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Light Reflection and Refraction Exercise Solutions | KalviGavani

NCERT solutions for CBSE Class 10 Science Ch 10 — mirror & lens formulae, Snell's law, ray diagrams, all intext and exercise Qs solved step by step.

Updated 15 June 202625 min read

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These are complete NCERT Solutions for CBSE Class 10 Science Chapter 10 — Light: Reflection and Refraction. For every numerical problem, the solution follows this format:

Formula→ Substitution → Calculation → Answer with units

Attempt each problem yourself before reading the solution. In your board exam, every step of a numerical carries marks — always write the formula first, then substitute values with correct signs, then calculate, and finally state the answer with units and sign. Ray diagrams are described in text so you can reproduce them on paper.

Key Concepts (Quick Recap)

Reflection of Light

  • Law of Reflection: (i) Angle of incidence = Angle of reflection (angle i = angle r). (ii) The incident ray, reflected ray, and normal all lie in the same plane.
  • Concave mirror: Converging mirror — parallel rays converge at the focus after reflection. Reflected surface is on the inner (concave) side of the sphere.
  • Convex mirror: Diverging mirror — parallel rays appear to diverge from the focus after reflection. Reflected surface is on the outer (convex) side.
  • Principal focus (F): Point where parallel rays converge (concave) or appear to diverge from (convex) after reflection.
  • Focal length (f): Distance from pole (P) to principal focus. f = R/2, where R = radius of curvature.
  • Pole (P): Centre of the reflecting surface of the mirror.
  • Centre of curvature (C): Centre of the sphere of which the mirror is a part. Distance PC = R.

Image Formation by Concave Mirror (for different object positions)

Object PositionImage PositionNature of ImageSize
At infinityAt FReal, invertedHighly diminished (point)
Beyond CBetween F and CReal, invertedDiminished
At CAt CReal, invertedSame size
Between C and FBeyond CReal, invertedMagnified
At FAt infinityReal, invertedHighly magnified
Between P and FBehind mirrorVirtual, erectMagnified

Image Formation by Convex Lens (for different object positions)

Object PositionImage PositionNature of ImageSize
At infinityAt F (other side)Real, invertedHighly diminished (point)
Beyond 2FBetween F and 2FReal, invertedDiminished
At 2FAt 2FReal, invertedSame size
Between F and 2FBeyond 2FReal, invertedMagnified
At FAt infinityReal, invertedHighly magnified
Between F and optical centreOn same side as objectVirtual, erectMagnified

Mirror Formula and Sign Convention

  • New Cartesian Sign Convention: All distances measured from the pole (P) of the mirror. Distances measured in the direction of incident light are positive; opposite to incident light are negative. Heights measured upward from principal axis are positive; downward are negative.
  • Mirror formula: 1/v + 1/u = 1/f
  • Magnification (mirror): m = -v/u = height of image / height of object = h'/h

Refraction of Light

  • Refraction: Bending of light at the interface of two optically different media due to change in speed of light.
  • Law of Refraction (Snell's Law): n1 sin i = n2 sin r, or sin i / sin r = n21 (refractive index of medium 2 w.r.t. medium 1).
  • Refractive index: n = c/v = speed of light in vacuum / speed of light in medium. Also n = sin i / sin r.
  • Light bends towards the normal when going from rarer to denser medium (speed decreases).
  • Light bends away from the normal when going from denser to rarer medium (speed increases).

Lens Formula and Power

  • Lens formula: 1/v - 1/u = 1/f
  • Magnification (lens): m = v/u = h'/h
  • Power of a lens: P = 1/f (f in metres); SI unit = dioptre (D). Convex lens: P is positive. Concave lens: P is negative.
  • Convex lens (converging): f is positive; thicker at the centre; used in cameras, magnifying glasses, projectors, to correct hypermetropia.
  • Concave lens (diverging): f is negative; thinner at the centre; used to correct myopia.

Intext Questions

Page 168

Q1. Define the principal focus of a concave mirror.
Answer

The principal focus of a concave mirror is the point on the principal axis at which rays of light, travelling parallel to the principal axis, converge (meet) after reflection from the mirror. It is a real point located in front of the mirror at a distance equal to the focal length from the pole.

Q2. The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Formulaf = R/2
Substitutionf = 20/2
Calculationf = 10
Answerf = 10 cm
Q3. Name the mirror that can give an erect and enlarged image of an object.
Answer

A concave mirror gives an erect and enlarged image when the object is placed between the pole (P) and the principal focus (F). The image formed is virtual, erect, and magnified.

