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Refraction of Light & Lenses Class 10 Science PPT (Ch 9)

Class X Science 🖥️ PPT Slides Ch 9: Light: Reflection and Refraction
⬇ Download PDF Updated 07 Nov 2026
Refraction of Light and LensesClass 10 Science · Chapter 9 · Light (refraction, lenses, lens formula and power) · Solved numericals and practice questions
6sections
22MCQs
15Q&A
15diagrams
19PPT slides
Chapter notes on refraction of light and lenses, written from our PPT, with every numerical worked out and checked, followed by practice questions. Tap an option in the MCQ section to check it at once, and tap Check answer anywhere for the written answer. To get the full PPT, use the Download PDF button at the top of this page.

Section 1  ·  Refraction of light

What is refraction?

Refraction is the bending of light when it passes from one transparent medium to another. The light changes its direction at the boundary. This is why a pencil in a glass of water appears bent.

A pencil in a glass of water that appears bent, with light changing direction at the water surface
Slide 2: what is refraction?

Refraction around us

A coin at the bottom of a bowl of water appears raised, a swimming pool looks shallower than it is, and a lemon in a glass of water looks bigger. All these happen because of refraction of light.

A coin appearing raised in a bowl of water, a shallow-looking pool and a lemon looking bigger in a glass
Slide 3: refraction around us

Why does light bend?

Light travels at different speeds in different media: about 3 × 10⁸ m/s in air (and vacuum) and about 2 × 10⁸ m/s in glass. Like a car moving from a road onto mud, when light passes from a fast medium to a slow one it changes direction at the boundary, so it bends.

A car moving from a fast road onto slow mud compared with light moving from air at 3 times 10 to the 8 metres per second into glass at 2 times 10 to the 8
Slide 4: why does light bend?

Which way does light bend?

Passing fromBendsAngles
Rarer to denser (air to glass)Towards the normalAngle of refraction r is less than angle of incidence i (i > r)
Denser to rarer (glass to air)Away from the normalAngle of refraction r is greater than angle of incidence i (i < r)
A ray bending towards the normal when going from air into glass and away from the normal when going from glass into air
Slide 5: which way does light bend?

Laws of refraction

  1. The incident ray, the refracted ray and the normal to the interface at the point of incidence all lie in the same plane
  2. Snell’s law: for light of a given colour and for the given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant:
    sin i ÷ sin r = constant (n)

Refractive index

The refractive index (n) of a medium is the ratio of the speed of light in vacuum to the speed of light in that medium:

n = c ÷ v = (speed of light in vacuum) ÷ (speed of light in the medium)

Here c = 3 × 10⁸ m/s. A higher n means the medium is optically denser and light travels more slowly in it.

MediumRefractive indexSpeed of light (c ÷ n)
Air1.0003about 3 × 10⁸ m/s
Water1.33about 2.26 × 10⁸ m/s
Glass1.52about 1.97 × 10⁸ m/s
Diamond2.42about 1.24 × 10⁸ m/s
A ladder of refractive indices: air 1.0003, water 1.33, glass 1.52 and diamond 2.42, with n equal to c over v
Slide 7: refractive index

Worked example: Snell’s law

Light goes from air into glass of refractive index 1.5, with an angle of incidence of 30°. Then sin r = sin 30° ÷ 1.5 = 0.5 ÷ 1.5 = 0.333, so r ≈ 19.5°, which is less than 30°, because the light bends towards the normal.

Refraction through a rectangular glass slab

When a ray passes through a glass slab, it bends towards the normal on entering and away from the normal on leaving. The emergent ray is parallel to the incident ray (angle of incidence i = angle of emergence e), but it is shifted sideways. This shift is called the lateral displacement.

A ray passing through a rectangular glass slab, with the emergent ray parallel to the incident ray and the lateral displacement marked
Slide 8: refraction through a glass slab

Section 2  ·  Lenses

Two types of lenses

Convex lensConcave lens
ShapeThick in the middle (thicker at the centre than at the edges)Thin in the middle
Action on lightConverging lens: it brings parallel rays togetherDiverging lens: it spreads parallel rays apart
Focal lengthPositiveNegative
A convex lens converging parallel rays to a point and a concave lens diverging them
Slide 9: two types of lenses

Know your lens

  • Optical centre (O): the central point of the lens; a ray through it passes undeviated
  • Principal axis: the line through the optical centre joining the centres of curvature
  • Principal focus: the point where parallel rays converge (convex) or appear to diverge from (concave). A lens has two foci (F₁ and F₂), one on each side, whereas a mirror has only one
  • Focal length (f): the distance from the optical centre to the principal focus

