Name: ______________________ Class: _________________ Date: _________ ID: A

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Physics Ch 13 Quiz

Multiple ChoiceIdentify the letter of the choice that best completes the statement or answers the question.

____ 1. What is the wavelength of microwaves of 3.0 × 109 Hz frequency?a.  0.10 m b.  0.20 m c.  0.060 m d.  0.050 m

____ 2. The relationship between frequency, wavelength, and speed holds for light waves becausea.  different forms of electromagnetic radiation travel at different speeds. b.  light travels in straight lines. c.  all forms of electromagnetic radiation travel at a single speed in a vacuum. d.  light travels slower in a vacuum than in air.

____ 3. The farther light is from a source,a.  the more spread out light becomes. b.  the more bright light becomes. c.  the more condensed light becomes. d.  the more light is available per unit area.

____ 4. Which portion of the electromagnetic spectrum is used to identify fluorescent minerals?a.  infrared waves b.  gamma rays c.  ultraviolet light d.  X rays

____ 5. When a straight line is drawn perpendicular to a flat mirror at the point where an incoming ray strikes the mirror’s surface, the angles of incidence and reflection are measured from the normal anda.  the angle of incidence can be greater than or less than the angle of reflection. b.  the angles of incidence and reflection are equal. c.  the angle of incidence is greater than the angle of reflection. d.  the angle of incidence is less than the angle of reflection.

____ 6. If a virtual image is formed 10.0 cm along the principal axis from a convex mirror with a focal length of –15.0 cm, what is the object’s distance from the mirror?a.  6.0 cm b.  3.0 cm c.  12 cm d.  30 cm

____ 7. When two parallel mirrors are placed so that their reflective sides face each other, ____ images form. This is because the image in one mirror becomes the ____ for the other mirror.a.  reduced, virtual image b.  enlarged, focal point c.  multiple, object d.  inverted, center of curvature

____ 8. What is the frequency of an electromagnetic wave with a wavelength of 1.0 ´ 105 m?a.  3.3 × 10−4 Hz b.  3.0 × 103 Hz c.  3.0 × 1013 Hz d.  1.0 × 1013 Hz

____ 9. The image of an object in a flat mirror is alwaysa.  the same size as the object. b.  smaller than the object. c.  independent of the size of the object. d.  larger than the object.

____ 10. Which portion of the electromagnetic spectrum is used in a microscope?a.  infrared waves b.  ultraviolet light c.  gamma rays d.  visible light

____ 11. What is the frequency of infrared light of 1.0 × 10−4 m wavelength?

a.  3.0 × 104 Hz b.  3.0 × 10−13 Hz c.  3.0 × 102 Hz d.  3.0 × 1012 Hz

____ 12. A concave mirror with a focal length of 10.0 cm creates a real image 30.0 cm away on its principal axis. How far from the mirror is the corresponding object?a.  20 cm b.  5.0 cm c.  7.5 cm d.  15 cm

Name: ______________________ ID: A

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____ 13. 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?a.  12.0 cm b.  1.40 cm c.  17.0 cm d.  7.14 cm

____ 14. If you are reading a book and you move twice as far away from the light source, how does the brightness at the new distance compare with that at the old distance? It isa.  one-eighth as bright. b.  twice as bright. c.  one-fourth as bright. d.  one-half as bright.

____ 15. In a vacuum, electromagnetic radiation of short wavelengthsa.  travels faster than radiation of long wavelengths. b.  travels as fast as radiation of long wavelengths. c.  travels slower than radiation of long wavelengths. d.  can travel both faster and slower than radiation of long wavelengths.

____ 16. Which portion of the electromagnetic spectrum is used in a television?a.  gamma waves b.  radio waves c.  infrared waves d.  X rays

____ 17. An object is 29 cm away from a concave mirror’s surface along the principal axis. If the mirror’s focal length is 9.50 cm, how far away is the corresponding image?a.  36 cm b.  29 cm c.  14 cm d.  12 cm

____ 18. When red light is compared with violet light,a.  both have the same wavelength. b.  both travel at the same speed. c.  both have the same frequency. d.  red light travels faster than violet light.

____ 19. If you know the wavelength of any form of electromagnetic radiation, you can determine its frequency becausea.  all wavelengths travel at the same speed. b.  the speed of light varies for each form. c.  the speed of light increases as wavelength increases. d.  wavelength and frequency are equal.

