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Class X Physics - Chapter 1

Light: Reflection and Refraction

1.1 Introduction to Light

Light: Light is a form of electromagnetic radiation that makes vision possible. It travels in straight lines and exhibits both wave and particle properties.

Properties of Light:

Example: When sunlight passes through a window, it travels in straight lines, creating sharp shadows of objects placed in its path.

1.2 Reflection of Light

Reflection: The bouncing back of light when it strikes a surface is called reflection of light.

Laws of Reflection:

First Law: The incident ray, reflected ray, and the normal to the surface at the point of incidence all lie in the same plane.

Second Law: The angle of incidence is equal to the angle of reflection.
∠i = ∠r
where ∠i = angle of incidence, ∠r = angle of reflection

Important Terms:

Term Definition Symbol
Incident Ray The ray of light falling on the surface -
Reflected Ray The ray of light bouncing back from the surface -
Normal A line perpendicular to the surface at point of incidence N
Angle of Incidence Angle between incident ray and normal ∠i
Angle of Reflection Angle between reflected ray and normal ∠r
Diagram: Laws of Reflection
                   Incident Ray
                        ↘
                         ↘  ∠i
                          ↘
            ________________↘________________ Surface
                            |  ↗
                            |N   ↗ ∠r
                            |     ↗
                            |      ↗
                                Reflected Ray
MCQ 1: The angle of incidence is 30°. What is the angle of reflection?
(A) 15°
(B) 30°
(C) 60°
(D) 90°
Answer: (B) 30°

1.3 Spherical Mirrors

Spherical Mirror: A mirror whose reflecting surface is a part of a hollow sphere is called a spherical mirror.

Types of Spherical Mirrors:

1. Concave Mirror (Converging Mirror):

A spherical mirror whose reflecting surface is curved inwards (towards the center of the sphere) is called a concave mirror.

2. Convex Mirror (Diverging Mirror):

A spherical mirror whose reflecting surface is curved outwards (away from the center of the sphere) is called a convex mirror.

Important Terms for Spherical Mirrors:

Term Definition Symbol Unit
Centre of Curvature Centre of the sphere of which mirror is a part C -
Radius of Curvature Radius of the sphere of which mirror is a part R m
Pole Centre of the spherical mirror P -
Principal Axis Line joining pole and centre of curvature - -
Principal Focus Point on principal axis where parallel rays converge/appear to diverge F -
Focal Length Distance between pole and principal focus f m
Aperture Effective diameter of the mirror - m
Relation: R = 2f
where R = radius of curvature, f = focal length
Example: If the focal length of a concave mirror is 15 cm, then its radius of curvature is R = 2f = 2 × 15 = 30 cm.
MCQ 2: The radius of curvature of a spherical mirror is 40 cm. Its focal length is:
(A) 10 cm
(B) 20 cm
(C) 40 cm
(D) 80 cm
Answer: (B) 20 cm

1.4 Image Formation by Spherical Mirrors

Sign Convention for Spherical Mirrors:

Mirror Formula:

1/f = 1/v + 1/u
where f = focal length, v = image distance, u = object distance

Magnification Formula:

m = -v/u = h₂/h₁
where m = magnification, h₁ = object height, h₂ = image height
Magnification: The ratio of height of image to height of object is called magnification. It tells us how many times the image is magnified with respect to the object.

Image Formation by Concave Mirror:

Object Position Image Position Nature of Image Size
At infinity At focus F Real, inverted Highly diminished
Beyond C Between C and F Real, inverted Diminished
At C At C Real, inverted Same size
Between C and F Beyond C Real, inverted Enlarged
At F At infinity Real, inverted Highly enlarged
Between F and P Behind mirror Virtual, erect Enlarged
Example: An object is placed 30 cm from a concave mirror of focal length 10 cm. Find the position and nature of the image.

Given: u = -30 cm, f = -10 cm
Using mirror formula: 1/f = 1/v + 1/u
1/(-10) = 1/v + 1/(-30)
-1/10 = 1/v - 1/30
1/v = -1/10 + 1/30 = -3/30 + 1/30 = -2/30 = -1/15
v = -15 cm

Answer: Image is formed 15 cm in front of the mirror, real and inverted.
MCQ 3: An object is placed between the focus and pole of a concave mirror. The image formed is:
(A) Real and inverted
(B) Virtual and erect
(C) Real and erect
(D) Virtual and inverted
Answer: (B) Virtual and erect

1.5 Image Formation by Convex Mirror

Key Point: Convex mirrors always form virtual, erect, and diminished images regardless of the object position.
Object Position Image Position Nature of Image Size
At infinity At focus F Virtual, erect Highly diminished
Anywhere in front Between F and P Virtual, erect Diminished
Example: An object 4 cm high is placed 20 cm from a convex mirror of focal length 15 cm. Find the position, size and nature of the image.

