CBSE Class 10 Science Chapter 9 Notes — Light: Reflection and Refraction

Reflection of Light

Laws of Reflection

1. The angle of incidence = the angle of reflection (∠i = ∠r) 2. The incident ray, reflected ray, and the normal at the point of incidence all lie in the same plane

These laws apply to ALL reflecting surfaces — plane mirrors, curved mirrors, and rough surfaces.

Terms for Spherical Mirrors

Concave vs Convex Mirror

Image Formation by Concave Mirror

Mirror Formula and Magnification

Mirror formula: $$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$

Linear magnification: $$m = \frac{h'}{h} = -\frac{v}{u}$$

Where:

• v = image distance (from pole)
• u = object distance (from pole)
• f = focal length
• h' = image height, h = object height

Sign Convention (New Cartesian):

• All distances measured from the pole
• Distances in the direction of incident light = positive
• Distances against the direction of incident light = negative
• Heights above principal axis = positive; below = negative

Worked numerical: An object is placed 15 cm in front of a concave mirror of focal length 10 cm. Find image distance.

Refraction of Light

Refraction = bending of light when it passes from one transparent medium to another, due to a change in speed.

Laws of Refraction (Snell's Law)

1. The incident ray, refracted ray, and normal at the point of incidence are all in the same plane 2. sin i / sin r = constant (for a given pair of media and given wavelength of light)

This constant is the refractive index (n) of medium 2 with respect to medium 1.

n = sin i / sin r = speed of light in medium 1 / speed of light in medium 2

Refractive Index

• Shows how much light slows down in a medium compared to vacuum
• n = c / v (c = speed of light in vacuum = 3×10⁸ m/s; v = speed in medium)
• Water: n ≈ 1.33; Glass: n ≈ 1.5; Diamond: n ≈ 2.42

High refractive index = light slows down more = bends more

Refraction through a Glass Slab

• Light bends toward normal when entering (going from rarer to denser medium)
• Light bends away from normal when exiting (going from denser to rarer medium)
• Emergent ray is parallel to incident ray but displaced laterally (lateral displacement)

Spherical Lenses

Convex vs Concave Lens

Image Formation by Convex Lens

Lens Formula and Magnification

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

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

Power of a lens: $$P = \frac{1}{f} \text{ (in metres)}$$

Unit of power: dioptre (D) = 1 m⁻¹

• Convex lens: positive power (e.g., +2 D)
• Concave lens: negative power (e.g., −1.5 D)
• Combined power of two lenses in contact: P = P₁ + P₂

Worked numerical: A convex lens of focal length 20 cm forms an image of an object placed 30 cm from the lens. Find image distance.

Board Exam Tips

• Numericals (5–8 marks): Mirror and lens formula numericals are consistently in boards. Practice at least 10 numericals each.
• Always write sign convention clearly before substituting values
• Ray diagram for mirrors: draw at least 2 of the standard rays (through F, through C, parallel to PA)
• Ray diagram for lenses: draw at least 2 of the standard rays (through optical centre, parallel to PA)
• Power = 1/f in metres — convert cm to m before calculating power
• "A convex mirror is used as rear-view mirror" — explain WHY (always forms erect, diminished image; gives wider field of view)
• Magnification sign tells you: negative m = inverted (real) image; positive m = erect (virtual) image

📚 Lakhmir Singh Science Class 10 — Light Chapter with Numericals

📝 Oswaal CBSE Question Bank — Physics Numericals Practice

NCERT Exercise Solutions

Intext Questions

Page 168, Q1: Define the principal focus of a concave mirror.

Page 168, Q2: The radius of curvature of a spherical mirror is 20 cm. What is its focal length?

Page 168, Q3: Name a mirror that can give an erect and enlarged image of an object.

Page 176, Q1: Find the focal length of a convex lens which forms a real image of magnification –1 when the object distance is 30 cm.

Using lens formula: 1/f = 1/v – 1/u = 1/30 – 1/(–30) = 1/30 + 1/30 = 2/30 f = 15 cm

The focal length of the convex lens is 15 cm.

Chapter-End Exercise

Q1. A concave mirror produces a 3× magnified real image of an object placed at 10 cm in front of it. Where is the image located?

The image is located 30 cm in front of the mirror (on the same side as the object). Since it's negative, it confirms a real image.

Using mirror formula to verify: 1/f = 1/v + 1/u = 1/(–30) + 1/(–10) = –1/30 – 3/30 = –4/30 f = –7.5 cm (concave mirror ✓)

Q2. 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.

Mirror formula: 1/v + 1/u = 1/f 1/v = 1/f – 1/u = 1/15 – 1/(–10) = 1/15 + 1/10 = 2/30 + 3/30 = 5/30 = 1/6 v = +6 cm

Positive v means the image forms behind the mirror (virtual). Magnification: m = –v/u = –6/(–10) = +0.6 (positive → erect, less than 1 → diminished)

Image: Virtual, erect, diminished, located 6 cm behind the mirror.

Q3. A ray of light travelling in air enters obliquely into water. Does the light ray bend towards or away from the normal? Why?

Reason: Water is an optically denser medium than air (higher refractive index, n ≈ 1.33). When light goes from a less dense medium (air) to a denser medium (water), it slows down. According to Snell's law, slowing causes bending toward the normal — the angle of refraction is less than the angle of incidence.

Q4. Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass? (Speed of light in vacuum = 3 × 10⁸ m/s)

Speed of light in glass = 2 × 10⁸ m/s

Q5. Distinguish between real and virtual images.