CONVEX LENSES 18 1 Lenses n A convex

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CONVEX LENSES

CONVEX LENSES

18. 1 Lenses n A convex lens (or a converging lens) converges parallel light

18. 1 Lenses n A convex lens (or a converging lens) converges parallel light rays passing through it. Various shapes of convex lenses

Terms for describing lenses n Optical centre is the centre of a lens.

Terms for describing lenses n Optical centre is the centre of a lens.

n Principal axis is the line passing through the optical centre and perpendicular to

n Principal axis is the line passing through the optical centre and perpendicular to the lens.

n For a convex lens, the principal focus is the point to which the

n For a convex lens, the principal focus is the point to which the parallel rays converge after passing through the lens.

n Focal length is the distance between the principal focus and the optical centre.

n Focal length is the distance between the principal focus and the optical centre.

n Focal plane is the plane passing through the focus and normal to the

n 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

18. 2 Properties of images formed by lenses Construction rules for drawing ray diagrams n For convex lenses Rule 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

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

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

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

n Symbols for a convex lens

n Symbols for a convex lens

Drawing ray diagram to locate the image. 1. Draw 2 light rays from the

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

image

n 1. Object within F¢: n The image is formed on the same side

n 1. Object within F¢: n The image is formed on the same side of the lens. n Nature of the image: erect, magnified, virtual Image formed by convex lens

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

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

n 3. Object between F¢ and 2 F¢: n The image is formed on

n 3. Object between F¢ and 2 F¢: n The image is formed on the other side of the lens beyond 2 F. n Nature of the image: inverted, magnified, real

n 4. Object at 2 F¢: n The image is formed on the other

n 4. Object at 2 F¢: n The image is formed on the other side of the lens at 2 F. n Nature of the image: inverted, same size as the object, real

n 5. Object beyond 2 F¢: n The image is formed on the other

n 5. Object beyond 2 F¢: n The image is formed on the other side of the lens between F and 2 F. n Nature of the image: inverted, diminished, real

Nature of images formed by convex lenses n 6. Distant object: n The image

Nature of images formed by convex lenses n 6. Distant object: n The image is formed on the focal plane on the other side of the lens. n Nature of the image: inverted, diminished, real

Do experiment 18. 2 (book P. 174)

Do experiment 18. 2 (book P. 174)

Linear magnification of images Linear magnification n Dimensionless (No unit) n Definition: linear magnification

Linear magnification of images Linear magnification n Dimensionless (No unit) n Definition: linear magnification m = height of the image height of the object

 • Also, ABC ~ A’B’C’ linear magnificat ion m = image distance v

• Also, ABC ~ A’B’C’ linear magnificat ion m = image distance v object distance u

 linear magnification m = height of the image height of the object If

linear magnification m = height of the image height of the object If the image is magnified, m _____1. > If the image is same size as object, m __ 1 = If the image is diminished, m _____ 1 <

Notes Worked examples - 1

Notes Worked examples - 1

15 the size of the image = _____ cm the image distance = 30

15 the size of the image = _____ cm the image distance = 30 cm The image is erect, virtual, magnified

Worked examples - 1 Draw the refracted rays for p, q and r. p

Worked examples - 1 Draw the refracted rays for p, q and r. p r q 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

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

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 = 2 f Image is formed at 2 F, real, inverted and same size.

3. F F

3. F F

3. Image distance = 30 cm magnification = 10 cm / 10 cm =

3. Image distance = 30 cm magnification = 10 cm / 10 cm = 1

4. An object of height 20 cm is placed 30 cm in front of

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

Lens, object, F scale 2 typical rays image 20 cm 10 cm O I

Lens, object, F scale 2 typical rays image 20 cm 10 cm O I F F

(b) What are the properties of the image? Real, inverted, magnified. (c ) Find

(b) What are the properties of the image? Real, inverted, magnified. (c ) Find the magnification. m = 40 / 20 = 2

5. An object of height 10 cm is placed 10 cm in front of

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 10 cm I F F

5 cm 10 cm I F F

(b) What are the properties of the image? The image is virtual, erect and

(b) What are the properties of the image? The image is virtual, erect and magnified. (c ) Find the magnification. m = 30/10 = 3

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

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

7. (a) I (b) m = 5 /2. 5= 2

7. (a) I (b) m = 5 /2. 5= 2

8 (a) real (b) On the other side of the lens. I O Object

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

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 = (ii) Real and inverted

9 (a) (iii) I F (iv) The image distance is 30 cm on the

9 (a) (iii) I F (iv) The image distance is 30 cm on the other side of the lens. The focal length of the lens is 7. 5 cm.

