convex lenses. 18.1 lenses a convex lens (or a converging lens) converges parallel light rays...

CONVEX LENSES

18.1 Lenses A convex lens (or a converging lens) converges

parallel light rays passing through it.

Various shapes of convex lenses

Terms for describing lenses

Optical centre is the centre of a lens.

Principal axis is the line passing through the

optical centre and perpendicular to the lens.

For a convex lens, the principal focus is the

point to which the parallel rays converge after

passing through the lens.

Focal length is the distance between the

principal focus and the optical centre.

Focal plane is the plane passing through

the focus and normal to the principal axis.

focal planes

18.2 Properties of images formed by lenses

Construction rules for drawing ray diagrams

For convex lensesRule 1:

The direction of the light ray

passing through the optical centre of

a convex lens remains unchanged.

Rule 2:

The light ray parallel to the

principal axis will pass through the

principal focus F after refracted by

the convex lens.

Rule 3:

The light ray passing through the

principal focus F becomes parallel

to the principal axis after refracted

by the convex lens.

1.     In each of the following cases, a ray is incident on a convex lens. Draw the refracted ray.

(a)

(b)

(c)

(d)

(e)

(f)

Symbols for a convex lens

Drawing ray diagram to locate the image.

1. Draw 2 light rays from the tip of the object to the lens and draw their corresponding refracted rays.

2. Extend the refracted rays. The intersection point is the position of the image.

image

1. Object within F: The image is formed on the same side of the lens.

Nature of the image: erect, magnified, virtual

Image formed by convex lens

2. Object at F: The image is formed at infinity.

3. Object between F and 2F: The image is formed on the other side of the lens

beyond 2F.

Nature of the image: inverted, magnified, real

4. Object at 2F: The image is formed on the other side of the lens at

2F.

Nature of the image: inverted, same size as the

object, real

5. Object beyond 2F: The image is formed on the other side of the lens

between F and 2F.

Nature of the image: inverted, diminished, real

Nature of images formed by convex lenses

6. Distant object: The image is formed on the focal plane on the other

side of the lens.

Nature of the image: inverted, diminished, real

Do experiment 18.2 (book P. 174)

Linear magnification of imagesLinear magnification

Dimensionless (No unit)

Definition:

object theof height

image theof height magnificationlinear m

u

vm

distanceobject

distance image ion magnificatlinear

•Also,

object theof size

image theof sizem

u

vAB

BA

u

v

AB

BA

ABC ~ A’B’C’

height of the object

height of the image magnificationlinear m

If the image is magnified, m _____1.If the image is same size as object, m __ 1If the image is diminished, m _____ 1

>=

<

NotesWorked examples - 1

30 cm

erect, magnifiedvirtual,

the size of the image = __________ cm

the image distance =

The image is

15

Worked examples - 1

Draw the refracted rays for p, q and r.

p

qr

All the refracted rays will pass through / seem to come from image.

2. Locate the image first by drawing two typical rays.

All the refracted rays will pass through / seem to come from image.

3. An object of height 10 cm is placed 30 cm in front of a convex lens. The focal length of the lens is 15 cm. Draw a ray diagram to find the linear magnification of the image.

Rough work:

f = 15 cm, u = 30 cm, u = 2fImage is formed at 2 F, real, inverted and same size.

3.

F F

3. Image distance =

magnification = =

30 cm

10 cm / 10 cm

1

4. An object of height 20 cm is placed 30 cm in front of a convex lens of focal length 20 cm. (a) Show graphically the formation of the

image by using a suitable scale.

Rough work:

f = 20 cm, u = 30 cm, f < u < 2f

Image is formed behind 2f, real, inverted and magnified.

scale

10 cm

20 c

m

Lens, object, F

O F F

2 typical rays

I

image

(b) What are the properties of the image?

(c ) Find the magnification.

Real,inverted, magnified.

m = 40 / 20 = 2

5.   An object of height 10 cm is placed 10 cm in front of a convex lens of focal length 15 cm.

(a) Show graphically the formation of the image by using a suitable scale.

f = 15 cm, u = 10 cm, u < f

Image is virtual, erect, magnified.

5 cm

F F

10 cm

I

(b) What are the properties of the image?

(c ) Find the magnification.

The image is virtual, erect and magnified.

m = 30/10 = 3

6.    Draw the refracted rays of p, q, r and s.

I

7. (a)

I

(b) m = 5 /2. 5= 2

8 (a) real

(b) On the other side of the lens.

I

O

Object is 90 cm from the lens.

9. (a) A convex lens is placed 10 cm in front of an object of height 2 cm. AP is the principal axis of the lens.