Three rules for lens ray diagrams

RayPath after the lens
1. A ray parallel to the principal axisPasses through the (second) principal focus F₂ (convex), or appears to come from it (concave)
2. A ray passing through the optical centreGoes straight, without any deviation
3. A ray passing through the first principal focus F₁Emerges parallel to the principal axis

Images formed by a convex lens

Position of the objectPosition of the imageSizeNature
At infinityAt the focus F₂Highly diminished, point-sizedReal and inverted
Beyond 2F₁Between F₂ and 2F₂DiminishedReal and inverted
At 2F₁At 2F₂Same sizeReal and inverted
Between F₁ and 2F₁Beyond 2F₂EnlargedReal and inverted
At F₁At infinityHighly enlargedReal and inverted
Between F₁ and OOn the same side as the objectEnlargedVirtual and erect (a magnifying glass)
Images by a convex lens when the object is at infinity, beyond 2F1 and at 2F1
Slide 12: images by a convex lens, part 1
Images by a convex lens when the object is between F1 and 2F1, at F1 and between F1 and the lens
Slide 13: images by a convex lens, part 2

Image by a concave lens

A concave lens always forms a virtual, erect and diminished image, between the optical centre and the focus on the same side as the object, wherever the object is placed. That is why it makes small images.

A concave lens always forming a virtual, erect and diminished image of an object placed anywhere
Slide 14: image by a concave lens

Lens formula and sign convention

1/v − 1/u = 1/f
  • u = object distance, v = image distance and f = focal length, all measured from the optical centre
  • New Cartesian sign convention: distances measured to the left of the lens (in the direction opposite to the incident light) are negative, and distances to the right are positive
  • The object is on the left, so u is always negative
  • Convex lens: f is positive. Concave lens: f is negative
The lens formula 1 over v minus 1 over u equals 1 over f with the sign convention for a lens
Slide 15: the lens formula and sign convention

Magnification

m = h₂ ÷ h₁ = v ÷ u

where h₁ is the height of the object and h₂ the height of the image. m > 1 means the image is enlarged and m < 1 means it is diminished. A positive m means the image is erect and virtual, and a negative m means it is inverted and real.

The magnification formula m equals h2 over h1 equals v over u, with an ant enlarged and a tree diminished
Slide 16: magnification

Power of a lens

P = 1 ÷ f (f in metres)
  • The SI unit of power is the dioptre (D). 1 D is the power of a lens whose focal length is 1 metre
  • A convex lens has positive power and a concave lens has negative power. A doctor’s prescription of +2.0 D means a convex lens of focal length 50 cm
  • When lenses are placed in contact, the combined power is P = P₁ + P₂
The power of a lens P equals 1 over f in metres, measured in dioptres, with a doctor’s prescription of plus 2.0 D
Slide 17: power of a lens

Lenses around us

Lenses are used in magnifying glasses, spectacles, cameras, microscopes and telescopes. Our own eye has a convex lens.

A magnifying glass, spectacles, a camera, a microscope, a telescope and an eye, all using lenses
Slide 18: lenses around us

