Chapter 15 Pretest Light and Refraction

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Chapter 15 Pretest Light and Refraction
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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
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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
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2. When a light ray passes from water (n 1.333)
into diamond (n 2.419) at an angle of 45, its
path isA. bent toward the normal. B. bent away
from the normal.C. parallel to the normal. D.
not bent.
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2. When a light ray passes from water (n 1.333)
into diamond (n 2.419) at an angle of 45, its
path isA. bent toward the normal. B. bent away
from the normal.C. parallel to the normal. D.
not bent.
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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
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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
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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 transmissionC.
partial reflection, partial transmission D.
partial reflection, total transmission
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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 transmissionC.
partial reflection, partial transmission D.
partial reflection, total transmission
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1. How is the index of refraction calculated?
How is light refracted as it speeds up? How is
light refracted as it slows down?
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• 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

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• When light speeds up it bends away from the
  normal.
• When light slows down it bends toward the normal.

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2. Why do you see wet spots on the road on a
hot day?
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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 observers eyes. The
  blue sky appears to be on the ground and looks
  like it is reflected in water.

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3. Explain why total internal reflection occurs.
Why are prisms used as optical reflectors? Why
are diamonds so bright?
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• 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.

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• At this angle of incidence, no light is
  transmitted, 100 of the light is reflected. This
  is total internal reflection.

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• Prisms are used as reflectors because total
  internal reflection is 100. No mirrored surface
  is as efficient.

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• 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.

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4. Explain how a prism disperses light.
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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.

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5. Why do stars twinkle?
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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.

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6. Why does the atmosphere make our days 4
minutes longer?
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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.

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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?
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• First draw a line parallel to the principle axis
  which refracts through the focal point.

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• Then draw a line through the focal point which
  refracts parallel. Where they cross is the image.

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• This image is real and inverted (case 3). We use
  the equations to find the actual distance and
  size of the image.

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• 1/f 1/ do 1/di
• 1/5 1/ 10 1/di
• di 10 cm

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hi / ho di / do hi / 3 cm 10 cm / 10
cm hi 3 cm, mag -1
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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?
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• First draw lines from each end of the object
  parallel to the principle axis which refract
  through the focal point.

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• First draw lines from each end of the object
  parallel to the principle axis which refract
  through the focal point.

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• Then draw lines from each end through the optical
  center of the lens. Where they cross forms the
  ends of the image.

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• This image is virtual and upright . We use the
  equations to find the actual distance and size of
  the image.

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• 1/f 1/ do 1/di
• 1/-4 1/ 7 1/di
• di -2.54 cm

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hi / ho di / do hi / 4 cm 2.54 cm /
7 cm hi 1.45 cm, mag 0.36
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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
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• First draw a line parallel to the principle axis
  which refracts through the focal point.

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• First draw a line parallel to the principle axis
  which refracts through the focal point.

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• Then draw a line through the optical center of
  the lens. Where they cross is the image.

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• This image is virtual and upright (case 6). We
  use the equations to find the actual distance and
  size of the image.

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• 1/f 1/ do 1/di
• 1/8 1/ 3 1/di
• di -4.8 cm

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hi / ho di / do hi / 5 cm 4.8 cm / 3
cm hi 8 cm, mag 1.6
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10. In what two ways can a convex lens be used to
produce an image that is larger than the object?
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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.

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11. How does the production of images with
mirrors compare with the production of images
with lenses?
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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.

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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?
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• 1/f 1/ do 1/di
• 1/39 1/51 1/di
• di 166 cm

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M di / do M 166 cm / 51 cm mag
-3.25
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• First draw a line parallel to the principle axis
  which refracts through the focal point.

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• First draw a line parallel to the principle axis
  which refracts through the focal point.

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• Then draw a line through the optical center of
  the lens. Where they cross is the image.

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• Or, you could draw a line through the other focal
  point which refracts parallel. All lines cross at
  the image.

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