New image F F (b) Move the screen towards the lens. I

New image F F (b) Move the screen towards the lens. I

(c ) When the lens was close to the object, virtual image is formed.

(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.

10. A lens is held close over a graph paper. magnified (a) Convex lens

10. A lens is held close over a graph paper. magnified (a) Convex lens (b) Virtual, erect (c) m = 2

F (d) (ii) 15 cm (iii) 30 cm

F (d) (ii) 15 cm (iii) 30 cm

11. Referring to the figure below, an image I is formed when an object

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).

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

13. An object (letter “P” card) which is 5 cm in height, is placed

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. convex lens translucent screen (a) Is the image real or virtual? The image is real. Answer

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

(b) When a boy is at position (i) X and then (ii) Y, what will he see? b (i) If the boy is at position X, he will see a letter ___ (ii) If the boy is at position Y, he will see a letter ____. convex lens d translucent screen Answer

(c) Draw a ray diagram to determine the image distance and magnification. Use the

(c) Draw a ray diagram to determine the image distance and magnification. Use the scale shown in the figure. Answer 5 cm

6 x 10 cm = _____. 60 cm (c) Image distance = _____    

6 x 10 cm = _____. 60 cm (c) Image distance = _____     Height of image 10 2 Magnification=───── =───= _______. 5 Height of object 5 cm

18. 3 Measuring the focal length of a convex lens Projecting the image of

18. 3 Measuring the focal length of a convex lens Projecting the image of a distant object n Focal length = distance between the lens and the screen

Using a lens-mirror combination n Focal length = distance between the lens and the

Using a lens-mirror combination n Focal length = distance between the lens and the screen Experiment 18. 4

18. 5 The lens formula (REAL is POSITIVE)

18. 5 The lens formula (REAL is POSITIVE)

Using the lens formula, do worked examples: 1, 3, 4, 5 and 9. 15

Using the lens formula, do worked examples: 1, 3, 4, 5 and 9. 15 minutes

P. 4 Notes – 1 Negative sign Means that the image is virtual. The

P. 4 Notes – 1 Negative sign Means that the image is virtual. The image distance is 30 cm.

The image’s size is 15 cm The image is virtual, erect and magnified.

The image’s size is 15 cm The image is virtual, erect and magnified.

P. 4 Notes – 3 The image distance is 30 cm.

P. 4 Notes – 3 The image distance is 30 cm.

P. 4 Notes – 4 The image is real, inverted and magnified. m =

P. 4 Notes – 4 The image is real, inverted and magnified. m = 60/30 = 2

P. 4 Notes – 5 The image is virtual, erect and magnified. m =

P. 4 Notes – 5 The image is virtual, erect and magnified. m = 30/10 = 3

P. 4 Notes – 9 Magnification = v / u 6/2 = v/ u

P. 4 Notes – 9 Magnification = v / u 6/2 = v/ u v = 3 u = 30 cm

Notes P. 14

Notes P. 14

P. 14 Notes v = -60 cm, f = 30 cm

P. 14 Notes v = -60 cm, f = 30 cm

P. 14 Notes m = 60/20 = 3 3 = hi / ho ho

P. 14 Notes m = 60/20 = 3 3 = hi / ho ho = 15/3 cm = 5 cm

P. 14 Notes The image is real, inverted and same size as object. m

P. 14 Notes The image is real, inverted and same size as object. m = 10/10 = 1

P. 15 Notes Convex lens Virtual, erect and magnified m = 40/8 hi =

P. 15 Notes Convex lens Virtual, erect and magnified m = 40/8 hi = 5 x 2 cm = 10 cm Area of the image of stamp = 10 x 10 = 100 cm 2

Lens formula Book P. 202 Checkpoint 1, 3 P. 205 Exercise 1, 2, 5,

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

Checkpoint (p. 202) 1. (a) Positive (b) Negative (c) Positive (d) Negative

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

3. 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 cm 2.