Q4. Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Answer

A convex mirror always forms an erect, diminished, and virtual image regardless of the object distance. It provides a wider field of view compared to a plane mirror — the driver can see a much larger area of the traffic behind the vehicle. Since the image is always within the focal length behind the mirror, it is compact and convenient.

Page 174

Q1. Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Formulaf = R/2
Substitutionf = 32/2
Calculationf = 16
Answerf = +16 cm (positive for convex mirror)
Q2. A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located?
Answer

The image is real, so it is inverted. Magnification of a real inverted image is negative.

Formulam = -v/u
Substitutionm = -3, u = -10 cm → -3 = -v/(-10)
Calculation-3 = v/10 → v = -30
Answerv = -30 cm — the image is located 30 cm in front of the mirror, on the same side as the object. It is real, inverted, and three times magnified.

Ray diagram: Draw a concave mirror with P, F, C marked on the principal axis. Place the object at 10 cm from P (between P and F). Draw a ray parallel to the principal axis — it reflects through F. Draw a ray through C — it reflects back along the same path. The reflected rays actually converge at 30 cm in front of the mirror, forming a real, inverted, magnified image.

Page 176

Q1. A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
Answer

The light ray bends towards the normal. Water is optically denser than air (refractive index of water = 1.33, air = 1.00). When light travels from a rarer medium (air) to a denser medium (water), its speed decreases and it bends towards the normal at the point of incidence.

Q2. Light enters from air to glass which has refractive index 1.50. What is the speed of light in glass?
Formulan = c/v → v = c/n
Substitutionv = (3 x 10^8) / 1.50
Calculationv = 3 x 10^8 / 1.50 = 2 x 10^8
AnswerSpeed of light in glass = 2 x 10^8 m/s
Q3. Define (a) refractive index, (b) absolute refractive index.
Answer

(a) Refractive index of medium 2 with respect to medium 1 (n21) = speed of light in medium 1 / speed of light in medium 2 = sin i / sin r. It measures how much light bends when passing from one medium to another.

(b) Absolute refractive index of a medium = speed of light in vacuum / speed of light in that medium = c/v. It is the refractive index with respect to vacuum (or air, approximately).

Page 184

Q1. Find the power of a concave lens of focal length 2 m.
FormulaP = 1/f (f in metres)
Substitutionf = -2 m (concave lens, f is negative) → P = 1/(-2)
CalculationP = -0.5
AnswerP = -0.5 D (negative power because it is a diverging lens)

Chapter-End Exercise

Q1. Which one of the following materials cannot be used to make a lens?
Answer

(a) Water (b) Glass (c) Plastic (d) Clay

Answer(d) Clay. A lens must be made of a transparent material that allows light to pass through it. Clay is opaque — light cannot pass through it. Water, glass, and plastic are all transparent and can be shaped into lenses.
Q2. The image formed by a concave mirror is observed to be virtual, erect, and larger than the object. Where should be the position of the object?
Answer

(a) Between the principal focus and the 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 of the mirror and its principal focus. When the object is placed between P and F of a concave mirror, the reflected rays diverge and appear to come from behind the mirror, forming a virtual, erect, and magnified image.

Ray diagram: Draw a concave mirror. Mark P, F, C on principal axis. Place object between P and F (close to the mirror). Draw a ray parallel to the principal axis — after reflection it passes through F. Draw a ray directed towards C — after reflection it retraces its path. Extend both reflected rays backward (behind the mirror) with dotted lines. They meet behind the mirror, forming a virtual, erect, magnified image.

Q3. Where should an object be placed in front of a convex lens to get a real and inverted image of equal size?
Answer

The object should be placed at 2F (the centre of curvature) — at a distance of 2f from the optical centre of the convex lens. At this position, the image is also formed at 2F on the other side, and it is real, inverted, and exactly the same size as the object.

Q4. A spherical mirror and a thin spherical lens each have a focal length of -15 cm. What type of mirror and lens are these?
Answer
  • Mirror with f = -15 cm: In the New Cartesian sign convention, a negative focal length means the focus is in front of the mirror (on the same side as the object). This is a concave mirror.
  • Lens with f = -15 cm: A negative focal length for a lens means it is a diverging lens. This is a concave lens.
Q5. No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be:
Answer

(a) Plane (b) Concave (c) Convex (d) Either plane or convex

Answer(d) Either plane or convex. A plane mirror always forms a virtual, erect image of the same size. A convex mirror always forms a virtual, erect, and diminished image regardless of object distance. A concave mirror can form an inverted (real) image for most positions, so it does not satisfy the condition.
Q6. Which of the following lenses would you prefer to use while reading small letters found in a dictionary?
Answer