Given: u = -20 cm, f = +15 cm, h₁ = +4 cm
Using mirror formula: 1/f = 1/v + 1/u
1/15 = 1/v + 1/(-20)
1/v = 1/15 + 1/20 = 4/60 + 3/60 = 7/60
v = 60/7 ≈ 8.57 cm

Magnification: m = -v/u = -8.57/(-20) = +0.43
Image height: h₂ = m × h₁ = 0.43 × 4 = 1.72 cm

Answer: Image is 8.57 cm behind the mirror, 1.72 cm high, virtual and erect.

1.6 Uses of Spherical Mirrors

Uses of Concave Mirrors:

Uses of Convex Mirrors:

MCQ 4: Convex mirrors are used as rear-view mirrors in vehicles because they:
(A) Form magnified images
(B) Form real images
(C) Give wider field of view
(D) Form inverted images
Answer: (C) Give wider field of view

1.7 Refraction of Light

Refraction: The bending of light when it passes from one transparent medium to another is called refraction of light.

Cause of Refraction:

Refraction occurs due to change in speed of light when it passes from one medium to another of different optical density.

Laws of Refraction (Snell's Laws):

First Law: The incident ray, refracted ray, and the normal to the surface at the point of incidence all lie in the same plane.

Second Law: The ratio of sine of angle of incidence to sine of angle of refraction is constant for a given pair of media.
sin i / sin r = constant = μ
or μ = sin i / sin r
where μ = refractive index of second medium with respect to first

Important Terms:

Term Definition Symbol Unit
Angle of Incidence Angle between incident ray and normal ∠i degree (°)
Angle of Refraction Angle between refracted ray and normal ∠r degree (°)
Refractive Index Ratio of speed of light in vacuum to speed in medium μ or n No unit
Optical Density Property that determines light bending capacity - -
Absolute Refractive Index: The refractive index of a medium with respect to vacuum (or air) is called absolute refractive index.
Formula: μ = c/v
where c = speed of light in vacuum, v = speed of light in medium
Example: Speed of light in water is 2.25 × 108 m/s. Find refractive index of water.

Given: v = 2.25 × 108 m/s, c = 3 × 108 m/s
Solution: μ = c/v = (3 × 108)/(2.25 × 108) = 1.33

Answer: Refractive index of water = 1.33

Relative Refractive Index:

μ₁₂ = μ₂/μ₁ = v₁/v₂
where μ₁₂ = refractive index of medium 2 with respect to medium 1
MCQ 5: The refractive index of diamond is 2.42. The speed of light in diamond is:
(A) 1.24 × 108 m/s
(B) 1.86 × 108 m/s
(C) 2.42 × 108 m/s
(D) 7.26 × 108 m/s
Answer: (A) 1.24 × 108 m/s

1.8 Refraction by Spherical Lenses

Lens: A transparent optical element with at least one curved surface that refracts light to form images.

Types of Lenses:

1. Convex Lens (Converging Lens):

A lens that is thicker at the middle and thinner at the edges. It converges parallel rays of light.

2. Concave Lens (Diverging Lens):

A lens that is thinner at the middle and thicker at the edges. It diverges parallel rays of light.

Important Terms for Lenses:

Term Definition Symbol Unit
Principal Focus Point where parallel rays converge/appear to diverge F -
Focal Length Distance between optical centre and principal focus f m
Optical Centre Centre of the lens O -
Principal Axis Line passing through optical centre and principal focus - -
Aperture Effective diameter of the lens - m

Sign Convention for Lenses:

Lens Formula:

1/f = 1/v - 1/u
where f = focal length, v = image distance, u = object distance

Magnification Formula:

m = v/u = h₂/h₁
where m = magnification, h₁ = object height, h₂ = image height
Example: An object is placed 15 cm from a convex lens of focal length 10 cm. Find the position and magnification of the image.

Given: u = -15 cm, f = +10 cm
Using lens formula: 1/f = 1/v - 1/u
1/10 = 1/v - 1/(-15)
1/10 = 1/v + 1/15
1/v = 1/10 - 1/15 = 3/30 - 2/30 = 1/30
v = 30 cm

Magnification: m = v/u = 30/(-15) = -2

Answer: Image is at 30 cm on the opposite side, magnification = -2 (real, inverted, enlarged)

1.9 Image Formation by Lenses

Image Formation by Convex Lens:

Object Position Image Position Nature of Image Size
At infinity At focus F₂ Real, inverted Highly diminished
Beyond 2F₁ Between F₂ and 2F₂ Real, inverted Diminished
At 2F₁ At 2F₂ Real, inverted Same size
Between 2F₁ and F₁ Beyond 2F₂ Real, inverted Enlarged
At F₁ At infinity Real, inverted Highly enlarged
Between F₁ and O Same side as object Virtual, erect Enlarged

