Chapter 19

Lenses (Sample)

A. Key Examples of Exam-type Questions

i Problem-solving strategyHow lenses produce images:

Steps1. principal axis2. convex or concave lens3. scale, object size and distance4. principal focus5. at least 2 light rays

6. if convergent refracted rays:real image I at intersection

7. if divergent refracted rays:virtual image I at intersection ofextended rays

Note:• Arrows should be used to indicate the directions of travel of the rays.• Use dotted lines for extended rays.• Use solid lines for real images and dotted lines for virtual images.

DSE-12-1B Q7DSE-13-1B Q8

Example 1An object PQ of height 4 cm is placed 2 cm in front of a lens L of focal length6 cm as shown. An erect and magni ied image is formed.

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(a) What kind of lens is L? Explain your answer. (2 marks)

(b) Complete the ray diagram on the igure and mark the image as P ′Q ′.Hence ind the linear magni ication of the image. (3 marks)

(c) (i) On the igure, complete the light ray r to show how it travelsafter passing through L. (1 mark)

(ii) Draw a light ray to show how Tom can see P , the head of theobject, through L. (1 mark)

. Solution

(a) L is a convex lens. (1A)

Because only a convex lens can form an erect and magni ied image.(1A)

(b) 1. Find the image distance from the intersection of the extended rays.2. Find m using the ratio of the image distance to the object distance.

The diagram: (2A)

Lenses (Sample) 3

The linear magni ication m = 32 = 1.5. (1A) Two correct rays to ind P ′Q ′: 2A

Mind the use of dotted lines. Dotted linesare used for extended rays behind theobject. (Reference: HKDSE report 2013)

(c) (i) See the above diagram. (1A) Note that the refracted ray should betraced back from Q ′ , instead of P ′ .(Reference: HKDSE report 2012)

(ii) See the above diagram. (1A) Tom is looking at the image, so the lightray appears to diverge from P ′ instead ofP .

î What-if How does the linear magni ication change when the object isapproaching the lens? Ans: decreases

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AL-07-1A Q5Example 2Mike investigates the imageformation of a convex lens byplacing an illuminated object anda screen on each side of the lens. Agraph of y , distance between theobject and the screen against x ,object distance is obtained.

(a) What is the focal length of the lens? (Hint: consider x = 20 cm) (2 marks)

(b) The object distance is now 60 cm.(i) what is the linear magni ication of the image? (1 mark)

(ii) what is the nature of the image? (2 marks)

(c) Find y when x = 5 cm. (2 marks)

(d) Mike thinks that the centre of the image will be dimmer if the centre ofthe lens is covered by a small black paper. Comment on his statement.

(2 marks)

. Solution(a) When x = 20 cm, the image distance is 40−20 = 20 cm. (1M)

Construct a ray diagram for the situation,the focal length f = 20/2 = 10 cm (1A)

(b) (i) The image distance is 72− 60 = 12 cm. Therefore, the linearmagni ication is 12

60= 0.2. (1A)

(ii) inverted, real, diminished (2A) According to HKDSE report 2012, youcan only get 2 marks even through thereare 3 image natures.(c) Applying 1

f= 1

u+ 1

v, we have 1

v= 1

f− 1

u= 1

10− 1

5=−0.1 ⇒ v =−10 cm

(1M)

So, y = 10−5 = 5 cm. (1A) Take v as +ve when you calculate the valueof y .(d) He is incorrect (1A)

When part of the lens is covered, part of the light rays is blocked. Therest of the light rays would travel through the uncovered part andfocus on the screen. As the light rays reaching the screen spread evenlyon the screen, the whole image will be dimmer, instead of the centre ofthe image. (1A) According to the HKALE report 2007,

quite a number of candidates made thesimilar mistake as Mike made.

î What-if If x = 12 cm, what is the image distance and the linear magni icationof the image? Ans: 60 cm; 5

(The answer can be easily obtainedwithout using the lens formula.)

Lenses (Sample) 5

DSE-12-1B Q7DSE-14-1B Q6

Example 3Peter carries out an experiment to determine the focal length of a convexlens with the set-up shown.