(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. To read small letters, we need a magnifying glass, which is a convex lens. The magnification is greater when the focal length is shorter (power P = 1/f is higher). A convex lens of f = 5 cm has power = 1/0.05 = +20 D, giving much greater magnification than one of f = 50 cm (P = +2 D). A concave lens always produces diminished images and is not suitable.
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.
Answer
  • Range of object distance: 0 to 15 cm (i.e., between the pole P and the principal focus F). The object must be placed at a distance less than 15 cm from the mirror.
  • Nature of image: Virtual and erect (since it is formed behind the mirror by the apparent intersection of reflected rays extended backward).
  • Size of image: Larger (magnified) than the object.

Ray diagram: Draw a concave mirror. Mark P, F (at 15 cm from P), and C on the principal axis. Place the object between P and F (say at 10 cm from P). Draw a ray from the top of the object parallel to the principal axis — after reflection, it passes through F. Draw a ray from the top of the object towards the centre of curvature C — it reflects back along its own path. Extend both reflected rays backward (behind the mirror) using dotted lines. The dotted lines meet behind the mirror, forming a virtual, erect, and magnified image.

Q8. Name the type of mirror used in the following situations:
Answer

(a) Headlights of a car: Concave mirror. The bulb (light source) is placed at the focus of the concave mirror. Light rays from the bulb, after reflection, emerge as a parallel beam that travels a long distance, illuminating the road ahead.

(b) Side/rear-view mirror of a vehicle: Convex mirror. It always forms a virtual, erect, and diminished image, providing a wider field of view so the driver can see more of the traffic behind.

(c) Solar furnace: Concave mirror. A large concave mirror concentrates (converges) sunlight at its focus, producing intense heat used for cooking or industrial processes.

Q9. One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object?
Answer

Yes, a complete image will still be formed. Each point on the object emits light rays in all directions, and the uncovered portion of the lens still refracts rays from every part of the object. The full image is formed, but it is less bright (dimmer) because fewer light rays pass through the lens. The position and size of the image remain the same; only the brightness decreases.

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 nature of the image formed.
GivenObject height h = +5 cm, Object distance u = -25 cm (negative by sign convention), Focal length f = +10 cm (convex/converging lens).

Finding image position (v):

Formula1/v - 1/u = 1/f
Substitution1/v - 1/(-25) = 1/10 → 1/v + 1/25 = 1/10
Calculation1/v = 1/10 - 1/25 = (5 - 2)/50 = 3/50
Answerv = 50/3 = +16.67 cm

The positive value of v confirms the image is formed on the other side of the lens (real image).

Finding magnification and image size:

Formulam = v/u
Substitutionm = (50/3) / (-25) = 50 / (3 x 25)
Calculationm = 50/75 = -2/3
Answerm = -2/3 (negative sign means inverted image)

Image height h' = m x h = (-2/3) x 5 = -3.33 cm (negative means image is inverted).

Summary: The image is formed at 16.67 cm on the other side of the lens. It is real, inverted, and diminished (3.33 cm tall, smaller than the 5 cm object).

Ray diagram: Draw a convex lens with optical centre O. Mark F at 10 cm and 2F at 20 cm on both sides. Place the object (5 cm arrow upward) at 25 cm from O (beyond 2F on the left side). Draw a ray from the top of the object parallel to the principal axis — after refraction it passes through F on the right side. Draw a ray through the optical centre O — it goes straight through undeviated. The two refracted rays converge at a point between F and 2F on the right side. The image is an inverted arrow between F and 2F, smaller than the object.

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.
Givenf = -15 cm (concave lens, focal length is negative), v = -10 cm (virtual image formed on the same side as the object for a concave lens).

Finding object distance (u):

Formula1/v - 1/u = 1/f
Substitution1/(-10) - 1/u = 1/(-15) → -1/10 - 1/u = -1/15
Calculation-1/u = -1/15 + 1/10 = (-2 + 3)/30 = 1/30
Answeru = -30 cm

The object is placed 30 cm in front of the lens (on the same side as the incoming light).

Ray diagram: Draw a concave lens with optical centre O. Mark F at 15 cm on both sides (virtual focus on the same side as object for a concave lens). Place the object (an upward arrow) at 30 cm from O on the left side. Draw a ray parallel to the principal axis from the top of the object — after refraction through the concave lens, it diverges as if coming from F on the left side. Draw a ray towards the optical centre O — it passes straight through. Extend the diverging refracted ray backward (dotted line) to meet the second ray. They appear to meet at 10 cm from O on the left side, forming a virtual, erect, diminished image.

