Reflection and Mirrors Review
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Reflection and Mirrors - Home || Printable Version || Questions with Links
Answers to Questions: All || #1-#19 || #20-#26 || #27-#32 || #33-#42
Part A: Multiple Choice
1. As the angle of incidence is increased for a ray incident on a reflecting surface, the angle between the incident and reflected rays ultimately approaches what value?
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a. zero
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b. 45 degrees
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c. 90 degrees
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d. 180 degrees
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Answer: D
The angle of incidence is the angle between the incident ray and the normal. As this angle approaches 90 degrees, the reflected ray also approaches a 90 degree angle with the normal; thus, the angle between the incident and reflected ray approach 180 degrees.
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2. If you stand three feet in front of a plane mirror, how far away would you see yourself in the mirror?
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a. 1.5 ft
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b. 3.0 ft
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c. 6.0 ft
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d. 12.0 ft
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Answer: C
If you stand 3 feet from the mirror, then your image is three feet on the other side of the mirror; this puts your image a total of six feet from you (3 feet to the mirror plus 3 more feet to the image).
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3. A concave mirror with a focal length of 10.0 cm creates a real image 30.0 cm away on its principal axis; the corresponding object is located how far from the mirror?
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a. 20.0 cm
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b. 15.0 cm
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c. 7.5 cm
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d. 5.0 cm
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Answer: B
Use the mirror equation:
1/di + 1/do = 1/f
where di = +30.0 cm and f = +10.0 cm.
Substitute and solve for do.
1/do + 1/(30.0 cm) = 1/(10.0 cm)
1/do = 1/(10.0 cm) - 1/(30.0 cm) = 2/(30.0 cm)
do = 15.0 cm
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4. A concave mirror forms a real image at 25.0 cm from the mirror surface along the principal axis. If the corresponding object is at a 10.0 cm distance, what is the mirror's focal length?
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a. 1.4 cm
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b. 16.7 cm
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c. 12.4 cm
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d. 7.1 cm
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Answer: 7.1 cm
Use the mirror equation:
1/di + 1/do = 1/f
where di = +25.0 cm and do = +10.0 cm.
Substitute and solve for f.
1/(10.0 cm) + 1/(25.0 cm) = 1/f
1/f = 14/(100. cm) or 0.14/cm
f = 100./14 cm = 7.1 cm
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5. If a virtual image is formed along the principal axis 10.0 cm from a concave mirror with the focal length 15.0 cm, what is the object distance from the mirror?
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a. 30.0 cm
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b. 10.0 cm
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c. 12.4 cm
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d. 6.0 cm
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Answer: D
Use the mirror equation:
1/di + 1/do = 1/f
where di = -10.0 cm and f = +15.0 cm. (Note that di is negative if the image is virtual.)
Substitute and solve for do.
1/do + 1/(-10.0 cm) = 1/(15.0 cm)
1/do = 1/(15.0 cm) - 1/(-10.0 cm) = 5.00/(30.0 cm)
do = 6.0 cm
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6. If a virtual image is formed 10.0 cm along the principal axis from a convex mirror of focal length -15.0 cm, what is the object distance from the mirror?
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a. 30.0 cm
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b. 10.0 cm
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c. 6.0 cm
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d. 3.0 cm
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Answer: A
Use the mirror equation:
1/di + 1/do = 1/f
where di = -10.0 cm and f = -15.0 cm. (Note that di is negative if the image is virtual.)
Substitute and solve for do.
1/do + 1/(-10.0 cm) = 1/(-15.0 cm)
1/do = 1/(-15.0 cm) - 1/(-10.0 cm) = -1/(30.0 cm)
do = 30.0 cm
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7. If a man's face is 30.0 cm in front of a concave shaving mirror creating an upright image 1.50 times as large as the object, what is the mirror's focal length?
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a. 12.0 cm
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b. 20.0 cm
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c. 70.0 cm
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d. 90.0 cm
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Answer: D
First find di.
If the do = 30.0 cm and the M = +1.50 (Note: M is + since the image is upright), then use M = -di/do to find that di = -45.0 cm (the - di value is consistent with the image being virtual - located behind the mirror).
Now use the mirror equation to find the focal length:
1/di + 1/do = 1/f
where di = -45.0 cm and do = +30.0 cm.
Substitute and solve for f.
1/(30.0 cm) + 1/(-45.0 cm) = 1/f
1/f = 1/(90.0 cm) or 0.0111/cm
f = 90.0 cm
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8. Which of the following best describes the image formed by a plane mirror?
- virtual, inverted and enlarged
- real, inverted and reduced
- virtual, upright and the same size as object
- real, upright and the same size as object
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Answer: C
When you look at your image in a plane mirror, you see an upright image; it is located on the other side of the mirror (and thus is virtual); finally, it has the same dimensions (height, width) as yourself (the object).
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9. Which of the following best describes the image formed by a concave mirror when the object is located somewhere between the focal point (F) and the center of curvature (C) of the mirror?
