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Chapter 15
Pretest
Light and
Refraction
1. Refraction is the term for the
bending of a wave disturbance
as it passes at an angle from
one _____ into another.
A. glass
B. medium
C. area
D. boundary
1. Refraction is the term for the
bending of a wave disturbance
as it passes at an angle from
one _____ into another.
A. glass
B. medium
C. area
D. boundary
2. When a light ray passes from
water (n = 1.333) into diamond
(n = 2.419) at an angle of 45°,
its path is:
A. bent toward the normal.
B. bent away from the normal.
C. parallel to the normal.
D. not bent.
2. When a light ray passes from
water (n = 1.333) into diamond
(n = 2.419) at an angle of 45°,
its path is:
A. bent toward the normal.
B. bent away from the normal.
C. parallel to the normal.
D. not bent.
3. A beam of light in air is
incident at an angle of 35° to the
surface of a rectangular block of
clear plastic
(n = 1.49).
What is the angle of refraction?
A. 42°
B. 23°
C. 55°
D. 59°
3. A beam of light in air is
incident at an angle of 35° to the
surface of a rectangular block of
clear plastic
(n = 1.49).
What is the angle of refraction?
A. 42°
B. 23°
C. 55°
D. 59°
4. Which of the following describes
what will happen to a light ray
incident on a glass-to-air boundary
at greater than the critical angle?
A. total reflection
B. total transmission
C. partial reflection, partial
transmission
D. partial reflection, total
transmission
4. Which of the following describes
what will happen to a light ray
incident on a glass-to-air boundary
at greater than the critical angle?
A. total reflection
B. total transmission
C. partial reflection, partial
transmission
D. partial reflection, total
transmission
1. How is the index of
refraction calculated?
How is light refracted
as it speeds up? How
is light refracted as it
slows down?
• Index of refraction = speed of
light in a vacuum divided by
speed of light in the substance
‘c’ in a vacuum
n = --------------------------‘c’ in the substance
• When light speeds up it
bends away from the
normal.
• When light slows down it
bends toward the normal.
2. Why do you
see ‘wet spots’
on the road on
a hot day?
2. Why do you see ‘wet spots’
on the road on a hot day?
• If the air close to the ground is
warmer than the air at higher
altitudes, light from the sky is
refracted upward into the observer’s
eyes. The blue sky appears to be on
the ground and looks like it is
reflected in water.
3. Explain why total
internal reflection
occurs. Why are
prisms used as optical
reflectors? Why are
diamonds so bright?
• As light moves into a medium
in which it moves faster, it
bends away from the normal.
As the angle of incidence
increases, the angle of
refraction reaches 90 degrees
before the angle of incidence
does.
• At this angle of incidence, no
light is transmitted, 100% of
the light is reflected. This is
total internal reflection.
• Prisms are used as reflectors
because total internal
reflection is 100%. No mirrored
surface is as efficient.
• Diamonds are bright because the
critical angle for total internal
reflection in diamond is so small
that most of the light that enters
the diamond is reflected back out.
The critical angle is small for
diamond because the speed of
light in diamond is so much slower
than it is in air.
4. Explain
how a prism
disperses light.
4. Explain how a prism
disperses light.
• Different colors of light are
refracted different amounts.
A prism refracts light twice in
the same direction. Each bend
splits the colors up a little
more, producing a spectrum.
5. Why do
stars twinkle?
5. Why do stars twinkle?
• The atmosphere distorts the
light from stars because of
differences in the density of
air. This distortion is seen
as twinkling.
6. Why does the
atmosphere make
our days 4 minutes
longer?
6. Why does the atmosphere
make our days 4 minutes
longer?
• The atmosphere refracts sunlight
toward the surface of the earth. This
allows the sun to be seen after it has
passed below the horizon and before
it moves above the horizon. This adds
about 4 minutes to each day.
7. A 3 cm object is placed 10
cm in front of a convex lens with
a focal length of 5 cm. Draw a
ray diagram and calculate the
location, magnification , and
size of the image formed. What
is the type and orientation of the
image?
• First draw a line parallel to the principle
axis which refracts through the focal point.
• Then draw a line through the focal point which
refracts parallel. Where they cross is the image.
• This image is real and inverted (case 3).
We use the equations to find the actual
distance and size of the image.
1/f =
1/ do + 1/di
1/5 =
1/ 10 + 1/di
di = 10 cm
hi / h o = d i / do
hi / 3 cm = 10 cm / 10 cm
hi = 3 cm, mag = -1
8. A 4 cm object is placed 7 cm
in front of a concave lens with a
focal length of -4 cm. Draw a ray
diagram and calculate the
location, magnification , and size
of the image formed. What is
the type and orientation of the
image?
• First draw lines from each end of the object
parallel to the principle axis which refract
through the focal point.
• First draw lines from each end of the object
parallel to the principle axis which refract
through the focal point.
• Then draw lines from each end through the
optical center of the lens. Where they
cross forms the ends of the image.
• This image is virtual and upright . We use
the equations to find the actual distance
and size of the image.
1/f =
1/ do + 1/di
1/-4 = 1/ 7 + 1/di
di = -2.54 cm
hi / h o = d i / do
hi / 4 cm = 2.54 cm / 7 cm
hi = 1.45 cm, mag = 0.36
9. A 5 cm object is placed 3 cm
in front of a convex lens with a
focal length of 8 cm. Draw a ray
diagram and calculate the
location, magnification, and size
of the image formed. What is
the type and orientation of the
image? diagram
• First draw a line parallel to the principle
axis which refracts through the focal point.
• First draw a line parallel to the principle
axis which refracts through the focal point.
• Then draw a line through the optical center
of the lens. Where they cross is the image.
• This image is virtual and upright (case 6). We
use the equations to find the actual distance and
size of the image.
1/f =
1/ do + 1/di
1/8 =
1/ 3 + 1/di
di = -4.8 cm
hi / h o = d i / do
hi / 5 cm = 4.8 cm / 3 cm
hi = 8 cm, mag = 1.6
10. In what two ways
can a convex lens be
used to produce an
image that is larger
than the object?
10. In what two ways can a
convex lens be used to produce
an image that is larger than the
object?
•Case 4 and case 6.
11. How does the
production of images
with mirrors compare
with the production of
images with lenses?
11. How does the production of
images with mirrors compare
with the production of images
with lenses?
• Convex mirrors produce images like
concave lenses.
• Concave mirrors produce images like
convex lenses.
12. An object is placed along
the principle axis of a thin
converging lens that has a focal
length of 39 cm. If the distance
from the object to the lens is
51 cm, what is the distance
from the image to the lens?
1/f =
1/ do + 1/di
1/39 = 1/51 + 1/di
di = 166 cm
M = di / do
M = 166 cm / 51 cm
mag = -3.25
• First draw a line parallel to the principle
axis which refracts through the focal point.
• First draw a line parallel to the principle
axis which refracts through the focal point.
• Then draw a line through the optical center
of the lens. Where they cross is the image.
• Or, you could draw a line through the other focal
point which refracts parallel. All lines cross at the
image.