NCERT Solutions Class 10 Science Chapter 10 Light Reflection and Refraction

NCERT Solutions Class 10 Science Chapter 10, "Light – Reflection and Refraction," provides students with complete, step-by-step conceptual clarity for one of the highest-scoring chapters in the CBSE Class 10 Science board examination. This chapter forms the foundation of geometric optics, covering the fundamental behavior of light, ray diagrams for spherical mirrors and lenses, sign conventions, the mirror and lens formulas, refractive indices, and the optical power of lenses. Master these detailed NCERT Science Class 10 solutions to tackle textbook exercises, conceptual questions, and numerical problems with 100% accuracy.

Chapter Overview

In CBSE Class 10 Science, the "Natural Phenomena" unit carries a substantial weightage of 12 marks, out of which Chapter 10 (Light – Reflection and Refraction) typically accounts for 7 to 10 marks in the annual board paper. Understanding ray optics is critical because CBSE frequently tests ray diagrams, numerical calculations with Cartesian sign conventions, and real-life applications.

The key topics covered in this chapter include:

• Reflection of Light: Laws of reflection, regular vs. diffused reflection.
• Spherical Mirrors: Concave and convex mirrors, definitions of Pole ($P$), Centre of Curvature ($C$), Principal Focus ($F$), Radius of Curvature ($R$), and Focal Length ($f$).
• Ray Diagrams for Spherical Mirrors: Image formation for various object positions, characteristics of real vs. virtual images.
• New Cartesian Sign Convention: Rules for measuring distances along and perpendicular to the principal axis.
• Mirror Formula and Magnification: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ and $m = \frac{h'}{h} = -\frac{v}{u}$.
• Refraction of Light: Cause of refraction, Snell's Law of refraction, and optical density.
• Refractive Index: Absolute refractive index ($n = c/v$) and relative refractive index ($n_{21} = v_1/v_2$).
• Refraction through Spherical Lenses: Convex (converging) and concave (diverging) lenses, standard ray paths, and complete image formation tables.
• Lens Formula and Magnification: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ and $m = \frac{h'}{h} = +\frac{v}{u}$.
• Power of a Lens: Definition, unit (Dioptre, $\text{D}$), and combinations of thin lenses ($P = P_1 + P_2 + \dots$).

Comprehensive Step-by-Step Exercise Solutions

Below are fully solved NCERT in-text questions and chapter-end exercise problems presented with standard CBSE board examination presentation formats.

In-Text Questions (Page 168 & 171)

Question 1: Define the principal focus of a concave mirror.

Answer:

The principal focus of a concave mirror is a specific point on its principal axis where rays of light travelling parallel to the principal axis converge after reflecting from the mirror surface. It is denoted by the letter $F$.

Question 2: The radius of curvature of a spherical mirror is 32 cm. What is its focal length?

Solution:

• Given: Radius of curvature, $R = 32\text{ cm}$
• To Find: Focal length, $f$
• Formula: $f = \frac{R}{2}$
• Substitution & Calculation: $f = \frac{32\text{ cm}}{2} = 16\text{ cm}$
• Answer: The focal length of the spherical mirror is $16\text{ cm}$.

Question 3: Name a mirror that can give an erect and enlarged image of an object.

Answer:

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

Question 4: Why do we prefer a convex mirror as a rear-view mirror in vehicles?

Answer:

Convex mirrors are preferred as rear-view (wing) mirrors in automobiles due to two critical physical advantages:

1. They always form an erect, virtual, and diminished image of objects behind the vehicle, regardless of their distance.
2. They are curved outwards, providing a much wider field of view than a plane mirror of the same size, allowing the driver to view a larger traffic area.

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In-Text Questions (Page 176 & 184)

Question 5: 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?

Solution:

• Given:
  • Object distance, $u = -10\text{ cm}$ (by sign convention)
  • Magnification, $m = -3$ (negative because the image is real and inverted)
• To Find: Image distance, $v$
• Formula: $m = -\frac{v}{u}$
• Substitution & Calculation:

  $-3 = -\frac{v}{-10}$

  $-3 = \frac{v}{10} \implies v = -30\text{ cm}$

• Answer: The image is located at a distance of $30\text{ cm}$ in front of the concave mirror (on the same side as the object).

