Lenses
• Knowing how curved mirrors reflect light can
help you understand how lenses affect light
rays.
• A lens is a curved piece of transparent
material. Light refracts as it passes through a
lens, causing the light rays to bend.
Basic Lens Shapes
•A converging lens is a
lens that is thickest in the
middle and causes
incident parallel rays to
converge at a single
point.
•A diverging lens is a lens
that is thinnest in the
middle and causes
incident parallel rays to
spread apart after
refraction.
Diverging (Concave) Lenses
• A concave lens is thinner and flatter in the
middle than around the edges.
• Light passing through the thicker, more curved
areas of the lens will bend more than light
that passes through the flatter, thinner area in
the middle.
• Rays of light are spread out (diverged) after
passing through the lens.
Optical Center
Secondary Principal Focus
(F’)
Converging (Convex) Lenses
• A convex lens is thinker in the middle than
around the edges.
• This thicker middle causes the refracting light
rays to come together (converge)
Optical Center
Secondary Principal Focus
(F’)
Three Rules for a Converging Lens
1. A ray parallel to the principal axis is refracted
through the principal focus.
2. A ray through the secondary principal focus is
refracted parallel to the principal axis.
3. A ray through the optical center continues
straight through without being refracted.
Using Ray Diagrams with Converging
Lenses
• Draw the ray parallel to the principal axis.
Draw the refracted ray so that it passes
through the principal focus.
• Draw a ray from the top of the object through
the middle of the lens. This ray is undeviated.
Where the rays meet, that is where the image
is.
• You can also draw the incident ray
passing through F’ on the way to the
lens. The result will be a parallel ray to
the principal axis.
Locating Images – Object beyond 2F’
•Image is
•Smaller
•Inverted
•Between F and 2F
•Real
Locating Images – Object at 2F’
•Image is
•Same size
•Inverted
• At 2F
•Real
Locating Images – Object between 2F’
and F’
•Image is
•Larger
•Inverted
•Beyond 2F
•Real
Locating Images – Object AT F’
•Image is
•No clear image due to
emergent rays being
parallel
Locating Images – Between F’ and Lens
•Image is
•Larger
•Upright
•Behind the lens
•Virtual
Three Rules for a Diverging Lens
1. A ray parallel to the principal axis is refracted
as if it had come through the principal focus.
2. A ray that appears to pass through the
secondary principal focus is refracted parallel
to the principal axis.
3. A ray through the optical center continues
straight through without being refracted.
•A diverging lens always produces
the same image no matter where the
object is.
•Smaller
•Upright
•On the same side as the object
•Virtual
The Lens Equation
• There are two ways to determine the
characteristics of images formed by lenses:
with ray diagrams or with algebra!
Lens Terminology
• do = distance from the
object to the optical
center
• di = distance from the
image to the optical center
• ho = height of object
• hi = height of image
• f is the focal length of the
lens; distance from the
optical center (f’ is the
same length as f)
Thin Lens Equation
•Object distance (do) is always positive
•Image distance (di) is positive for real images and
negative for virtual images.
•F is positive for a converging lens and negative for a
diverging lens
Sample Question
• A converging lens has a focal length of 34. A
candle is located 96 cm from the lens. What
type of image is formed and where is it
located?
•do = 96 cm
•f= 34 cm
•2f = 68 cm
•The object is further than 2f.
According to the rules, the
image should be smaller,
inverted, between f’ and 2f’,
and it is real.
• 1/f = 1/34 = 0.0294
• 1/do = 1/96 = 0.0104
• 1/di = 1/f – 1/do
0.0294 – 0.0104
0.0190
• 1/di = 1/0.0190 = 52.6 cm
Example 2
• A pencil is located 18 cm from the lens. The
image is real and located 24 cm away from the
optical center. What is the focal length of the
lens?
• 1/do = 1/18 = 0.0556
• 1/di = 1/24 = 0.0417
• 1/f = 1/do + 1/di
0.0556 + 0.0417
0.0973
• 1/f = 1/0.0973 =10.2 cm
Example 3
• A diverging lens (image is upright and virtual)
has a focal length of 25 cm. A virtual marble
has a distance of 12 cm in front of the lens.
What is the distance of the object?
• 1/f = -1/25 = -0.0400 (- because of virtual)
• 1/di = -1/12 = -0.0833
• 1/do = 1/f - 1/di
-0.0400 – (-0.0833)
0.0433
• 1/do = 1/0.0433 =23.09 cm
The Magnification Equation
• When you compare the size of an image with
the size of an object, you are determining the
magnification of the lens.
•Object height and image heights
are positive when measured
upright and negative when
measured downward
•Magnification is positive for an
upright image and negative for an
inverted image.
•Magnification is dimensionless.
Sample Question
• A toy has a height of 8.4 cm. It stands in front
of a converging lens. The image is real,
inverted, and has a height of 23 cm. What
isthe magnification?
•
•
•
•
•
ho = 8.4 cm
hi = - 23 cm (inverted)
M = hi/ho
M = -23 / 8.4
M = -2.7
Sample Question 2
• A brick is placed 7.2 cm from the optical
center. An upright, virtual image with a
magnification of 2.3 is noticed. What is the di?
•
•
•
•
•
•
do = 7.2 cm
M = 2.3 cm
M = -di/do
di = -M x do
di = -2.3 x 7.2
di = -16.56 cm (image is on the same side as the object,
16.6 cm from the optical lens.