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a. virtual, upright and enlarged
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b. real, inverted and reduced
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c. virtual, upright and reduced
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d. real, inverted and enlarged
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Answer: D
For concave mirrors, when the object is located anywhere between F and C, the image is real, inverted, enlarged in size, and located beyond C. You should get this very result if you were to draw a ray diagram.
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10. Which of the following best describes the image formed by a concave mirror when the object is at a distance further than the center of curvature (C) of the mirror?
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a. virtual, erect and enlarged
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b. real, inverted and reduced
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c. virtual, upright and reduced
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d. real, inverted and enlarged
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Answer: B
For concave mirrors, when the object is located anywhere beyond C, the image is real, inverted, reduced in size, and located between F and C. You should get this very result if you were to draw a ray diagram.
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11. Which of the following best describes the image formed by a concave mirror when the object distance from the mirror is less than the focal length (f)?
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a. virtual, upright and enlarged
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b. real, inverted and reduced
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c. virtual, upright and reduced
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Answer: A
For concave mirrors, when the object is located anywhere inside the F, the image is virtual, upright, enlarged in size, and located on the opposite side of the mirror. You should get this very result if you were to draw a ray diagram.
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12. Which of the following best describes the image formed by a convex mirror when the object distance from the mirror is less than the absolute value of the focal length (f)?
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a. virtual, upright and enlarged
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b. real, inverted and reduced
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c. virtual, upright and reduced
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d. real, inverted and enlarged
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Answer: C
For convex mirrors, regardless of where the object is located, the image is virtual, upright, reduced in size, and located on the opposite side of the mirror. You should get this very result if you were to draw a ray diagram.
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13. Use of a parabolic mirror, instead of one made of a circular arc surface, can be used to reduce the occurrence of which of the following effects?
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a. spherical aberration
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b. mirages
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c. chromatic aberration
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d. light scattering
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Answer: A
The problem of spherical aberration are caused by the inability of the outer portions of a concave spherical mirror to reflect and focus light to the same image locations as other portions of the mirror. A cure for the problem involves the use of a mirror with a different shape in the regions far from the principal axis; a parabolic mirror solves this problem.
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14. When the image of an object is seen in a plane mirror, the image is
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a. real and upright.
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b. real and inverted.
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c. virtual and upright.
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d. virtual and inverted.
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Answer: C
Look at yourself in a plane mirror and you see your image - it is upright. The image is located on the other side of the mirror since reflected rays diverge upon reflection; when mirrors produce images on the the opposite side of the mirror, the images are said to be virtual.
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15. When the image of an object is seen in a plane mirror, the distance from the mirror to the image depends on
- the wavelength of light used for viewing.
- the distance from the object to the mirror.
- the distance of both the observer and the object to the mirror.
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Answer: B
For plane mirrors, the image distance is the same as the object distance (di =-do). The only way to modify the image distance is to modify the object distance.
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16. If a man wishes to use a plane mirror on a wall to view both his head and his feet as he stands in front of the mirror, the required length of the mirror
- is equal to the height of the man.
- is equal to one half the height of the man.
- depends on the distance the man stands from the mirror.
- depends on both the height of the man and the distance from the man to the mirror.
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Answer: B
The portion of mirror required to view the full image of an object is always one-half the height of the object (for plane mirrors only). For the skeptics who believe C is the more believable answer, see the proof given on the Physics Classroom tutorial page.
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17. When the image of an object is seen in a concave mirror the image will
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a. always be real.
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b. always be virtual.
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c. be either real or virtual.
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d. will always be magnified.
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Answer: C
Concave mirrors can produce real images (if the object is beyond the focal point) and virtual images (if the object is located in front of the focal point). These images can be either magnified in size (if object is in front of F or between C and F), reduced in size (if object is located beyond C),or the same size as the object (if object is located at C).
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18. When the image of an object is seen in a convex mirror the image will
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a. always be real.
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b. always be virtual.
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c. may be either real or virtual.
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d. will always be magnified.
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Answer: B
Convex mirrors always (always) produce images which are virtual, upright, and reduced in size. Since reflected light rays always diverge from each other, these convex mirrors can never produce real images. Draw a ray diagram to convince yourself that this is true.
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19. Rays of light traveling parallel to the principal axis of a concave mirror will come together
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a. at the center of curvature.
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b. at the focal point.
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c. at infinity.
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d. at a point half way to the focal point.
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Answer: B
The focal point is the location where light rays traveling parallel to the principal axis converge. (Perhaps you remember the demo with the concave (converging) mirror and the pencil or dollar bill.)
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Navigate to:
Review Session Home - Topic Listing
Reflection and Mirrors - Home || Printable Version || Questions with Links
Answers to Questions: All || #1-#19 || #20-#26 || #27-#32 || #33-#42
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Visit: The Calculator Pad Home | Calculator Pad - Reflection and Mirrors
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