Topic
Displacement
Vectors
2
3 Kinematics
4 Graphs
5 Energy
6 Power
7 Springs
8 Shadows
9 Field of Vision
10 Colors
11 Concave mirrors
12 Convex mirrors
13 Refraction
14 Lenses
15 Optical Power
1
Slides Minutes
9
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18
Convex lenses are thicker in
the middle and thus they
converge light rays.
Concave lenses are thinner
in the middle and thus they
diverge light rays.
Click
Just as with concave mirrors, the characteristics of the image
formed by a converging lens depend upon the location of the object.
There are six "strategic" locations where an object may be placed.
For each location, the image will be formed at a different place and
with different characteristics. We will illustrate the six different
locations and label them as CASE-1 to CASE-6.
Case-1:
Case-2:
Case-3:
Case-4:
Object
Object
Object
Object
at infinity
just beyond 2 F’
at 2F’
between 2F’ and F’
Case-5: Object at F’
Case-6: Object within focal length (f)
Click
CASE-1 : Object at “infinity”
Infinity simply means
“far away”.
No image
Object
NOTE
Since the object is at “infinity”,
all rays arrive parallel.
No image formed
(All rays pass through F)
Click
CASE-2 :
Object
Object just beyond
NOTE
In order to establish
2F’
an image point, all
we need are two
Note-1
intersecting
rays.
Note-2
A ray thatNote-3
comes parallel
AAray
that
goes
through
ray
that goesthrough
throughF.the
F’
is refracted
vertex
goes right
through.
is refracted
parallel.
Image
Image is real (formed by refracted rays)
Inverted (upside down)
Reduced (smaller than object)
Located between F and 2F
This ray is extra
in locating the
image.
Click
CASE-3 :
Object at 2F’
Object
Again:
In order to establish
an image point, all
we need are two
intersecting rays.
Image
Image is real (formed by refracted rays)
Inverted (upside down)
Same size as object
Located at 2F
This ray is extra.
Click
CASE-4 :
Object between 2F’ and F’
Object
Image is real (formed by refracted rays)
Inverted (upside down)
Magnified (larger than object)
Located beyond 2F
Image
Click
CASE-5 :
Object at F’
Object
No image
No image is formed
(rays refract parallel)
Click
CASE-6 :
Object is within focal length
Image
Image is virtual
Object
(formed by extended rays)
Upright
Magnified
Located on same side as object
Click
Lenses Slide: 14. 1
An object whose height is 0.10 m is placed 1.0 m from the
focal point of a converging lens whose focal length is 0.50 m.
Determine the height of the image.
Given
Calculation of object distance
Calculation of image distance
The negative sign indicates inversion.
Click
Lenses Slide: 14. 2
Which of the following lenses corrects
the eye defect known as myopia?
A
B
C
D
Click
Lenses Slide: 14. 3
Nearsightedness
Which type of lens is used to correct hyperopia?
Thicker in the middle than at the ends.
A) Plane
B) Plano-concave
C) Concavo-convex
D) Convex-concave
Myopia (farsightedness) is corrected with a Concavo-concave lens.
Thicker at the ends than in the middle.
Click
Lenses Slide: 14. 4
Lenses Slide:
14. 5
This ray should
go thru F since it
emerges parallel
to the principal axis.
parallel
incident
ray
parallel
refracted
ray
This ray should go
thru V since it is
refracted without
being bent.
This ray should go
thru F since it
arrives parallel to
the principal axis.
Click
Lenses Slide: 14. 6
An object that is 4.0 cm tall is placed 7.0 cm from a converging
lens that has a focal length of 2.0 cm.
Calculate the height of the image formed by this lens.
The negative sign
indicates inversion
and may be ignored.
The image is 1.6 cm tall.
Lenses Slide: 14. 7
As illustrated below, an object is placed infront of a diverging lens.
Image
Which of the following statements correctly describes the
characteristics of the image formed.
A) Virtual, upright, smaller and on the same side as the object
B) Real, upright, smaller and on the same side as the object
C) Virtual, inverted, larger and on the opposide side of the object
D) Real, inverted, larger and on the opposide side of the object
Click
Lenses Slide: 14. 8
Center of lens, V
F’
V
F
Converging lens
Draw a ray connecting the object to the image
in order to establish the center of the lens, V.
Lenses Slide: 14. 9
Karen wants to draw a mural of her dog on a wall. Using a projector
she displays a picture of her dog on the wall. The focal length of the
projector lens is 35 mm. She needs to position the projector so that
the real image of the picture of the dog has a magnification of 40.
How far from the wall must the projector lens be?
Note that since the image is real, it is inverted
and the magnification factor is negative.
Lenses Slide: 14. 10
Construct a ray diagram to locate the image formed by the lens
illustrated below.
This is the symbol for a converging lens
Top (object) point
Bottom (image) point
Image
Bottom (object) point
Top (image) point
Click
… and good luck!