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.
Givenu = -10 cm, f = +15 cm (convex mirror, focal length is positive).

Finding image position (v):

Formula1/v + 1/u = 1/f
Substitution1/v + 1/(-10) = 1/15 → 1/v - 1/10 = 1/15
Calculation1/v = 1/15 + 1/10 = (2 + 3)/30 = 5/30 = 1/6
Answerv = +6 cm

The positive value of v means the image is formed 6 cm behind the mirror (virtual image).

Finding magnification:

Formulam = -v/u
Substitutionm = -6/(-10) = 6/10
Calculationm = +0.6
Answerm = +0.6 (positive means erect, 0.6 means diminished)

Summary: The image is formed 6 cm behind the convex mirror. It is virtual, erect, and diminished (smaller than the object, 0.6 times the object size).

Ray diagram: Draw a convex mirror. Mark P at the pole, F at 15 cm behind the mirror, and C at 30 cm behind the mirror on the principal axis. Place the object (an upward arrow) at 10 cm in front of the mirror. Draw a ray parallel to the principal axis — after reflection it appears to diverge from F behind the mirror. Draw a ray directed towards C behind the mirror — it reflects back along its own path. Extend both reflected rays backward behind the mirror (dotted lines). They meet at 6 cm behind P, forming a virtual, erect, diminished image.

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

Magnification m = +1 has two implications:

  • Positive sign (+): The image is erect — it has the same orientation as the object (not inverted).
  • Magnitude = 1: The image is the same size as the object — the height of the image equals the height of the object (h' = h).
Together, this means a plane mirror produces an image that is erect and of exactly the same size as the object. This also tells us the image distance equals the object distance (v=u) — the image is as far behind the mirror as the object is in front.
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.
Givenh = +5 cm, u = -20 cm, R = 30 cm.

Finding focal length:

Formulaf = R/2
Substitutionf = 30/2
Answerf = +15 cm (positive for convex mirror)

Finding image position (v):

Formula1/v + 1/u = 1/f
Substitution1/v + 1/(-20) = 1/15 → 1/v - 1/20 = 1/15
Calculation1/v = 1/15 + 1/20 = (4 + 3)/60 = 7/60
Answerv = 60/7 = +8.57 cm

The positive value of v means the image is formed 8.57 cm behind the mirror (virtual image).

Finding magnification and image size:

Formulam = -v/u
Substitutionm = -(60/7)/(-20) = 60/(7 x 20)
Calculationm = 60/140 = 3/7 = +0.43
Answerm = +3/7 (positive means erect, 3/7 means diminished)

Image height h' = m x h = (3/7) x 5 = 15/7 = 2.14 cm

Summary: The image is formed 8.57 cm behind the convex mirror. It is virtual, erect, and diminished with a height of 2.14 cm.

Ray diagram: Draw a convex mirror. Mark P, F (at 15 cm behind mirror), C (at 30 cm behind mirror). Place the object (5 cm arrow) at 20 cm in front of the mirror. Draw a ray parallel to the principal axis — after reflection it appears to diverge from F behind the mirror. Draw a ray directed towards C — it reflects back along its path. Extend both reflected rays backward behind the mirror. They meet at 8.57 cm behind P, forming a virtual, erect, diminished image of height 2.14 cm.

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.
Givenh = +7 cm, u = -27 cm, f = -18 cm (concave mirror, focal length is negative).

Finding image position (v):

Formula1/v + 1/u = 1/f
Substitution1/v + 1/(-27) = 1/(-18) → 1/v - 1/27 = -1/18
Calculation1/v = -1/18 + 1/27 = (-3 + 2)/54 = -1/54
Answerv = -54 cm

The negative value of v means the image is formed 54 cm in front of the mirror (real image, on the same side as the object). The screen must be placed 54 cm from the mirror.

Finding magnification and image size:

Formulam = -v/u
Substitutionm = -(-54)/(-27)
Calculationm = 54/(-27) = -2
Answerm = -2 (negative means inverted, 2 means magnified)

Image height h' = m x h = (-2) x 7 = -14 cm (negative means inverted).

Summary: The screen should be placed 54 cm in front of the mirror. The image is real, inverted, and magnified with a size of 14 cm (twice the object size).

Ray diagram: Draw a concave mirror. Mark P, F at 18 cm, and C at 36 cm on the principal axis. Place the object (7 cm arrow upward) at 27 cm from P (between F and C). Draw a ray parallel to the principal axis — it reflects through F. Draw a ray through C — it reflects back along its own path. The reflected rays converge at 54 cm in front of the mirror, forming a real, inverted, magnified image of 14 cm height. Since the object is between F and C, the image forms beyond C on the same side.

