Types of Lenses
Locating images using ray diagrams
Calculations involving Lenses
Recall mirrors have one
reflective & one opaque surface.
Lenses on the other hand have
no opaque surface.
So instead of light being
reflected as in mirrors,
lights in lens will be refracted
(as it goes through from air to
the lens and back to air)
Converging lens:
•Thickest in the middle
•Parallel incident rays will be refracted and
converge at a single point
Diverging lens:
•Thinnest in the middle
•Parallel incident rays will be refracted and
diverge (spread apart)
In reality, as light goes from AIR to LENS then
out to AIR again, you see 2 refractions (bending
of light).
To simplify it, we only show 1 refraction through
central line.
Incident ray
Emergent ray
Different terminology than the ones used for
mirrors.
Optical Centre
F’
Secondary
Principal focus
O
Principal axis
F
Principal focus
Different terminology than the ones used for
converging lens.
A ray parallel to principal axisA ray through the secondary
principal focus (F’) is refracted
is refracted through the
parallel to principal axis
principal focus (F)
Principal axis
2F’



F’
F = principal focus
F’ = secondary principal focus
O = optical centre
O
F
2F
A ray through Optical
centre (O) continues
straight through without
being refracted.

If Image is on the OPPOSITE SIDE of the lens
from the object = REAL Image

If Image is on the SAME SIDE as object =
Virtual Image
This is the opposite of
mirrors! Don’t get them
mixed up.
VIRTUAL
Image
Object
REAL
Image

Use 2 of the rules to find the image
2F’
F’
O
F
2F

Use the templates given to investigate and
complete the following table:
Beyond 2F’
Size
Attitude
Location
Type
At 2F’
Between
2F’ & F’
At F’
Between
F’ & lens
When object is beyond 2F’, the image will be:
S – smaller
A – inverted
L – between F and 2F
T – Real
2F’
F’
O
F
2F
When object is at 2F’, the image will be:
S – same size
A – inverted
L – at 2F
T – Real
2F’
F’
O
F
2F
When object is between 2F’ and F’, the image will be:
S – larger
A – inverted
L – beyond2F
T – Real
2F’
F’
O
F
2F
When object is at F’
NO IMAGE – lines are parallel
2F’
F’
O
F
2F
When object is between F’ and
the lens, the image will be:
S – larger
A – UPRIGHT
L – same side as object
T – VIRTUAL
2F’
F’
Extend your refracted ray
BACKWARDS to locate
the image
O
F
2F
Remember: The only time you’ll get VIRTUAL
image with converging lens is when the object is
between F’ and O
Extend your refracted ray
BACKWARDS to locate
virtual image
2F’
F’
O
F
2F
Beyond 2F’
At 2F’
Between
2F’ & F’
At F’
Size
Smaller
Same size
Larger
Attitude
Inverted
Inverted
Inverted
Upright
Location
Between 2F At 2F
and F
Beyond 2F
Same side
as object
Type
Real
Real
Virtual
Real
NO IMAGE
Between
F’ & lens
Larger
A ray parallel to principal axis A ray that APPEARS TO PASS
is refracted AS IF IT HAD COME through the secondary
through the principal focus (F) principal focus (F’) is
refracted parallel to p.a
Then EXTEND refracted ray
backwards!
Principal axis
2F
F
O
F’
2F’
A ray through Optical
centre (O) continues
straight through
Note the DIFFERENCE IN PLACEMENT of F and F’ in diverging lens
Use 2 of the rules to locate the image:
Principal axis
2F
F
O
F’
2F’

Use the templates given to investigate and
complete the following table:
Beyond 2F
Size
Attitude
Location
Type
At 2F
Between
2F & F
At F
Between
F & lens
Always the SAME image characteristics no matter
where the object is located:
ANY location
Size
Smaller
Attitude
Upright
Location
Same side as object
Type
Virtual
1 + 1 = 1
do di
f
do
di
Image
Object 2F’
O
F’
f
F
f
2F
1 + 1 = 1
do di
f
do is always +
di is + for real image
– for virtual image
f is + for converging lens
– for diverging lens
1 + 1 = 1
do di
f
do is always +
di is + for real image
– for virtual image
f is + for converging lens
– for diverging lens
Sample Problem:
A converging lens has a focal
length of 17 cm.
A candle is located 48 cm from the
lens.
What type of image will be formed
and where will it be located?
1 + 1 = 1
do di
f
do is always +
di is + for real image
– for virtual image
f is + for converging lens
– for diverging lens
Given:
f = 17cm (converging lens so +)
do = 48 cm (do always +)
Required:
a) type of image?
b) d i ?
Analysis:
Rearrange thin lens equation
1 = 1 - 1
di
f
do
1 + 1 = 1
do di
f
do is always +
di is + for real image
– for virtual image
f is + for converging lens
– for diverging lens
Solution:
1 = 1
– 1
.
di
17cm
48cm
1 = 0.038 cm-1
di
Use inverse function: di = 26 cm
Since di is positive, it’s real image.
Statement:
The image is real image and is
located 26 cm from the lens,
opposite to the object
M = hi = – di
ho
do
do
di
ho
Object 2F’
Image
O
F’
f
F
f
2F
hi
M = hi = – di
ho
do
h0 and h1 are:
+ when measured upward
– when measured downward
M is:
+ for upright image
– for inverted image
M = hi = - di
Sample Problem :
A toy of height 8.4 cm is balanced in
front of converging lens.
h0 and h1 are:
+ when upward
– when downward
An inverted, real image of height 23
cm is noticed on the other side of the
lens.
M is:
+ for upright image
– for inverted image
What is the magnification of the lens?
ho
do
M = hi = - di
ho
do
h0 and h1 are:
+ when upward
– when downward
M is:
+ for upright image
– for inverted image
Given:
ho = 8.4 cm (upward, so +)
hi = – 23 cm (inverted, so – )
Required: M?
Analysis:
Use M = hi
ho
M = hi = - di
ho
do
h0 and h1 are:
+ when upward
– when downward
M is:
+ for upright image
– for inverted image
Solution:
M = hi = – 23 cm
ho
8.4 cm
M = – 2.7
Statement:
The lens has a magnification of – 2.7