Tung Wah Group of Hospitals Kap Yan Directors’ College
F3 Physics Extra Exercise 3
Lenses
Answers and explanations are provided at the end of this file.
Multiple Choice
1. The image of an object formed is smaller than the object when the object is placed 21 cm from a convex
lens. Placed at a distance of 19 cm, the image becomes slightly larger than the object. What is the
approximate focal length of the lens?
A. 20 cm
B. 10 cm
C. 5 cm
D. 18 cm
2. X is a point on the focal plane of a convex lens. A ray XY is
incident on the lens. Which line is the correct emergent ray after
passing through the lens?
A. YA
C. YC
B. YB
D. YD
3. The focal length of the convex lens shown is 15 cm. With suitable adjustment of the position of the screen,
a clear image is formed on it. The image looks like
4. In the figure shown, a light ray PQ is incident on a concave lens. Which line is the correct emergent ray
after passing through the lens?
A. I
B. II
C. III
D. IV
5. In the diagram, the image I of an object O is produced by a
lens. What is the nature and position of this lens?
A.
B.
C.
D.
Convex lens at Q
Concave lens at R
Convex lens at S
Concave lens at T
6. Two lenses A and B are made of the same material. A is a thin lens while B is a thick lens. Which of the
following statements is/are correct?
(1) Both lenses A and B are converging lens.
(2) Lens A has a shorter focal length than B.
(3) Both lenses can form virtual images.
A. (1) only
C. (1) and (2) only
B. (1) and (3) only
D. (1), (2) and (3)
A
B
7. A boy views an airplane in sky through a lens. If he sees an inverted image, which of the following
statements is/are correct?
(1) The lens is a convex lens.
(2) The image is diminished.
(3) The image is real.
A. (1) only
B. (1) and (3) only
C. (2) and (3) only
D. (1), (2) and (3)
Structured Questions
1. The following set-up is used to study the image formation of a convex lens. A letter F is acted as an object.
The object is placed at various distances from the lens, the image distance is then measured.
(a) Sketch the image as seen by the eye.
(2 marks)
(b) When the object distance u is 40 cm, the image distance v is also 40 cm. Find the focal length of the
convex lens.
(2 marks)
2. The figure shows an object OO' placed in front of a convex lens whose foci are F and F'.
(a) Draw a ray diagram to show how the image is formed.
(3 marks)
(b) Complete the paths of the ray O'A (emitted from O') and the beam OB (emitted from O).
(3 marks)
(c) Describe the CHANGE in the nature (real/virtual, erect/inverted) of the image when the object is
moved away from the lens to a position beyond F.
(2
marks)
3. In the figure shown, an object AB is placed in front of a concave lens with foci F and F’. P, Q, R and S are
rays incident on the lens.
(a) Draw the refracted rays of P, Q, R and S and the image of AB.
(b) State the nature (real/virtual, erect/inverted) of the image.
(5 marks)
(2 marks)
(c) Find the height of the image if the object is of height 10 cm.
(2 marks)
(d) Can this image be focused on
(2 marks)
(i) a screen?
(ii) the retina of an eye?
Answers:
Multiple Choice
1-7 B B D A D B D
Explanations to MC
1. Now the image changes from slightly smaller to slightly larger image, the object must moves from a
position slightly farther from 2f (f = focal length) to closer to 2f. Thus the object distance must be within 19 cm
to 21 cm and 2f must be lying between 19 cm to 21 cm.
2. If the incident ray passes through focus F, the refracted ray will be parallel (i.e. ray C). If the incident ray
passes through 2F, the refracted ray is ray D.
3. The object lies between F and 2F, so the image is real, inverted and magnified.
4. Concave lens diverges light rays, so III and IV are impossible. Imagine an object at 2F giving the light ray
PQ (as shown), draw the image (by yourself) and using the principle (tip to tip), it can be found that I is
correct.
5. Erect and diminished image must be due to a concave lens. Object distance must be longer than image
distance because the image is diminished. So the answer is D.
7. Inverted image implies that the lens is a convex lens because concave lens always gives erect image.
Structured Questions
1. (a)
inversion)
(b) (i) 2f = 40
f = 20 cm
(ii) Image height / 2 = 36/45
Image height = 1.6 cm
(2A, 1A for inverted image but no lateral
(1M)
(1A)
(1M)
(1A)
2. (a) and (b)
(c) When the object is moved to a point beyond F, the image will change from virtual to real, and from
erect to inverted.
3. (a) (b)
(b) Virtual and erect
(c) Height of the image = 3.3 cm (from figure) (3 cm acceptable)
(d) (i) The image cannot be focused on a screen because it is a virtual image.
(ii) The image can be focused on the retina of an eye because the lens of the eye focuses light rays on
the retina.