If an image of height 6 cm is formed on a screen,

(i) find the magnification of the image,

Magnification = '

'

image s height

object s height

63

2m

(ii) Real and inverted

9 (a) (iii)

IF

(iv) The image distance is 30 cm on the other side of the lens. The focal length of the lens is 7.5 cm.

(b) Move the screen towards the lens.

IF F

New image

(c ) When the lens was close to the object, virtual image is formed. Virtual image can only be seen by eyes and cannot be formed on screen.

(a) Convex lens

(b) Virtual, erect

(c) m = 2

10. A lens is held close over a graph paper.

magnified

(d) (ii) 15 cm

F

(iii) 30 cm

11. Referring to the figure below, an image I is formed when an object is placed on the left hand side of a convex lens. Draw two light rays to locate the position of the object as O.

Answer

12. In the following figure, sketch the refracted rays and locate the image (I).

Answer

I

(a)(a) Is the image real or virtual?

The image is real.

13. An object (letter “P” card) which is 5 cm in height, is placed at 30 cm in front of a convex lens. A clear image is formed on the screen. The focal length of the lens is 20 cm.

Answer

convex lenstranslucent

screen

(b)(b) When a boy is at position (i)(i) X and then (ii)(ii) Y, what will he see?

(i)(i) If the boy is at position X, he will see a letter __

_

(ii)(ii) If the boy is at position Y, he will see a letter ___

_.

Answer

b

dconvex

lens

translucent screen

Answer

(c)(c) Draw a ray diagram to determine the image dist

ance and magnification. Use the scale shown in t

he figure.

5 cm

(c) (c) Image distance = __________ = __________.

　　　 Magnification ＝───── ＝───＝ _______.

6 x 10 cm 60 cm

Height of image

Height of object

105

2

5 cm

18.3 Measuring the focal length of a convex lens

Projecting the image of a distant object

Focal length = distance between the lens and the

screen

Using a lens-mirror combination

Focal length = distance between the lens and the screen

Experiment 18.4

18.5 The lens formula

The lens formula

fvu

111 (REAL is POSITIVE)

Using the lens formula, do worked examples:

1, 3,4, 5 and 9.

15 minutes

P. 4 Notes – 1

fvu

111

30150

5150

1510110

1

15

1115

11

10

1

v

v

v

v

The image distance is 30 cm.

Negative signMeans that the image is virtual.

cmh

cm

h

h

h

u

vm

i

i

o

i

15

510

30

The image’s size is 15 cm

The image is virtual, erect and magnified.

P. 4 Notes – 3

fvu

111

3030

130

12130

1

15

1115

11

30

1

v

v

v

v

The image distance is 30 cm.

130

30

m

m

u

vm

P. 4 Notes – 4

fvu

111

6060

1130

1

20

1120

11

30

1

vv

v

v

The image is real, inverted and magnified.m = 60/30 = 2

P. 4 Notes – 5

fvu

111

30150

5110

1

15

1115

11

10

1

vv

v

v

The image is virtual, erect and magnified.m = 30/10 = 3

P. 4 Notes – 9fvu

111

5.7

30

41

1

30

1

10

1

f

f

f

Magnification = v / u 6/2 = v/ u v = 3u = 30 cm

Notes P. 14

P. 14 Notes

fvu

111

2060

3160

1

30

1130

1

60

11

uu

u

u

v = -60 cm, f = 30 cm

P. 14 Notes

m = 60/20 = 3

3 = hi / ho

ho = 15/3 cm = 5 cm

P. 14 Notes

fvu

111

cmvv

v

v

1010

1110

1

5

115

11

10

1

The image is real, inverted and same size as object.m = 10/10 = 1

P. 15 Notes

fvu

111

4080

218

1

10

1110

11

8

1

vv

v

v

Convex lens

Virtual, erect and magnified

m = 40/8hi = 5 x 2 cm = 10 cm Area of the image of stamp= 10 x 10= 100 cm2

Lens formula

Book

P. 202 Checkpoint 1, 3

P. 205 Exercise 1, 2, 5, 6, 8, 9

Checkpoint (p.202)

1. (a) Positive (b) Negative (c) Positive (d) Negative

3.

205

11

4

1

111

vv

fvu

The image distance is 20 cm.Since the sign of v is negative, the image is virtual.

P. 205 Exercise

1. A. v = 30 cm f = 10 cm2.

1510

1

30

11

111

uu

fvu

3010

11

15

1

111

vv

fvu

5.

(a)The image distance is 30 cm. The linear magnification of the image is 15

30 = 2

(b) Since v is positive, the image is real.

67.1610

1

25

11

111

uu

fvu

6. (a)

The object distance is 16.7 cm.The linear magnification of the image is 67.16

25

= 1.5.The height of the image is 2 × 1.5 = 3 cm.