Image Formation by Concave Lens:

Key Point: Concave lenses always form virtual, erect, and diminished images regardless of object position.
Object Position Image Position Nature of Image Size
At infinity At focus F₁ Virtual, erect Highly diminished
Anywhere in front Between O and F₁ Virtual, erect Diminished
MCQ 6: An object is placed at 2F₁ of a convex lens. The image formed is:
(A) At F₂, real and diminished
(B) At 2F₂, real and same size
(C) Beyond 2F₂, real and enlarged
(D) Virtual and enlarged
Answer: (B) At 2F₂, real and same size

1.10 Power of a Lens

Power of a Lens: The ability of a lens to converge or diverge light rays is called power of lens. It is defined as reciprocal of focal length.
P = 1/f
where P = power of lens, f = focal length in metres

Unit of Power:

SI Unit: Dioptre (D)
Definition: 1 Dioptre = 1 m⁻¹
Note: Power is positive for convex lens and negative for concave lens

Power of Combination of Lenses:

P = P₁ + P₂ + P₃ + ...
where P₁, P₂, P₃ are powers of individual lenses
Example 1: A convex lens has focal length 25 cm. Find its power.

Given: f = 25 cm = 0.25 m
Solution: P = 1/f = 1/0.25 = 4 D

Answer: Power = +4 D
Example 2: Two lenses of power +2 D and -1.5 D are placed in contact. Find the power and focal length of the combination.

Given: P₁ = +2 D, P₂ = -1.5 D
Solution: P = P₁ + P₂ = 2 + (-1.5) = 0.5 D
f = 1/P = 1/0.5 = 2 m = 200 cm

Answer: Power = +0.5 D, focal length = 200 cm
MCQ 7: The power of a lens is -2.5 D. Its focal length is:
(A) 40 cm
(B) -40 cm
(C) 25 cm
(D) -25 cm
Answer: (B) -40 cm

1.11 Human Eye and Vision

Human Eye: The human eye is a natural optical instrument that forms real, inverted images on the retina, which are interpreted by the brain as erect images.

Parts of Human Eye:

Part Function
Cornea Transparent front layer; provides most of eye's focusing power
Iris Colored part; controls amount of light entering eye
Pupil Opening in iris; size varies with light intensity
Eye Lens Transparent biconvex lens; provides variable focusing
Ciliary Muscles Control curvature and focal length of eye lens
Retina Light-sensitive layer; contains rod and cone cells
Optic Nerve Carries visual information to brain

Important Distances:

Distance Value Significance
Near Point 25 cm Minimum distance for clear vision
Far Point Infinity Maximum distance for clear vision
Distance of distinct vision 25 cm Most comfortable reading distance
Power of Accommodation: The ability of eye lens to adjust its focal length by changing curvature with the help of ciliary muscles is called power of accommodation.
Example: When we look at distant objects, ciliary muscles relax, eye lens becomes thin, and focal length increases. When we look at nearby objects, ciliary muscles contract, eye lens becomes thick, and focal length decreases.

1.12 Defects of Vision and their Correction

1. Myopia (Short-sightedness or Near-sightedness):

Myopia: A person can see nearby objects clearly but cannot see distant objects clearly. Images of distant objects are formed in front of the retina.

Causes:

Correction:

Remedy: Use of concave lens of appropriate power
Formula: P = -1/d
where d = distance of far point in metres
Example: A myopic person can see clearly up to 2 m. What power of lens is required to correct the defect?

Given: Far point = 2 m
Solution: P = -1/d = -1/2 = -0.5 D

Answer: Power of corrective lens = -0.5 D (concave lens)

2. Hypermetropia (Long-sightedness or Far-sightedness):

Hypermetropia: A person can see distant objects clearly but cannot see nearby objects clearly. Images of nearby objects are formed behind the retina.

Causes:

Correction:

Remedy: Use of convex lens of appropriate power
Formula: P = 1/0.25 - 1/d
where d = distance of near point in metres
Example: A hypermetropic person cannot see objects closer than 50 cm. What power of lens is required?

Given: Near point = 50 cm = 0.5 m
Solution: P = 1/0.25 - 1/0.5 = 4 - 2 = 2 D

Answer: Power of corrective lens = +2 D (convex lens)

3. Presbyopia:

Presbyopia: Gradual weakening of ciliary muscles and diminishing flexibility of eye lens with age. Both near and distant objects appear blurred.

