Exercise 3 (Lenses) Suggested answers
1. (a) convex lens
(b) virtual, erect
(c) (ii) Image distance = 24 cm, focal length = 12 cm
(d) The image size and image distance both decrease.
(e) A real image is formed in regions where the student cannot observe a clear image, e.g. the image is
formed in front of or behind the retina, or the image is formed between the near point and the eye.
2. (a) erect and virtual
(b) 12 cm from the lens
(c) 24 cm in front of the lens
3. (a) Convex lens.
It is because image is caught on the screen, so it must be real image.
Only convex lens can produce real image.
(OR concave lens would only produce virtual image.)
(b) Height of image = 5 cm x (100/20) = 25 cm
Area of image = 25 x 25 = 625 cm2
(c)
focal length = 16.7 cm
(d) (i) The slide should move towards the lens.
The screen should move away from the lens.
(ii) The object distance should not be less than the focal length 16.7 cm, otherwise the no real image
can be formed on the screen.
4. (a) 1/12 + 1/5 = 1/f
f = 3.53 cm
5 cm
image distance
=
= 0.417
object distance 12 cm
(b)
Magnification =
(c)
If we view the screen through the lens, its image will be formed at the
position of the object.
So, when the object distance is 5 cm, the image distance will be 12 cm.
Distance moved by the lens = 12  5 = 7 cm
12 cm
image distance
Magnification =
=
= 2.4
object distance
5 cm
5. (a) The lens is a convex lens since the image is real.
v
(b) By m = ,
u
v = mu = 4u
Put v = 4u and f = 20 into the lens formula,
1 1 1
 =
u v f
1 1
1
=

u 4u 20
u = 25
 The object distance is 25 cm.
Image distance = mu = 4  25 = 100 cm
6. (a) The lens is a convex lens because
(b)
*
images that can be projected onto a screen must be real.
*
the image can be captured by a screen.
*
image formed by concave lenses are always virtual and cannot be captured by a screen.
*
Only convex lens can form a real image or a magnified image.
(ii) Magnification = v/u = 100/25 = 4
(iii) (1) The projector should be moved away from the screen (or the image distance should be
increased). The lens-to-slide distance (OR the object distance) should be reduced for re-focusing the image
onto the screen.
(2) Since the focal length is larger than the object distance (41-55 mm), the image will be
virtual/cannot be captured by the screen. So Karen’s proposal does not work.
(c) Karen is correct, i.e. the lens should be moved up.
OR
MC
1-5 D B D D D
-----------
6-10
DDCDB
11-15 A D C D B
16-17 C C
1. Only when the object is placed at the focus of the lens will the image coincide with the object. It is
because the refracted beam is parallel to the principal axis and after reflection from the plane mirror, the
parallel beam converges to the focus again.
2. Light ray (from LHS) parallel to the principal axis will be refracted as S.
Light ray (from LHS) passing through F will be refracted as R.
Light ray (from LHS) passing through like the figure will be refracted as Q.
Alternatively, imagine an object placed between the focus and the lens with the bottom at the
intersection point between the ray and the principal axis. Then construct a ray diagram to find the image.
Use the principle that ‘bottom of object to bottom of image’, work out the path of light ray.
4. Erect and diminished image produced must be due to a concave lens. Now the magnification is 1/3, so
the image distance / object distance = 1/3, the lens must therefore be at D.
5. v/u = height of image / height of object = 25/100 = 1/4
u = 2m, => v = 1/2 = 0.5 m
Both convex lens and concave lens can produce diminished image.
For convex lens, diminished image must be real, so v is positive.
1/2 + 1/0.5 = 1/f
=> f = 0.4 m
For concave lens,
1/2 + 1/ -0.5 = 1/f => f = -0.67 m
6. (1). Moving the lamp closer to the lens A would increase the image distance, so it is possible to capture
the image.
(2). Adding a diverging lens would move the image to the right.
(3). From 1/u + 1/v = 1/f, increasing f decreases 1/f, and hence 1/v decreases, and v increases.
8. The following combinations of lenses can give the desired effect (parallel incident beam becomes
parallel narrower beam)
A concave lens placed on LHS and a convex lens placed on RHS can also give parallel beam but must
be wider than before.
13. Since OY > OX, the image is magnified. Thus the object must lie between f and 2f where f is the focal
length of the lens. So statement 1 is correct.
If the point light source is shifted away from the lens, separation OY would decrease.
14. Statement 3, draw a light ray parallel to the principal axis and then draw the refracted ray passing
through the image head, extend this line to meet the principal axis at a point. This is the focus.
Alternatively, use 1/20 + 1/-10 = 1/f => f = -20 cm
15. 1/20 + 1/16 = 1/f
=> f = 8.89 cm
17. When O moves towards L, the image moves away from L. The ratio v / u is increasing, i.e. the image
gets larger and larger.