Light - Reflection and Refraction (Part - II) – Class 10 Notes, Formulae, and Numerical

Chapter: Light – Reflection and Refraction (Part - 2: Refraction of Light)

(For Part - 1 click here )

Class: 10 Science

Target: CBSE Board Exams | NTSE | Olympiads

Introduction to Refraction of Light

- Refraction of light is a very important concept not only in Class 10 Board Exams but also in real life. Whenever light changes its medium it bend, this phenomenon is called refraction.

Important Keywords:

Refraction of Light
Laws of Refraction
Refractive index
Lens
Ray Diagrams
Image formation by lens
Lens Formula
Magnification of lens
Power of lens
Solved Numerical

Definition of Refraction of Light:

The phenomenon of bending of ray of light when enter from one transparent medium to another transparent medium is called Refraction of Light.

Laws of Refraction:

1. The incident ray, the refracted ray, and the normal all lie in the same plane.
2. The ratio of the sine of the angle of incidence to the sine of angle of refraction is constant for a given pair of media. (Snell’s Law):

sin(i)/sin(r) = μ

Refractive Index (μ):

It is the ratio of speed of light in vacuum (c) to the speed of light in medium (v)
It tells how much the light bends when it enters a new medium.

μ = Speed of light in vacuum (c) / Speed of light in medium (v)

Higher the refractive index, denser the medium.

*Medium*	*Speed of Light in Medium (approx.)*	*Refractive Index (n)*
Vacuum	3.00 × 10⁸ m/s	1.000
Air	2.998 × 10⁸ m/s	1.0003 (≈ 1)
Water	2.25 × 10⁸ m/s	1.33
Glass	2.00 × 10⁸ m/s	1.5 (varies with type)

Note : Refraction from Different Media:
Rarer medium to Denser medium (Ex. - Air to Glass): Toward the Normal, Speed Decreases.
Denser medium to Rarer medium (Ex. - Glass to Air) : Away from the Normal, Speed Increases.
Relative Refractive Index:

Definition:

The Relative Refractive Index of medium 2 with respect to medium 1 is defined as the ratio of the absolute refractive index of medium 2 to that of medium 1:
n₂₁= n₂ / n₁

Where:
n₂₁ = refractive index of medium 2 with respect to medium 1
n₁ = refractive index of first medium
n₂ = refractive index of second medium
Relative Refractive index in terms of speed of light:
n₂₁= v₁ / v₂
Where:
v₁= speed of light in medium 1
v₂= speed of light in medium 2

Solved Numerical:

1) The absolute refractive index of water is 1.33 and that of glass is 1.50. Find the refractive index of glass with respect to water.

Solution:
n_glass/water = n_glass / n_water

= 1.50 / 1.33
≈ 1.127
2) The speed of light in air is 3 × 10⁸ m/s and in a certain medium is 2 × 10⁸ m/s. Find the refractive index of that medium with respect to air.

Solution:
n_medium/air = c_air / v_medium

= (3 × 10⁸) / (2 × 10⁸)
= 1.5
3) The refractive index of medium A with respect to medium B is 1.25, and that of medium B with respect to medium C is 1.4. Find the refractive index of medium A with respect to medium C.

Solution:
n_AC = n_AB × n_BC

= 1.25 × 1.4
= 1.75
4) The refractive index of medium X with respect to medium Y is 1.6. Find the refractive index of medium Y with respect to medium X.
Solution:
n_YX = 1/ n_XY
= 1 / 1.6
≈ 0.625

Refraction of Light through Rectangular Glass Slab:

The emergent ray is parallel to the incident ray, but laterally displaced.
Light bends towards the normal when entering the glass (air → glass).
It bends away from the normal when leaving the glass (glass → air).

Lateral Displacement: The sideways shift of the emergent ray from the incident ray due to refraction.

Applications of Refraction:

                                       - Lenses in microscopes and telescopes
                                       - Vision correction using glasses
                                       - Magnifying glasses
                                       - Optical instruments and fiber optics
                                       - Mirage and apparent depth in water

The device use to study the Refraction of light is termed as Lens.