Correction:

Remedy: Bifocal lenses
Upper part: Concave lens for distant vision
Lower part: Convex lens for near vision
MCQ 8: A myopic person has a far point of 1.5 m. The power of corrective lens required is:
(A) +1.5 D
(B) -1.5 D
(C) +0.67 D
(D) -0.67 D
Answer: (D) -0.67 D

1.13 Dispersion of Light

Dispersion: The splitting of white light into its constituent colors when it passes through a prism is called dispersion of light.

Spectrum:

Spectrum: The band of colors obtained on a screen when white light is dispersed by a prism is called spectrum.
Colors of Spectrum (VIBGYOR):
Violet → Indigo → Blue → Green → Yellow → Orange → Red
Wavelength order: Violet (shortest) to Red (longest)
Frequency order: Red (lowest) to Violet (highest)

Cause of Dispersion:

Different colors have different wavelengths and hence different refractive indices in the same medium. Violet light bends most, red light bends least.

Example: When sunlight passes through water droplets in the atmosphere, dispersion occurs, creating a rainbow. Red appears on the outer edge and violet on the inner edge.

Recombination of Spectrum:

When the dispersed colors are passed through another identical prism placed in inverted position, they recombine to form white light again.

MCQ 9: Which color of light has the maximum refractive index in glass?
(A) Red
(B) Green
(C) Blue
(D) Violet
Answer: (D) Violet

1.14 Atmospheric Refraction

Atmospheric Refraction: The refraction of light through the earth's atmosphere is called atmospheric refraction. It occurs due to gradual change in refractive index of air with altitude.

Optical Density of Air:

Air near the earth's surface is denser (higher refractive index) than air at higher altitudes (lower refractive index).

Phenomena due to Atmospheric Refraction:

1. Twinkling of Stars:

2. Apparent Position of Stars:

3. Advanced Sunrise and Delayed Sunset:

Example: Due to atmospheric refraction, we can see the sun even when it is slightly below the horizon. The sun's rays get bent towards the normal as they enter denser layers of atmosphere.

1.15 Scattering of Light

Scattering of Light: When light falls on small particles, it gets scattered in different directions. This phenomenon is called scattering of light.

Rayleigh Scattering:

Condition: Size of particles << wavelength of light
Law: Scattering ∝ 1/λ⁴
Result: Shorter wavelengths (blue) scatter more than longer wavelengths (red)

Applications of Scattering:

1. Blue Color of Sky:

2. Red Color of Sun at Sunrise/Sunset:

3. White Clouds:

Example: When you look at the sky through colored glasses:
• Red glasses: Sky appears dark (blue light blocked)
• Blue glasses: Sky appears bright (blue light passes through)
MCQ 10: The color of sky is blue due to:
(A) Absorption of blue light
(B) Scattering of blue light
(C) Reflection of blue light
(D) Refraction of blue light
Answer: (B) Scattering of blue light

1.16 Important Formulas Summary

Mirrors:

Mirror Formula: 1/f = 1/v + 1/u
Magnification: m = -v/u = h₂/h₁
Relation: R = 2f

Lenses:

Lens Formula: 1/f = 1/v - 1/u
Magnification: m = v/u = h₂/h₁
Power: P = 1/f (in metres)
Combined Power: P = P₁ + P₂ + P₃ + ...

Refraction:

Snell's Law: μ = sin i / sin r
Refractive Index: μ = c/v
Relative R.I.: μ₁₂ = μ₂/μ₁ = v₁/v₂

Eye Defects:

Myopia: P = -1/d (d = far point)
Hypermetropia: P = 1/0.25 - 1/d (d = near point)

Practice Questions

MCQ 11: The focal length of a convex mirror is 20 cm. Its radius of curvature is:
(A) 10 cm
(B) 20 cm
(C) 40 cm
(D) 80 cm
Answer: (C) 40 cm
MCQ 12: A ray of light passes from air into water. The angle of incidence is 60° and angle of refraction is 40°. The refractive index of water is:
(A) 1.35
(B) 1.5
(C) 0.74
(D) 1.33
Answer: (A) 1.35 [μ = sin 60°/sin 40° = 0.866/0.643 ≈ 1.35]
MCQ 13: Which lens is used to correct presbyopia?
(A) Concave lens
(B) Convex lens
(C) Bifocal lens
(D) Cylindrical lens
Answer: (C) Bifocal lens
MCQ 14: The speed of light in a medium is 2 × 108 m/s. The refractive index of the medium is:
(A) 1.5
(B) 2.0
(C) 0.67
(D) 1.33
Answer: (A) 1.5 [μ = c/v = 3×10⁸/2×10⁸ = 1.5]
MCQ 15: The magnification of a convex mirror is always:
(A) Greater than 1
(B) Less than 1 and positive
(C) Less than 1 and negative
(D) Equal to 1
Answer: (B) Less than 1 and positive

End of Chapter 1

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