Question 6: Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass? The speed of light in vacuum is $3 \times 10^8\text{ m/s}$.

Solution:

• Given:
  • Refractive index of glass, $n = 1.50 = \frac{3}{2}$
  • Speed of light in vacuum, $c = 3 \times 10^8\text{ m/s}$
• To Find: Speed of light in glass, $v$
• Formula: $n = \frac{c}{v} \implies v = \frac{c}{n}$
• Substitution & Calculation:

  $v = \frac{3 \times 10^8\text{ m/s}}{1.50} = 2.0 \times 10^8\text{ m/s}$

• Answer: The speed of light in glass is $2 \times 10^8\text{ m/s}$.

Question 7: Find the focal length of a lens of power $-2.0\text{ D}$. What type of lens is this?

Solution:

• Given: Power of lens, $P = -2.0\text{ D}$
• To Find: Focal length ($f$) and lens type
• Formula: $P = \frac{1}{f\text{ (in metres)}} \implies f = \frac{1}{P}$
• Substitution & Calculation:

  $f = \frac{1}{-2.0}\text{ m} = -0.5\text{ m} = -50\text{ cm}$

• Answer: The focal length of the lens is $-0.5\text{ m}$ (or $-50\text{ cm}$). Since the power and focal length are negative, it is a concave (diverging) lens.

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NCERT Chapter-End Exercise Solutions

Question 8 (MCQ): Which one of the following materials cannot be used to make a lens?

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

Answer: (d) Clay
Explanation: A lens requires a transparent medium through which light can easily pass and undergo refraction. Clay is an opaque material and does not transmit light; hence, it cannot be used to make an optical lens.

Question 9 (MCQ): 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 ($2F_1$)
Explanation: When an object is placed at $2F_1$ in front of a convex lens, a real, inverted image of identical size ($m = -1$) is formed on the other side at $2F_2$.

Question 10 (MCQ): A spherical mirror and a thin spherical lens have each a focal length of $-15\text{ cm}$. 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
Explanation: By the New Cartesian Sign Convention, the focal length is negative for both a concave mirror and a concave (diverging) lens.

Question 11: A concave lens has focal length of 15 cm. At what distance should the object from the lens be placed so that it forms an image at 10 cm from the lens? Also, find the magnification produced by the lens.

Solution:

• Given:
  • Focal length of concave lens, $f = -15\text{ cm}$
  • Image distance for concave lens, $v = -10\text{ cm}$ (concave lens forms virtual image on same side)
• To Find: Object distance ($u$) and Magnification ($m$)
• Formula:

  Lens Formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \implies \frac{1}{u} = \frac{1}{v} - \frac{1}{f}$

  Magnification: $m = \frac{v}{u}$

• Substitution & Calculation:

  $$\frac{1}{u} = \frac{1}{-10} - \frac{1}{-15} = -\frac{1}{10} + \frac{1}{15}$$

  Taking LCM of 10 and 15 ($= 30$):

  $$\frac{1}{u} = \frac{-3 + 2}{30} = \frac{-1}{30} \implies u = -30\text{ cm}$$

  Now calculating magnification:

  $$m = \frac{v}{u} = \frac{-10\text{ cm}}{-30\text{ cm}} = +\frac{1}{3} \approx +0.33$$

• Answer: The object should be placed at a distance of $30\text{ cm}$ in front of the lens. The magnification is $+0.33$, indicating a virtual, erect, and diminished image (one-third the object's height).

Question 12: 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.