P. 205 Exercise 1. A. v = 30 cm f = 10 cm 2.

5. (a)The image distance is 30 cm. The linear = 2 magnification of the

5. (a)The image distance is 30 cm. The linear = 2 magnification of the image is (b) Since v is positive, the image is real.

6. (a) The object distance is 16. 7 cm. The linear magnification of the

6. (a) The object distance is 16. 7 cm. The linear magnification of the image is = 1. 5. The height of the image is 2 × 1. 5 = 3 cm.

6 (b) The object distance is 7. 14 cm. The linear magnification of the

6 (b) The 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.

8. A sharp magnified image is formed. (a) The object distance is 12. 9

8. A sharp magnified image is formed. (a) The object distance is 12. 9 cm. The linear magnification of the image is = 3. 5.

(b)(i) a b b a The lens is moved by 45 − 12. 86

(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 ≈ 0. 286.

(b (ii) The linear magnification of the image is ≈ 0. 286.

9. (a) The data is tabulated below. (1/u) / 0. 167 0. 143 0.

9. (a) The data is tabulated below. (1/u) / 0. 167 0. 143 0. 125 0. 111 0. 100 cm− 1 (1/v) / 0. 164 0. 189 0. 208 0. 222 0. 233 cm− 1

(b) The y-intercept is 0. 337 cm− 1. The focal length f is ≈

(b) The y-intercept is 0. 337 cm− 1. The focal length f is ≈ 2. 97 cm.

Revision Book P. 173 Checkpoint 1, 3 P. 180 Checkpoint 1, 3 P. 188

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

P. 173 Check. Point 1. R 3. (a) (b) (c) Impossible Possible Impossible. Note

P. 173 Check. Point 1. R 3. (a) (b) (c) Impossible Possible 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.

Checkpoint (p. 180) 1. A 2. The ray diagram is shown below. The height

Checkpoint (p. 180) 1. A 2. The ray diagram is shown below. The height of the image is 4 cm. The image distance is 15 cm.

3.

3.

P. 188 Checkpoint 1. A 2. (a) Incorrect. The image becomes larger as an

P. 188 Checkpoint 1. A 2. (a) Incorrect. The image becomes larger as an object moves from 2 F to F. (b) Correct (c) Incorrect. A concave lens form virtual images only. (d) Incorrect

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

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

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

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

P. 189 Exercise 1. A convex lens of focal length f is used as

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

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

5. The image is erect, magnified and virtual.

5. The image is erect, magnified and virtual.

7. The focal length is 6 × 2 = 12 cm.

7. The focal length is 6 × 2 = 12 cm.

8. (a) (b) The image becomes dimmer.

8. (a) (b) The image becomes dimmer.

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

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

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

P. 196 checkpoints 1 - 4

1. height of the image height of linear the object magnification (a) (b) (c)

1. height of the image height of linear the object magnification (a) (b) (c) 4 cm 8 cm (d) (e) (f) image distance object distance 2 5 cm 0. 5 40 cm 2 1 cm 6 cm 4 2. 67 cm 3 16 cm 0. 5 5 cm

2. An object is placed twice the focal length away from a convex lens.

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

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

4. (a) The linear magnification of the image = = 1. 5. (b) The

4. (a) The linear magnification of the image = = 1. 5. (b) The linear magnification of the image decreases. The object height is 4 cm

P. 197 Exercise 3, 5 , 6

P. 197 Exercise 3, 5 , 6

3. An object of height 3 cm is placed 6 cm in front of

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?

5. (a) The image is diminished. Since the lens used is convex, the image

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.

5. (c) The focal length of the lens is 7 cm.

5. (c) The focal length of the lens is 7 cm.

6. (a) L 1 can form a real image and is thus a convex

6. (a) L 1 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.

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

(c) Let v and u be the image distance and the object distance respectively. v + u = 50 (1) 1. 5 v = 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)

6 (d)

(e) (i) The focal length of L 2 is larger. (ii) The image is

(e) (i) The focal length of L 2 is larger. (ii) The image is erect, magnified and virtual.

Revision Book P. 210 Chapter Exercise 1, 3, 4, 7, 8, 9, 10, 11,

Revision Book P. 210 Chapter Exercise 1, 3, 4, 7, 8, 9, 10, 11, 12, 14, 17, 18, 19, 21, 22, 23