Section 3  ·  Solved numericals

ProblemSolution
A convex lens has a focal length of 10 cm. A 2.0 cm tall object is placed 15 cm from it. Find the position, nature and size of the image.u = −15 cm, f = +10 cm. From 1/v − 1/u = 1/f: 1/v = 1/10 + 1/(−15) = 1/30, so v = +30 cm (on the other side of the lens). m = v/u = 30/(−15) = −2, so the image height = −2 × 2.0 = −4 cm. The image is real, inverted and enlarged (4 cm tall)
A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object? Find the magnification.f = −15 cm and v = −10 cm (virtual image on the same side). 1/u = 1/v − 1/f = −1/10 + 1/15 = −1/30, so u = −30 cm. m = v/u = (−10)/(−30) = +1/3: the image is virtual, erect and diminished
A 5 cm tall object is placed 25 cm from a convex lens of focal length 10 cm. Find the image.1/v = 1/10 − 1/25 = 3/50, so v = +50/3 ≈ 16.7 cm. m = v/u = (50/3)/(−25) = −2/3, so the image is about 3.3 cm tall, real, inverted and diminished
A convex lens forms a real, inverted image of a needle at 50 cm from the lens, the same size as the needle. Find the focal length and power of the lens, and where the needle is.Same size means the object is at 2F₁, so u = −50 cm and v = +50 cm. 1/f = 1/v − 1/u = 1/50 + 1/50 = 1/25, so f = +25 cm and P = 1/0.25 = +4 D. The needle is 50 cm from the lens
An object is 20 cm from a convex lens of f = 10 cm. Where is the image?u = −20 cm: 1/v = 1/10 − 1/20 = 1/20, so v = +20 cm and m = −1. The image is at 2F₂, the same size, real and inverted
An object is 30 cm from a concave lens of f = −20 cm. Where is the image and what is m?1/v = 1/(−20) + 1/(−30) = −5/60, so v = −12 cm and m = (−12)/(−30) = +0.4. The image is virtual, erect and diminished, 12 cm from the lens on the same side as the object
Find the power of lenses of focal length (a) +50 cm, (b) −25 cm, (c) +20 cm.(a) P = 1/0.5 = +2 D (convex). (b) P = 1/(−0.25) = −4 D (concave). (c) P = 1/0.2 = +5 D (convex)
Two lenses of powers +2.0 D and −0.5 D are in contact. Find the combined power and focal length.P = 2.0 + (−0.5) = +1.5 D, so f = 1/1.5 ≈ 0.67 m (66.7 cm), which is a converging combination
Find the speed of light in glass of refractive index 1.52 and in water of refractive index 1.33.v = c ÷ n. Glass: 3 × 10⁸ ÷ 1.52 ≈ 1.97 × 10⁸ m/s. Water: 3 × 10⁸ ÷ 1.33 ≈ 2.26 × 10⁸ m/s
A one-page revision map of refraction: bending of light, Snell’s law, refractive index, glass slab, convex and concave lenses, the lens formula, magnification and power
Slide 19: refraction, one-page revision

Section 4  ·  Practice: multiple choice questions

Tap an option to check it at once. Open Check answer for the reason.

1Refraction of light is: [1]

(a) the absorption of light(b) the bouncing back of light from a surface(c) the splitting of white light only(d) the bending of light when it passes from one medium to another
Check answer
(d) the bending of light when it passes from one medium to another
The light changes direction at the boundary because its speed changes.

2Light bends towards the normal when it passes from: [1]

(a) a rarer to a denser medium(b) vacuum to vacuum(c) glass to diamond only(d) a denser to a rarer medium
Check answer
(a) a rarer to a denser medium
In a denser medium light slows down, so i > r.

3Snell’s law states that for a given pair of media: [1]

(a) sin i × sin r is a constant(b) i + r is a constant(c) sin i ÷ sin r is a constant(d) i ÷ r is a constant
Check answer
(c) sin i ÷ sin r is a constant
This constant is the refractive index.

4The refractive index of a medium is: [1]

(a) the angle of incidence divided by the angle of emergence(b) the speed of light in vacuum divided by its speed in the medium(c) the density of the medium(d) the speed in the medium divided by the speed in vacuum
Check answer
(b) the speed of light in vacuum divided by its speed in the medium
n = c ÷ v.

5Which of these has the highest refractive index? [1]

(a) Glass (1.52)(b) Water (1.33)(c) Diamond (2.42)(d) Air (1.0003)
Check answer
(c) Diamond (2.42)
A higher n means light travels more slowly.

6In a rectangular glass slab, the emergent ray is: [1]

(a) along the normal(b) perpendicular to the incident ray(c) reflected back(d) parallel to the incident ray but laterally displaced
Check answer
(d) parallel to the incident ray but laterally displaced
The angle of emergence equals the angle of incidence.

7A convex lens is also called a: [1]

(a) converging lens(b) reflecting lens(c) plane lens(d) diverging lens
Check answer
(a) converging lens
It is thick in the middle.

8A lens has: [1]

(a) three principal foci(b) two principal foci(c) one principal focus(d) no principal focus
Check answer
(b) two principal foci
A mirror has only one focus.

9A ray parallel to the principal axis, after passing through a convex lens: [1]

(a) goes straight(b) becomes parallel to the axis(c) is reflected back(d) passes through the principal focus
Check answer
(d) passes through the principal focus
Rule 1 of the lens ray diagrams.

10A ray passing through the optical centre of a lens: [1]

(a) is absorbed(b) bends towards the axis(c) goes straight without deviation(d) passes through the focus
Check answer
(c) goes straight without deviation
Rule 2.