Common Mistakes to Avoid

1. Wrong sign for object distance (u): The most common error is forgetting that u is always negative in the New Cartesian convention. The object is always placed in front of the mirror or on the side from which light travels, so u < 0. Always write u = -value.

2. Mixing up mirror formula and lens formula: The mirror formula is 1/v + 1/u = 1/f (plus sign between 1/v and 1/u). The lens formula is 1/v - 1/u = 1/f (minus sign between 1/v and 1/u). Writing the wrong formula is the number one cause of wrong answers.

3. Confusing magnification formulas: For mirrors, m = -v/u (negative sign is essential). For lenses, m = v/u (no negative sign). Students often forget the negative sign in the mirror magnification formula, leading to wrong image nature.

4. Wrong sign for concave mirror focal length: By New Cartesian convention, a concave mirror has negative f and negative R (the focus and centre of curvature are in front of the mirror, opposite to the direction of reflected light). Many students incorrectly write f as positive for a concave mirror.

5. Not converting cm to m for power calculation: Power P = 1/f requires f in metres. If f = 20 cm, you must convert to f = 0.20 m before calculating P. A common mistake is to use f = 20 directly, giving P = 1/20 = 0.05 D instead of the correct P = 1/0.20 = 5 D.

6. Confusing real and virtual image properties: A real image is always inverted and can be projected on a screen; a virtual image is always erect and cannot be projected. Students sometimes state a real image as erect or a virtual image as inverted, which is incorrect for single mirror/lens systems.

7. Drawing ray diagrams with incorrect ray paths: The two standard rays for mirrors are: (i) a ray parallel to the principal axis reflects through F (concave) or appears to come from F (convex), and (ii) a ray through C reflects back on itself. Students often draw rays that do not follow these rules. Always use the two standard rays.

8. Forgetting that a convex mirror always forms a virtual, diminished, erect image: Regardless of where the object is placed, a convex mirror always produces the same type of image. Students sometimes incorrectly state that a convex mirror can produce a real or magnified image.

9. Sign error with concave lens image distance: For a concave lens, the image is always on the same side as the object (virtual), so v is negative. Students sometimes assign a positive v to a concave lens image, leading to wrong answers.

10. Confusing refractive index direction: n21 means the refractive index of medium 2 with respect to medium 1 (n21 = v1/v2 = sin i / sin r). Students sometimes reverse the media, getting the reciprocal of the correct answer.

Exam Checklist

Before writing your exam, verify each item:

  • [ ] I write the correct formula first (mirror: 1/v + 1/u = 1/f; lens: 1/v - 1/u = 1/f) before substituting values.
  • [ ] I always assign a negative sign to the object distance u.
  • [ ] I use the correct sign for f: negative for concave mirror and concave lens; positive for convex mirror and convex lens.
  • [ ] I convert focal length from cm to m before calculating the power of a lens (P = 1/f in metres).
  • [ ] I use m = -v/u for mirrors and m = v/u for lenses.
  • [ ] I interpret the sign of magnification correctly: positive = erect, negative = inverted.
  • [ ] I interpret the sign of v correctly: for mirrors, positive v = virtual (behind mirror), negative v = real (in front); for lenses, positive v = real (other side), negative v = virtual (same side as object).
  • [ ] I draw ray diagrams using the two standard rays and label P, F, C clearly.
  • [ ] I state the nature of the image explicitly: real/virtual, erect/inverted, magnified/diminished/same size.
  • [ ] I remember that a convex mirror always forms a virtual, erect, diminished image, and a concave lens always forms a virtual, erect, diminished image.

Internal Links

  • [Chapter 10 Guide — Light: Reflection and Refraction](/school/cbse/class/10/science/cbse-light-reflection-and-refraction)
  • [All Class 10 Science Solutions](/school/cbse/class/10/science/solutions)
  • [Class 10 Science Subject Guide](/school/cbse/class/10/science)

Frequently Asked Questions

Mirror formula: 1/v + 1/u = 1/f, where u = object distance, v = image distance, f = focal length. Sign convention (New Cartesian): all distances are measured from the pole (mirror) or optical centre (lens). Distances measured in the direction of incident light are positive; opposite direction are negative. For a concave mirror, f and R are negative. For a convex mirror, f and R are positive. Object distance u is always negative (object is in front of the mirror or on the side from which light comes).