Image formed by refraction
• Light rays are deflected by refraction through
media with different refractive indexes.
• An image is formed by refraction across flat or
curved interfaces and by passage through
lenses.
4.3 Lenses
Images formed by refraction
Images formed by a thin lens
Image formed by refraction through
a refracting surface.
Rotation
of the ray
at the interface
Light is
caused to
Converge.
Image formed by refraction through
a refracting surface.
Light is caused
to diverge in
a different
direction.
Virtual Image formed by refraction.
Real image formed by refraction.
Why is the pencil bent?
o
o
Converging Lenses
Fatter in the middle.
Cause light to converge toward the optic axis
Image of the tip
1
Parallel light though a converging
lens is focused at the focal point.
Diverging Lenses
Thinner in the middle
Cause light to diverge away from the optic axis
A real image is formed
Ray diagram for a converging
lenses
Ray tracing for lenses
• A line parallel to the lens axis passes
through the focal point
• A line through the center of the lens
passes through undeflected.
Image
Object
f
Images formed by a converging
lens
converging light
Real
Inverted
reduced
converging light
Question
How will an object viewed through a
converging lens appear as the lens is
brought closer to the object?
Real
Inverted
Enlarged
diverging light
Virtual
Upright
Enlarged
At the focal point the image changes
from real to virtual
2
Real Image
Inverted
Magnified
Real Image
Inverted
Virtual image
Upright
Virtual Image
Upright
Magnified
Parallel light though a diverging
lens appears to go through the
focal point.
A virtual image is formed.
Image formed by a diverging lens
Virtual
Upright
Reduced
3
Question
How will the image of an object formed by a
diverging lens change as the lens is
brought closer to the object?
Virtual Image
Upright
Reduced
Virtual image
Upright
Reduced
Virtual image
Upright
Reduced
Thin lens equation.
1 1 1
+ =
p q f
p and q are positive if light passes through
p is positive for real objects
f is positive for converging lenses
f is negative for diverging lenses
q is positive for real images
q is negative for virtual images.
Magnification
M=−
h'
q
=−
h
p
M positive- upright
M negative- inverted
for real image
q is positive – image is inverted
for virtual image
q is negative – image is upright
4
Example
Example
An object is placed 30 cm in front of a converging
lens with focal length 10 cm. Find the object distance
and magnification.
An object is placed 30 cm in front of a converging
lens with focal length 10 cm. Find the object distance
and magnification.
Ray diagram.
1 1 1
+ =
p q f
1 1 1
= −
q f p
Real image
q=
fp
(10)(30)
=
= 15cm
p−f
30 − 10
M=−
Example
35 mm
1.5 m
1 1 1
+ =
p q f
f=10cm
•
1 1 1
= −
q f p
M=−
Inverted
Reduced
Projector lens
An object is placed 30 cm in front of a diverging
lens with a focal length of -10 cm. Find the
image distance and magnification
fp
q=
p−f
q
15
=−
= −0.5
p
30
F
10 cm
( −10)(30)
=
= −7.5cm
30 − ( −10)
q
−7.5
=−
= 0.25
p
30
30 cm
Virtual image
Upright image
reduced
Suppose you want to project the image of a transparency 35
mm high on to a screen that is 1.5 m high using a lens with a
focal length of 10 cm. Where would you position the film? How
far from the lens would you place the screen?
.
Projector lens
35 mm
1.5 m
f=10cm
Find ratio of p and q.
Use magnification
Thin lens eq.
h'
q
M= =−
h
p
1 1 1
+ =
p q f
solve p = f ⎛⎜ 42.8 + 1 ⎞⎟ = 10 ⎛⎜ 43.8 ⎞⎟ = 10.23cm
.
⎝ 42.8 ⎠
⎝ 42.8 ⎠
h'
1.5
=−
= −42.8
h
35x10−3
q = 42.8p
1
1
1
+
=
p 42.8p f
just outside of f
q = 42.8p = 42.8(10.23) = 438cm
5