11For an object at infinity, a convex lens forms an image that is: [1]

(a) at the focus, highly diminished, real and inverted(b) at 2F₂ and the same size(c) at infinity(d) virtual and erect
Check answer
(a) at the focus, highly diminished, real and inverted
The image is point-sized.

12When an object is placed at 2F₁ of a convex lens, the image is formed: [1]

(a) at F₂, point-sized(b) at 2F₂, the same size, real and inverted(c) beyond 2F₂, enlarged(d) on the same side, virtual
Check answer
(b) at 2F₂, the same size, real and inverted
u = v = 2f and m = −1.

13A convex lens works as a magnifying glass when the object is: [1]

(a) at infinity(b) beyond 2F₁(c) between F₁ and the optical centre(d) at 2F₁
Check answer
(c) between F₁ and the optical centre
The image is virtual, erect and enlarged.

14A concave lens always forms an image that is: [1]

(a) virtual and enlarged(b) real and inverted(c) real and enlarged(d) virtual, erect and diminished
Check answer
(d) virtual, erect and diminished
It does so wherever the object is placed.

15The lens formula is: [1]

(a) 1/v + 1/u = 1/f(b) 1/v − 1/u = 1/f(c) u + v = f(d) v − u = f
Check answer
(b) 1/v − 1/u = 1/f
u, v and f are measured from the optical centre with sign convention.

16In the sign convention for lenses, the object distance u is: [1]

(a) always negative(b) always positive(c) zero(d) positive for a concave lens
Check answer
(a) always negative
The object is placed on the left of the lens.

17A negative magnification means the image is: [1]

(a) real and erect(b) virtual and erect(c) real and inverted(d) virtual and inverted
Check answer
(c) real and inverted
A positive m means erect and virtual.

18The power of a lens of focal length 20 cm is: [1]

(a) +0.2 D(b) +20 D(c) −5 D(d) +5 D
Check answer
(d) +5 D
P = 1/0.2 = 5 D for a convex lens.

19A convex lens of f = 10 cm has an object 15 cm in front of it. The image distance is: [1]

(a) 30 cm(b) 6 cm(c) 5 cm(d) 25 cm
Check answer
(a) 30 cm
1/v = 1/10 − 1/15 = 1/30.

20A doctor prescribes a lens of +2.0 D. This is a: [1]

(a) convex lens of focal length 2 cm(b) convex lens of focal length 50 cm(c) concave lens of focal length 2 cm(d) concave lens of focal length 50 cm
Check answer
(b) convex lens of focal length 50 cm
f = 1/P = 1/2 m = 50 cm, and positive power means convex.
ASSERTION-REASON

21Assertion (A): A pencil partly dipped in water appears bent at the water surface.
Reason (R): Light changes direction when it passes from water to air. [1]

(a) Both A and R are true, and R is the correct explanation of A(b) Both A and R are true, but R is not the correct explanation of A(c) A is true, but R is false(d) A is false, but R is true
Check answer
(a) Both A and R are true, and R is the correct explanation of A
Both statements are true, and the change in direction of light at the boundary (refraction) is the reason, so R explains A.
ASSERTION-REASON

22Assertion (A): A concave lens can form a real, inverted image of a real object.
Reason (R): A concave lens always forms a virtual, erect and diminished image. [1]

(a) Both A and R are true, and R is the correct explanation of A(b) Both A and R are true, but R is not the correct explanation of A(c) A is true, but R is false(d) A is false, but R is true
Check answer
(d) A is false, but R is true
A is false: a concave lens always forms a virtual, erect and diminished image. R is true.

Section 5  ·  Practice: short answers

EXAM QUESTION

Q1What is refraction of light? Why does it take place? [3]

Check answer
Refraction is the bending of light when it passes from one transparent medium to another. It takes place because the speed of light is different in different media: for example about 3 × 10⁸ m/s in air and about 2 × 10⁸ m/s in glass.
EXAM QUESTION

Q2State the laws of refraction. [3]

Check answer
1. The incident ray, the refracted ray and the normal all lie in the same plane. 2. Snell’s law: sin i ÷ sin r is a constant for a given pair of media (the refractive index n).
EXAM QUESTION

Q3Define refractive index. What does a higher refractive index mean? [3]

Check answer
The refractive index is the ratio of the speed of light in vacuum to the speed of light in the medium (n = c/v). A higher value means the medium is optically denser and light travels more slowly in it. For example diamond has n = 2.42 and water has 1.33.
EXAM QUESTION

