Problem Set 6
Due: see website for due date
Chapter 26: The Refraction of Light: Lenses
Questions: A, 11, 12
Problems: 12, 13, 20, 50, 54, 57, 59, 62
Question A: A student draws a ray diagram but
forgets to label the object, the image, or the type of
lens used. Using the diagram, explain whether the
lens is converging or diverging, which arrow
represents the object, and which represents the
image.
Q26.11: An object is situated to the left of a lens. A ray of light from the object is close to
and parallel to the principal axis of the lens. The ray passes through the lens. Which one
of the following statements is true? (a) The ray crosses the principal axis at a distance
from the lens equal to twice the focal length, no matter whether the lens is converging or
diverging. (b) The ray passes through the lens without changing direction, no matter
whether the lens is converging or diverging. (c) The ray passes through a focal point of
the lens, no matter whether the lens is converging or diverging. (d) The ray passes
through a focal point of the lens only if the lens is a diverging lens. (e) The ray passes
through a focal point of the lens only if the lens is a converging lens.
Q26.12: What type of single lens produces a virtual image that is inverted with respect to
the object? (a) Both a converging and a diverging lens can produce such an image. (b)
Neither a converging nor a diverging lens produces such an image. (c) A converging
lens (d) A diverging lens.
P26.12: A narrow beam of light from a laser travels through air (n = 1.00) and strikes
point A on the surface of the water (n = 1.33) in a lake. The angle of incidence is 55o.
The depth of the lake is 3.0 m. On the flat lake-bottom is point B, directly below point A.
(a) If refraction did not occur, how far away from point B would the laser beam strike the
lake-bottom? (b) Considering refraction, how far away from point B would the laser beam
strike the lake-bottom? Answer: 4.3 m, 2.3 m
P26.13: The drawing shows a coin resting on the bottom of a beaker
filled with an unknown liquid. A ray of light from the coin travels to
the surface of the liquid and is refracted as it enters into the air. A
person sees the ray as it skims just above the surface of the liquid.
How fast is the light traveling in the liquid?
Answer: 0.641c.
P26.20: The drawing shows a rectangular block of glass (n =
1.52) sur- rounded by liquid carbon disulfide (n = 1.63). A ray
of light is incident on the glass at point A with a 30.0o angle of
incidence. At what angle of refraction does the ray leave the
glass at point B? Answer: 51.9o
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P26.50: The owner of a van installs a rear-window lens that has a focal length of −0.300
m. When the owner looks out through the lens at a person standing directly behind the
van, the person appears to be just 0.240 m from the back of the van, and appears to be
0.34 m tall. (a) How far from the van is the person actually standing, and (b) how tall is
the person? Answer: 1.2 cm, 1.7 cm
P26.54: (a) For a diverging lens (f = 220.0 cm), construct a ray diagram to scale and find
the image distance for an object that is 20.0 cm from the lens. (b) Determine the
magnification of the lens from the diagram. Answer: −10.0 cm, 0.500
P26.57: A converging lens has a focal length of 88.00 cm. An object 13.0 cm tall is
located 155.0 cm in front of this lens. (a) What is the image distance? (b) Is the image
real or virtual? (c) What is the image height? Be sure to include the proper algebraic
sign. Answer: 204 cm, real, −17.1 cm
P26.59: The moon’s diameter is 3.48 × 106 m, and its mean distance from the earth is
3.85 × 108 m. The moon is being photographed by a camera whose lens has a focal
length of 50.0 mm. (a) Find the diameter of the moon’s image on the slide film. (b) When
the slide is projected onto a screen that is 15.0 m from the lens of the projector ( f =
110.0 mm), what is the diameter of the moon’s image on the screen?
Answer: 0.452 mm, 61.2 mm
P26.62: An object is placed in front of a converging lens in such a position that the lens
(f1 = 12.0 cm) creates a real image located 21.0 cm from the lens. Then, with the object
remaining in place, the lens is replaced with another converging lens (f 2 = 16.0 cm). A
new, real image is formed. What is the image distance of this new image?
Answer: 37.3 cm
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