{27} = \dfrac{-3+2}{54} = -\dfrac{1}{54} \]

\(v = -54\) cm. The screen should be placed 54 cm in front of the mirror.

\(m = -v/u = -(-54)/(-27) = -2\). \(h' = mh = -2 \times 7 = -14\) cm.

Nature: real, inverted, enlarged; image size 14 cm.

Q16.
Find the focal length of a lens of power −2.0 D. What type of lens is this?

\(P = 1/f\) ⇒ \(f = 1/P = 1/(-2.0) = -0.5\) m = −50 cm.

Negative focal length ⇒ concave (diverging) lens.

Q17.
A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?

\(f = 1/P = 1/1.5 = 0.667\) m = +66.7 cm ≈ 66.7 cm.

Positive focal length ⇒ convex (converging) lens. Such a lens is prescribed to correct hypermetropia (long-sightedness).

Frequently Asked Questions — NCERT Exercises & Intext Questions

How do I solve NCERT Class 10 Science Chapter 9 (Light - Reflection and Refraction) exercise questions for the CBSE board exam?

Solve NCERT Chapter 9 — Light - Reflection and Refraction — exercise questions by first reading the question carefully, writing down the given data, recalling the relevant concepts like reflection, refraction, mirror formula, and applying them step by step. This Part 4 covers every intext and end-of-chapter exercise from the NCERT textbook. Write balanced equations, label diagrams clearly and show each step — CBSE Class 10 board examiners award step marks even if the final answer has a small slip. Practising these solutions strengthens conceptual clarity and builds speed for the board exam.

Are the NCERT intext questions from Light - Reflection and Refraction important for the Class 10 board exam?

Yes, NCERT intext questions for Chapter 9 Light - Reflection and Refraction are highly important for the CBSE Class 10 Science board exam. Many board questions are directly lifted or only slightly modified from these intext questions, and they test the foundational concepts — reflection, refraction, mirror formula — that chapter-end questions build on. Attempt every intext question first, then move on to the exercises. This practice ensures complete NCERT coverage, which is the CBSE exam's primary source.

What types of questions from Light - Reflection and Refraction are asked in the CBSE Class 10 Science board exam?

The CBSE Class 10 board paper asks a mix of question types from Light - Reflection and Refraction: 1-mark MCQ and assertion-reason, 2-mark short answers, 3-mark explanations, 5-mark long answers with diagrams or derivations, and 4-mark competency-based / case-study questions. These test understanding of reflection, refraction, mirror formula, lens formula. Practising every NCERT exercise and intext question prepares you to answer all of these formats with confidence.

How many marks does Chapter 9 — Light - Reflection and Refraction — carry in the Class 10 Science CBSE paper?

Chapter 9 — Light - Reflection and Refraction — is part of the Class 10 Science syllabus and typically contributes 5–9 marks in the CBSE board paper, depending on the annual weightage. Questions are drawn from definitions, reasoning, numerical/descriptive problems and diagrams on topics like reflection, refraction, mirror formula. Solving the NCERT exercises in this part is essential because CBSE directly references NCERT for question design.

Where can I find step-by-step NCERT solutions for Chapter 9 Light - Reflection and Refraction Class 10 Science?

You can find complete, step-by-step NCERT solutions for Chapter 9 Light - Reflection and Refraction Class 10 Science on MyAiSchool. Every intext and end-of-chapter exercise question is solved with full working, labelled diagrams and CBSE-aligned mark distribution. Solutions highlight key points about reflection, refraction, mirror formula that examiners look for. This makes revision quick and exam-focused for Class 10 CBSE board students.

What is the best way to revise Light - Reflection and Refraction before the Class 10 Science board exam?

The best way to revise Light - Reflection and Refraction for the CBSE Class 10 Science board exam is a three-pass approach. First pass: skim the chapter and note down key terms like reflection, refraction, mirror formula in a one-page mind map. Second pass: solve every NCERT intext and exercise question without looking at the solution, then self-check. Third pass: attempt previous CBSE board questions and competency-based questions under timed conditions. This structured revision secures full marks for this chapter.

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