Q4Light goes from air into glass (n = 1.5) at an angle of incidence of 30°. Find the angle of refraction. [3]

Check answer
sin r = sin 30° ÷ 1.5 = 0.5 ÷ 1.5 = 0.333, so r ≈ 19.5°. It is less than 30° because the light bends towards the normal.
EXAM QUESTION

Q5Describe the refraction of light through a rectangular glass slab. [4]

Check answer
The ray bends towards the normal on entering the slab and away from the normal on leaving. The emergent ray is parallel to the incident ray (angle of emergence = angle of incidence), but it is shifted sideways, which is called lateral displacement.
EXAM QUESTION

Q6Differentiate between a convex lens and a concave lens. [3]

Check answer
A convex lens is thick in the middle and is a converging lens with positive focal length. A concave lens is thin in the middle and is a diverging lens with negative focal length. A convex lens can form real or virtual images, but a concave lens always forms a virtual, erect and diminished image.
EXAM QUESTION

Q7Define optical centre, principal focus and focal length of a lens. [4]

Check answer
The optical centre is the central point of the lens; a ray through it goes undeviated. The principal focus is the point where rays parallel to the principal axis converge (convex lens) or appear to diverge from (concave lens). The focal length is the distance between the optical centre and the principal focus.
EXAM QUESTION

Q8State the three rules used to draw ray diagrams for a lens. [3]

Check answer
A ray parallel to the principal axis passes through the principal focus on the other side. A ray through the optical centre goes straight. A ray through the first focus emerges parallel to the principal axis.
EXAM QUESTION

Q9Where must an object be placed so that a convex lens works as a magnifying glass? State the nature of the image. [3]

Check answer
Between the focus F₁ and the optical centre. The image is virtual, erect, enlarged and on the same side of the lens as the object.
EXAM QUESTION

Q10Write the lens formula and the formula for magnification. State the sign convention. [4]

Check answer
Lens formula: 1/v − 1/u = 1/f. Magnification: m = h₂/h₁ = v/u. Distances are measured from the optical centre; those to the left (against the incident light) are negative and those to the right are positive, so u is always negative, f is positive for a convex lens and negative for a concave lens.
EXAM QUESTION

Q11A 2 cm tall object is placed 15 cm from a convex lens of focal length 10 cm. Find the image position, nature and size. [4]

Check answer
1/v = 1/10 − 1/15 = 1/30, so v = +30 cm. m = v/u = −2, so the image is 4 cm tall. It is real, inverted and enlarged.
EXAM QUESTION

Q12A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object, and what is the magnification? [4]

Check answer
1/u = 1/v − 1/f = −1/10 + 1/15 = −1/30, so u = −30 cm. m = v/u = +1/3: the image is virtual, erect and diminished.
EXAM QUESTION

Q13A convex lens forms a real, inverted image of a needle at 50 cm from the lens, the same size as the needle. Find its focal length and power. [4]

Check answer
The object is at 2F₁, so u = −50 cm and v = 50 cm. 1/f = 1/50 + 1/50 = 1/25, so f = 25 cm and P = 1/0.25 = +4 D.
EXAM QUESTION

Q14Define the power of a lens and its unit. Find the power of a lens of focal length −25 cm. [3]

Check answer
Power is the reciprocal of the focal length in metres (P = 1/f); its unit is the dioptre (D). For f = −25 cm = −0.25 m, P = 1/(−0.25) = −4 D (a concave lens).
EXAM QUESTION

Q15What lens does a doctor prescribe with a power of +2.0 D? Find its focal length. [2]

Check answer
A convex lens, since the power is positive. f = 1/P = 1/2 = 0.5 m = 50 cm.

Section 6  ·  Frequently asked questions

What is refraction of light?
The bending of light when it passes from one transparent medium to another, caused by the change in the speed of light.
What is the difference between refraction and reflection?
In reflection, light bounces back into the same medium. In refraction, light passes into another medium and changes direction.
What is Snell’s law?
For a given pair of media, sin i ÷ sin r is a constant, equal to the refractive index.
What is the sign convention for a lens?
Distances are measured from the optical centre; those to the left are negative and those to the right are positive. The object distance u is always negative, f is positive for a convex lens and negative for a concave lens.
How is the power of a lens calculated?
P = 1 ÷ f, with f in metres. It is measured in dioptres (D), and is positive for a convex lens and negative for a concave lens.
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