Lens:

Definition:

- A spherical lens is a transparent optical device bounded by one or two curved surfaces (spherical in shape) that refract light rays to form an image.

- It is made of glass, plastic, or any transparent material.

Types of Spherical Lenses:

Spherical lenses are classified into two main types:

1) Convex Lens (Converging Lens)

- Thicker at the center and thinner at the edges.
- Converges parallel light rays to a point called the Principal Focus.
- Examples: Magnifying glass, camera lens, human eye lens.

2) Concave Lens (Diverging Lens)

- Thinner at the center and thicker at the edges.
- Diverges parallel light rays as if they come from a point called the Principal Focus.
- Examples: Spectacles for myopia, peepholes.

Terms Related to the Spherical Lens:

1. Optical Centre (O): The geometric center of the lens; a ray passing through it does not deviate.

2. Center of Curvature (C1 and C2): The spherical lens is a part of sphere center of this sphere is called centre of curvature.

3. Radius of Curvature (R1 and R2): The spherical lens is a part of sphere radius of this sphere is called raduys of curvature.
4. Principal Axis: Straight line joining the optical centre to the centers of curvature.
5. Principal Focus (F):
- Convex lens: Point where parallel rays converge after refraction.
- Concave lens: Point from which parallel rays appear to diverge after refraction.
6. Focal Length (f): Distance between optical centre and principal focus.
7. Aperture: Diameter of the circular boundary of the lens.

Ray Diagram For Lenses:

Important Ray Rules:
1. Ray parallel to principal axis → passes through or appears to come from the principal focus after refraction.
2. Ray passing through optical centre → goes undeviated.
3. Ray passing through principal focus (convex) or directed towards principal focus (concave) → emerges parallel to principal axis.

Image Formation by Spherical Lenses:

1) Convex Lens:

*Sr.No*	*Object Position*	*Image Position*	*Nature of Image*
1	At infinity	At F	Real, inverted, highly diminished
2	Beyond 2F	Between F and 2F	Real, inverted, diminished
3	At 2F	At 2F	Real, inverted, same size
4	Between F and 2F	Beyond 2F	Real, inverted, magnified
5	At F	At infinity	Real, inverted, highly magnified
6	Between F and O	Same side as object	Virtual, erect, magnified

2) Concave Lens:

- Always forms virtual, erect, and diminished images regardless of object position.

Lens Formula:

1/f = 1/v - 1/u

_Where,v = image distance
u = object distance
f = focal length

Magnification by a Lens:

m = v/u =h' /h

_Where,v = image distance
u = object distance
h' = image height
h = object height

Sign Convention:

All distances are measured from the optical centre of the lens.
Distances measured in the direction of incident light are positive.
Distances measured opposite to incident light are negative.
Heights measured upward from principal axis are positive.
Heights measured downward from principal axis are negative.

Power of Lens:

P = 100 / f(cm) or P = 1 / f(m)

Unit: dioptre (D)
1 D = power of a lens of focal length 1 metre.

Application of Lens:

- Spectacles (vision correction)
- Cameras
- Microscopes & telescopes
- Projectors
- Magnifying glasses

Solved Numerical:

Problem 1

A convex lens has focal length f = 20 cm. An object is placed at 60 cm from the lens. Find the image distance v and magnification m.

Solution:
Sign values: f = +20 cm, u = -60 cm.
(1/v) = (1/f) + (1/u) = (1/20) + (1/-60) = (3 - 1)/60 = 2/60 = 1/30.
So v = 30 cm.
Magnification m = v/u = 30 / -60 = -1/2 = -0.5.

Answer: v = +30 cm (real, on the right), m = -0.5 (image inverted, half the object height).

Problem 2

A concave lens has focal length f = -15 cm. An object is placed at 30 cm from the lens. Find the image distance and magnification.

Solution:
f = -15 cm, u = -30 cm.
(1/v) = (1/f) + (1/u) = (1/-15) + (1/-30) = -(2 + 1)/30 = -3/30 = -1/10.
So v = -10 cm.
Magnification m = v/u = -10 / -30 = +1/3 ≈ 0.333.