Solution:

• Given:
  • Object distance, $u = -10\text{ cm}$
  • Focal length of convex mirror, $f = +15\text{ cm}$
• To Find: Image distance ($v$) and Nature of image
• Formula: Mirror Formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \implies \frac{1}{v} = \frac{1}{f} - \frac{1}{u}$
• Substitution & Calculation:

  $$\frac{1}{v} = \frac{1}{15} - \frac{1}{-10} = \frac{1}{15} + \frac{1}{10}$$

  Taking LCM of 15 and 10 ($= 30$):

  $$\frac{1}{v} = \frac{2 + 3}{30} = \frac{5}{30} = \frac{1}{6} \implies v = +6\text{ cm}$$

  Magnification: $m = -\frac{v}{u} = -\frac{6}{-10} = +0.6$

• Answer: The image is formed at a distance of $6\text{ cm}$ behind the mirror ($v = +6\text{ cm}$). Since $v > 0$ and $m = +0.6$, the image is virtual, erect, and diminished.

Question 13: 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.

Solution:

• Given:
  • Object height, $h = +5.0\text{ cm}$
  • Object distance, $u = -20\text{ cm}$
  • Radius of curvature, $R = +30\text{ cm} \implies f = \frac{R}{2} = +15\text{ cm}$
• To Find: Image position ($v$), height ($h'$), and nature
• Formula: $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ and $m = \frac{h'}{h} = -\frac{v}{u}$
• Substitution & Calculation:

  $$\frac{1}{v} = \frac{1}{15} - \left(-\frac{1}{20}\right) = \frac{1}{15} + \frac{1}{20} = \frac{4 + 3}{60} = \frac{7}{60}$$

  $$v = \frac{60}{7}\text{ cm} \approx +8.57\text{ cm}$$

  Now calculating image height $h'$:

  $$h' = -\frac{v}{u} \times h = -\frac{60/7}{-20} \times 5.0 = \frac{3}{7} \times 5.0 = \frac{15}{7}\text{ cm} \approx +2.14\text{ cm}$$

• Answer: The image is formed at $8.57\text{ cm}$ behind the mirror. It is virtual, erect, and has a size of $2.14\text{ cm}$.

Question 14: 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 the nature of the image.

Solution:

• Given:
  • Object height, $h = +7.0\text{ cm}$
  • Object distance, $u = -27\text{ cm}$
  • Focal length of concave mirror, $f = -18\text{ cm}$
• To Find: Screen position ($v$), image height ($h'$), and nature
• Formula: $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ and $h' = -\frac{v}{u} \times h$
• Substitution & Calculation:

  $$\frac{1}{v} = \frac{1}{-18} - \frac{1}{-27} = -\frac{1}{18} + \frac{1}{27}$$

  Taking LCM of 18 and 27 ($= 54$):

  $$\frac{1}{v} = \frac{-3 + 2}{54} = -\frac{1}{54} \implies v = -54\text{ cm}$$

  Calculating image size:

  $$h' = -\frac{-54}{-27} \times 7.0 = -(2) \times 7.0 = -14.0\text{ cm}$$

• Answer: The screen must be placed at a distance of $54\text{ cm}$ in front of the concave mirror. The image formed is real, inverted, and magnified with a height of $14.0\text{ cm}$.

Question 15: A doctor has prescribed a corrective lens of power $+1.5\text{ D}$. Find the focal length of the lens. Is the prescribed lens diverging or converging?

Solution:

• Given: Power, $P = +1.5\text{ D}$
• To Find: Focal length ($f$) and type of lens
• Formula: $f = \frac{1}{P}$
• Substitution & Calculation:

  $$f = \frac{1}{+1.5}\text{ m} = \frac{10}{15}\text{ m} = +\frac{2}{3}\text{ m} \approx +0.67\text{ m} = +66.7\text{ cm}$$

• Answer: The focal length of the prescribed lens is $+0.67\text{ m}$ ($+66.7\text{ cm}$). Since the focal length is positive, it is a converging (convex) lens.