143.710

1

25

11

111

uu

fvu

143.7

25The object distance is 7.14 cm.The linear magnification of the image is

= 3.5.The height of the image is 2 × 3.5 = 7 cm.

6 (b)

8. A sharp magnified image is formed. (a)

86.1210

1

45

11

111

uu

fvu

The object distance is 12.9 cm.The linear magnification of the image is

86.12

45 = 3.5.

(b)(i)

a b

b a

The lens is moved by 45 − 12.86 ≈ 32.1 cm.

(b (ii) The linear magnification of the image is

45

86.12 ≈ 0.286.

(1/u) / cm−1

0.167 0.143 0.125 0.111 0.100

(1/v) / cm−1

0.164 0.189 0.208 0.222 0.233

9. (a) The data is tabulated below.

(b) The y-intercept is 0.337 cm−1.

The focal length f is 337.0

1

≈ 2.97 cm.

Revision

Book

P. 173 Checkpoint 1, 3

P. 180 Checkpoint 1, 3

P. 188 Checkpoint 1, 2 (a) (b), 3 (a) (b)

P. 189 Exercise 1, 3, 4, 5, 7, 8, 9, 10

P. 196 Checkpoint 1 - 4

P. 197 Exercise 3, 5, 6

1. R

3. (a) Impossible(b) Possible(c) Impossible. Note that the rays should

converge to a point on the focal plane and the ray passing through the optical centre will not be bent.

P. 173 CheckPoint

Checkpoint (p.180)1. A2. The ray diagram is shown below.

The height of the image is 4 cm.The image distance is 15 cm.

3.

1. A2.(a) Incorrect. The image becomes larger as an object moves from 2F to F.(b)Correct(c)Incorrect. A concave lens form virtual images only.(d) Incorrect

P. 188 Checkpoint

3. (a)The image is inverted, diminished and real.

(b) The image is erect, magnified and virtual.

P. 189 Exercise

1. A convex lens of focal length f is used as a magnifying glass. The object distance should be

D. smaller than f

3. An image is formed by a convex lens. Which of the following statements about the image formed are correct?

(1)Its nature depends on the object distance.(2)It must be erect.(3)It is formed on the focal plane if the

object is placed at infinity.

C. (1) and (3) only

4. (a)

(b)

(c)

(d)

5.

The image is erect, magnified and virtual.

7.

The focal length is 6 × 2 = 12 cm.

8. (a)

(b) The image becomes dimmer.

9.(a) The image forms on the same side as the object.

(b) Initially the object distance is smaller than the focal length. Now, the object distance is between the focal length and twice the focal length. Thus the image changes from erect to inverted and virtual to real.

10. (a) Lenses A and C are convex lenses.

(b) Lens A has a larger focal length.

(c) The nature does not change.

(d) The nature does not change.

P. 196 checkpoints

1 - 4

height of the image

height of the object

linear magnification

image distance

object distance

(a) 2 5 cm

(b) 4 cm 0.5 40 cm

(c) 8 cm 2(d) 1 cm 4(e) 6cm 3 2.67cm

(f) 16 cm 0.5 5 cm

1.

2. An object is placed twice the focal length away from a convex lens. The linear magnification of the image is A. greater than 1.B. equal to 1.C. smaller than 1

3.

The linear magnification of the image is 4 / 2 = 2

10

15

o

i

h

hm

cm 4

65.1

o

o

h

h

4.(a) The linear magnification of the image

= = 1.5.

The object height is 4 cm

(b) The linear magnification of the image decreases.

P. 197 Exercise

3, 5 , 6

3. An object of height 3 cm is placed 6 cm in front of a convex lens. A sharp image of 6 cm is caught by a screen placed on the other side of the lens. What is the image distance?

cmv

v

u

v

h

hm

1263

6o

i

5.(a) The image is diminished. Since the lens used is convex, the image is also inverted and real.

(b) The linear magnification of the image is 3/5 = 0.6.

The focal length of the lens is 7 cm.

5. (c)

6.(a) L1 can form a real image and is thus a convex lens.

(b)The linear magnification of the image is 2/3 = 0.6667 ≈ 0.667.

3

2

u

v

cm 20

505.2

v

v

(c) Let v and u be the image distance and the object distance respectively.

v + u = 50 (1)

1.5v = u (2)Substitute (2) into (1)

u = 1.5 × 20 = 30 cm.The image distance and the object distance are 20 cm and 30 cm respectively.

6 (d)

(e) (i) The focal length of L2 is larger.

(ii) The image is erect, magnified and virtual.

Revision

Book

P. 210 Chapter Exercise

1, 3, 4, 7, 8,

9, 10, 11, 12, 14,

17, 18, 19, 21, 22,

23