Answer: v = -10 cm (virtual image, left side), m = +0.333 (upright, one-third the height).

Problem 3

Object located at 2f for a lens of focal length f = 10 cm. Find image distance and magnification.

Solution:
f = +10 cm, u = -20 cm.
(1/v) = (1/10) + (1/-20) = (2 - 1)/20 = 1/20 ⇒ v = 20 cm.
Magnification m = 20 / -20 = -1.

Answer: v = +20 cm (image at 2f, real), m = -1 (inverted, same size).

Problem 4

A convex lens of focal length f = 25 cm receives parallel rays (object at infinity). Where is the image formed?

Solution:
For object at infinity, image is at focal point: v = f = +25 cm.

Answer: v = +25 cm (real image at focal plane).

Problem 5

A 5.0 cm tall object is placed 40 cm in front of a convex lens of focal length 10 cm. Find image distance, magnification and image height.

Solution:
f = +10 cm, u = -40 cm.
(1/v) = (1/10) + (1/-40) = (4 - 1)/40 = 3/40 ⇒ v = 40/3 ≈ 13.333 cm.
Magnification m = (40/3) / -40 = -1/3 ≈ -0.333.
Image height h' = m × h = -0.333 × 5.0 cm ≈ -1.667 cm.

Answer: v ≈ +13.33 cm (real), m ≈ -0.333 (inverted), h' ≈ 1.667 cm (inverted).

Problem 6 (Power of a lens)

Find the power (in diopters) of a lens whose focal length is 50 cm.

Solution:
Convert to meters: f = 0.50 m.
P = 1 / f(m) = 1 / 0.50 = 2.0 D.

Answer: P = +2.0 D.

Problem 7

A concave lens has f = -20 cm. An object is placed at 10 cm in front of it. Find image distance and magnification.

Solution:
f = -20 cm, u = -10 cm.
(1/v) = (1/-20) + (1/-10) = -(1 + 2)/20 = -3/20 ⇒ v = -20/3 ≈ -6.667 cm.
m = (-20/3) / -10 = +2/3 ≈ 0.6667.

Answer: v ≈ -6.67 cm (virtual image, left of lens), m ≈ +0.667 (upright, two-thirds size).

Real Also : Light-Reflection and Refraction (Part-1) class 10 Notes, Practice Numerical pdf download

Practice Numerical:

1) An object 4 cm high is placed at a distance of 10 cm from a convex lens of focal length 20 cm. Find the position, nature and size of the image. ( Answer : v = -20 cm, virtual and erect, H_i = 8 cm)

2) Calculate the focal length of a convex lens, which produces a virtual image at a distance of 50 cm of an object placed 20 cm in front of it. (Answer : f = 33.33 cm)

3) An object is placed at a distance of 100 cm from a converging lens of focal length 40 cm. What is the nature and position of the image? (Answer : Real and inverted, v=66.67 cm)

4) Determine how far an object must be placed in front of a converging lens of focal length 10 cm in order to produce an erect image of linear magnification 4. (Answer : u = 7.5 cm)

5) An object placed 4 cm in front of a converging lens produces a real image 12 cm from the lens. What is the magnification of the image? What is the focal length of the lens? Also draw the ray diagram to show the formation of the image. (Answer : m = -3 , f = 3cm)

6) A convex lens produces an inverted image magnified three times of an object at a distance of 15 cm from it. Calculate focal length of the lens. (Answer : f = 11.25 cm )

7) A small object is so placed in front of a convex lens of 5 cm focal length that a virtual image is formed at a distance of 25 cm. Find the magnification. (Answer : m = 6 )

8) An object is placed at a distance of 50 cm from a concave lens produces a virtual image at a distance of 10 cm in front of the lens. Draw a diagram to show the formation of image. Calculate focal length of the lens and magnification produced. (Answer : f = -12.5 cm , m = 0.2 )

9) 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. (Answer : u = 30 cm )

10) An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. Find the position and nature of the image. (Answer : v = 3 cm , Virtual and erect)

Rea Also: Light-Reflection and Refraction (Part-1) class 10 Notes, Practice Numerical pdf download