Important Formulas, Laws, and Ray Rules

Mastering the mathematical relationships in geometric optics is vital for scoring full marks. The key formulas and rules from this chapter are summarized below:

Concept / Law	Formula / Statement	Key Variables & Units
Focal Length & Radius	$$f = \frac{R}{2}$$	$R$ = Radius of curvature, $f$ = focal length ($\text{cm}$ or $\text{m}$)
Mirror Formula	$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$	$u$ = object distance, $v$ = image distance, $f$ = focal length
Mirror Magnification	$$m = \frac{h'}{h} = -\frac{v}{u}$$	$h'$ = image height, $h$ = object height ($m < 0$ real, $m > 0$ virtual)
Snell's Law of Refraction	$$\frac{\sin i}{\sin r} = \frac{n_2}{n_1} = n_{21}$$	$i$ = angle of incidence, $r$ = angle of refraction, $n$ = refractive index
Absolute Refractive Index	$$n = \frac{c}{v}$$	$c = 3 \times 10^8\text{ m/s}$, $v$ = speed of light in medium
Lens Formula	$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$	Note the minus sign between image and object distance reciprocals
Lens Magnification	$$m = \frac{h'}{h} = +\frac{v}{u}$$	Positive ratio for lenses (unlike mirrors which have a minus sign)
Power of a Lens	$$P = \frac{1}{f\text{ (in metres)}}$$	Expressed in Dioptres ($\text{D}$ where $1\text{ D} = 1\text{ m}^{-1}$)
Combination of Lenses	$$P_{\text{net}} = P_1 + P_2 + P_3 + \dots$$	Algebraic sum using respective positive/negative signs

New Cartesian Sign Convention Summary

1. The object is always placed to the left of the mirror or lens (incident light travels left to right).
2. All distances parallel to the principal axis are measured from the Pole ($P$) for mirrors, and from the Optical Centre ($O$) for lenses.
3. Distances measured in the direction of incident light (along $+x$-axis, to the right) are taken as positive ($+$).
4. Distances measured against the direction of incident light (along $-x$-axis, to the left) are taken as negative ($-$). Consequently, object distance ($u$) is always negative.
5. Distances measured upward and perpendicular to the principal axis (along $+y$-axis) are positive ($+$).
6. Distances measured downward and perpendicular to the principal axis (along $-y$-axis) are negative ($-$).

Common Mistakes and How to Avoid Them

• Confusing the Sign in Mirror vs. Lens Formulas:
  Mistake: Writing $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ for mirrors or $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ for lenses.
  Tip: Remember: Mirror has a $+$ in formula and $-$ in magnification ($m = -v/u$); Lens has a $-$ in formula and $+$ in magnification ($m = +v/u$).
• Forgetting to Convert Centimetres to Metres for Power:
  Mistake: Calculating $P = 1/20 = 0.05\text{ D}$ when $f = 20\text{ cm}$.
  Tip: Always convert focal length to metres first ($20\text{ cm} = 0.2\text{ m} \implies P = 1/0.2 = +5\text{ D}$) or use $P = \frac{100}{f\text{ (in cm)}}$.
• Missing Arrows on Ray Diagrams:
  Mistake: Drawing correct straight lines without directional arrowheads.
  Tip: CBSE marking schemes deduct half to one full mark if incident and reflected/refracted ray arrows are missing.
• Incorrect Focal Length Sign:
  Tip: For all concave optics (concave mirror and concave lens), $f$ is negative ($-$). For all convex optics (convex mirror and convex lens), $f$ is positive ($+$).

Board Exam Relevance and Question Patterns

In CBSE Board Examinations, Chapter 10 Light is structured into distinct question formats across Sections A, B, C, D, and E:

• 1-Mark Objective / Assertion-Reason Questions: Often test refractive index comparisons, unit conversions for dioptre, or identification of mirror types for headlights, solar furnaces, and dental mirrors.
• 2-Mark Conceptual Questions: Focus on Snell's law derivation, conditions for no refraction, why the sky or underwater objects shift positions, and why convex mirrors are used as security mirrors.
• 3-Mark Ray Diagrams & Standard Numericals: Drawing ray diagrams for concave mirrors (especially object between $F$ and $P$, or at $C$) and convex lenses (object between $F_1$ and $2F_1$, or at $2F_1$), accompanied by a 3-step numerical.
• 4 or 5-Mark Case-Based / Integrated Problems: Case studies on ophthalmic lenses, corrective glasses, telescope objectives, or multi-step numericals asking for position, size, nature, and verification through ray diagrams.

More NCERT Solutions and Practice

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