PHYS 2212 Module 14 Self Assessment Practice Problems

Module 14 Self Assessment Practice Problems

14.1
A mirror on the passenger side of your car is convex and has a radius of curvature with magnitude 18.0 cm.
(a) Another car is behind your car, 9.0 m from the mirror, and this car is viewed in the mirror by your passenger. If this car is 1.5 m tall, what is the height of the image
(b) The mirror has a warning attached that objects viewed in it are closer than they appear. Why is this so?
Answer: 1.48 cm

14.2
A light bulb is 3.00 m from a wall. You are to use a concave mirror to project an image of the bulb on the wall, with the image 3.50 times the size of the object.
(a) How far should the mirror be from the wall?
(b) What should its radius of curvature be?
Answer: (a) 4.2 m (b) 1.87 m

14.3
The diameter of Mars is 6794 km, and its minimum distance from the earth is 5.58 x 10⁷ km. When Mars is at this distance, find the diameter of the image of Mars formed by a spherical, concave telescope mirror with a focal length of 1.75 m.
Answer: 0.213 mm

14.4
You want to use a lens with a focal length of 38.0 cm to produce a real image of an object, with the image twice as long as the object itself.
(a) What kind of lens — converging or diverging — do you need?
(b) Where should the object be placed?
(c) Suppose you want a virtual image of the same object, with the same magnification — what kind of lens do you need?
(d) Where should the object be placed?
Answer: (a) converging (b) 57 cm (c) converging (d) 0.5f

14.5
An object is 16.0 cm to the left of a lens. The lens forms an image 36.0 cm to the right of the lens.
(a) What is the focal length of the lens?
(b) Is the lens converging or diverging?
(c) If the object is 8.00 mm tall, how tall is the image?
(d) Is it upright or inverted?
(e) Draw a principal-ray diagram.
Answer: (a) 11.1 cm (b) converging (c) -18 mm (d) inverted

14.8
A 1.20-cm-tall object is 50.0 cm to the left of a converging lens of focal length 40.0 cm. A second converging lens, this one having a focal length of 60.0 cm, is located 300.0 cm to the right of the first lens along the same optic axis.
(a) Find the location and height of the image (call it I₁) formed by the lens with a focal length of 40.0 cm.
(b) I₁ is now the object for the second lens. Find the location and height of the image produced by the second lens. This is the final image produced by the combination of lenses.
Answer: (a) 200 cm, -4.8 cm (b) 150 cm, +7.2 cm

14.9
A thin lens with a focal length of 6.00 cm is used as a simple magnifier.
(a) What angular magnification is obtainable with the lens if the object is at the focal point?
(b) When an object is examined through the lens, how close can it be brought to the lens? Assume that the image viewed by the eye is at the near point, 25.0 cm from the eye, and that the lens is very close to the eye.
Answer: (a) 4.17 (b) 4.84 cm

14.10
Contact lenses are placed right on the eyeball, so the distance from the eye to an object (or image) is the same as the distance from the lens to that object (or image). A certain person can see distant objects well, but his near point is 45.0 cm from his eyes instead of the usual 25.0 cm.
(a) Is this person nearsighted or farsighted?
(b) What type of lens (converging or diverging) is needed to correct his vision?
(c) If the correcting lenses will be contact lenses, what focal length lens is needed and what is its power in diopters?
Answer: (a) Farsighted (b) Converging (c) 56.25 cm, +1.78 D