(a) (i) Describe how Peter can determine the focal length f of the lens.Illustrate your answer with a diagram. (3 marks)

(ii) What is the linear magni ication of the image formed whenPeter determines the focal length of the lens? (1 mark)

(b) Can Peter measure the focal length of a concave lens using the aboveset-up? If yes, how? If not, why? (2 marks)

(c) When Peter replaces the convex lens with another convex lens ofshorter focal length, he notices that a diminished image is formed.(i) Why? Brie ly explain. (2 marks)

(ii) Compare the brightness of this image with the image mentionedin (a)(ii). (2 marks)

. Solution

(a) (i) The diagram: (2A) 1A for correct light rays to ind the image1A for correct position of imageThe light rays coming from the head of theobject should intersect at the head of theimage. (Reference: HKDSE report 2014)

The focal length of the lens is equal to the distance between thelens and the pin. (1A)

(ii) The magni ication of the image is 1. (1A)

(b) no (1A)

The pin and its image formed are not on the same side of the lens. (1A)

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(c) (i) By the lens formula, 1

f= 1

u+ 1

v. If f decreases, v decreases while

keeping u ixed. (1A) The lens is convex and the image is real,therefore both f and v are +ve.Therefore the linear magni ication decreases. (1A)

(ii) This image is brighter than the previous one (1A)

because the same amount of light spreads over the diminishedimage. (1A) The brightness of the image depends on

the light energy received per unit area.(Reference: HKDSE report 2012)

î What-if How does the measured value of f change if the mirror ismoved farther from the lens? Ans: no change

B. Commonmistakes made by Exam Sitters

• Focal plane and principal focus

see CE-10-2 Q14see DSE-14-1B Q6Parallel light rays converge to a pointon the focal plane after passing througha convex lens. They converge to theprincipal focus only if they are parallel tothe principal axis.

V Light rays from different points at in inity should focus on differentpoints on the focal length.

• Position of image

see CE-09-1 Q5see DSE-12-1B Q7Refracted light rays from the same pointof an object converges to, or appears todiverge from, the corresponding point ofthe image.

V Light from the head (tail) of the object focus on (or appear to divergefrom) the head (tail) of the image.

Lenses (Sample) 7

• Paths of light rays

V The light rays should pass through the image (not stop at it).

• Object and image positions

V Mind the positions of the object and the image, they may not lie on theprincipal axis.

• Position of the lens

see AL-05-2A Q9see DSE-12-1A Q22

V For a ixed object– image separation, there are two positions for thelens to form a sharp image. One gives a magni ied image and the otherone gives a diminished image. exception: f = d/2

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• Magni ied images

V A convex lens can produce a magni ied real image or a magni iedvirtual image. Pay attention to the nature of the image produced whenanswering related problems.

C. Reminders of Exam Syllabus

Convex and concave lenses

B only thin lenses are discussed• convex lens: central part thickerthan the edge

• concave lens: central part thinnerthan the edge

Terms

a lens has two principal foci

• optical centre C : centre of a lens• principal axis: line through C ; perpendicular to the lens• focal plane: parallel incident light rays converge to a point or appear to

diverge from a point of this plane• principal focus F : intersection of the focal plane and the principal axis• focal length f : distance C F

Lenses (Sample) 9

Real images

• light rays intersect at X

⇒ light rays come from the imageposition⇒ real image is formed at X

• Real image can be seen directly(inside a particular region).

• Our brains perceive that theobject is located at X .

Virtual images

• light rays seem to diverge from Y

⇒ no light ray comes from theimage position⇒ virtual image is formed at Y

• Virtual image can be seen directly(inside a particular region).

• Our brains perceive that theobject is located at Y .

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Capturing images with screens

• Real images can be captured by a translucent screen but virtual imagescannot.

Convex lens

rules (2) and (3) are inverse: considerreversibility of light

• ray-tracing rules:1. a light ray passes straight

through C

2. parallel to the principal axis→converge to F

3. through F ′

→ parallel to the principal axis• object O, object distance u, image I , image distance v

Lenses (Sample) 11

• nature of image:

u v erect/inverted real/virtual image size∞ v = f

inverted real

diminishedu > 2 f f < v < 2 f diminishedu = 2 f v = 2 f same size

f < u < 2 f v > 2 f magni iedu = f ∞ — — —u < f v > u erect virtual magni ied

Concave lens

• ray-tracing rules:1. a light ray passes straight

through C

2. parallel to the principal axis→appear to diverge from F ′

3. towards F

→ parallel to the principal axis

• nature of image:

� erect, diminished and virtual

The image must be locatedbetween the focal plane and thelens

Covering part of the lens

• If part of the lens is covered, the whole image can still be seen but itbecomes dimmer.

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Linear magni ication m

consider similar triangles• m = hiho

= v

u• for convex and concave lenses,

convex lens u > 2 f m < 1

u = 2 f m = 1

u < 2 f m > 1

concave lens all m < 1

Image size and brightness

• The larger the size of the image, the dimmer the brightness of the image. for a given object (same setting)

Lens formula

• 1

f= 1

u+ 1

v

• real-is-positive:� v (virtual image): −ve� f (concave lens): −ve� otherwise: +ve

f u v

convex (real image) + + +convex (virtual image) + + −concave − + −