____ 20. A highly polished finish on a new car provides a ____ surface for ____ reflection.a.  specular, diffused b.  smooth, specular c.  rough, diffused d.  rough, regular

____ 21. Snow reflects almost all of the light incident upon it. However, a single beam of light is not reflected in the form of parallel rays. This is an example of ____ reflection off a ____ surface.a.  diffuse, specular b.  regular, rough c.  diffuse, rough d.  regular, specular

____ 22. A convex mirror with a focal length of –20.0 cm has an object 30.0 cm in front of the mirror. What is the value of q for the corresponding image?a.  −12 cm b.  60 cm c.  12 cm d.  −60 cm

____ 23. What is the wavelength of an infrared wave with a frequency of 4.2 × 1014 Hz?

a.  7.1 × 10−7 m b.  1.4 × 10−6 m c.  1.4 × 106 m d.  7.1 × 105 m

____ 24. A mirror has an object located on its principal axis 40.0 cm from the mirror’s surface. A virtual image is formed 15.0 cm behind the mirror. What is the mirror’s focal length?a.  13 cm b.  −10.9 cm c.  2.38 cm d.  −24.0 cm

____ 25. For a spherical mirror, the focal length is equal to ____ the radius of curvature of the mirror.a.  the square of b.  one-half c.  one-third d.  one-fourth

ID: A

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Physics Ch 13 QuizAnswer Section

MULTIPLE CHOICE

1. ANS: AGiven

f = 3.0 × 109 Hz = 3.0 × 109 s −1

c = 3.00 × 108 m/s

Solution

Rearrange the wave speed equation,c = fλ, to isolateλ, and calculate.

λ =cf

=3.00 × 108 m/sÊËÁÁÁÁ

ˆ¯˜̃̃˜

3.0 × 109 s −1ÊËÁÁÁÁ

ˆ¯˜̃̃˜

= 0.10 m

DIF: IIIA OBJ: 13-1.2 2. ANS: C DIF: II OBJ: 13-1.3 3. ANS: A DIF: I OBJ: 13-1.4 4. ANS: C DIF: I OBJ: 13-1.1 5. ANS: B DIF: I OBJ: 13-2.2 6. ANS: D

Given

f = −15.0 cm

q = −10.0 cm

Solution

Rearrange the mirror equation,1p

+1q

=1f

,and solve for p.

1p

=1f−

1q

=1

−15.0 cm−

1−10.0 cm

= −1

15.0 cm+

1.515.0 cm

=0.5

15.0 cm

p = 30 cm

DIF: IIIB OBJ: 13-3.1 7. ANS: C DIF: II OBJ: 13-2.3

ID: A

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8. ANS: BGiven

λ = 1.0 × 105 m

c = 3.00 × 108 m/s

Solution

Rearrange the wave speed equation,c = fλ, to isolate f, and calculate.

f =cλ

=3.00 × 108 m/sÊËÁÁÁÁ

ˆ¯˜̃̃˜

1.0 × 105 mÊËÁÁÁÁ

ˆ¯˜̃̃˜

= 3.0 × 103 s −1 = 3.0 × 103 Hz

DIF: IIIA OBJ: 13-1.2 9. ANS: A DIF: I OBJ: 13-2.3 10. ANS: D DIF: I OBJ: 13-1.1 11. ANS: D

Given

λ = 1.0 × 10−4 m

c = 3.00 × 108 m/s

Solution

Rearrange the wave speed equation,c = fλ, to isolate f, and calculate.

f =cλ

=3.00 × 108 m/sÊËÁÁÁÁ

ˆ¯˜̃̃˜

1.0 × 10−4 mÊËÁÁÁÁ

ˆ¯˜̃̃˜

= 3.0 × 1012 s −1 = 3.0 × 1012 Hz

DIF: IIIA OBJ: 13-1.2

ID: A

3

12. ANS: DGiven

f = 10.0 cm

q = 30.0 cm

Solution

Rearrange the mirror equation,1p

+1q

=1f

,and solve for p.

1p

=1f−

1q

=1

10.0 cm−

130.0 cm

=3

30.0 cm−

130.0 cm

=2

30.0 cm

p = 15 cm

DIF: IIIB OBJ: 13-3.1 13. ANS: D

Given

p = 10.0 cm

q = 25.0 cm

Solution

Use the mirror equation,1p

+1q

=1f

,and solve for f.

1f

=1p

+1q

=1

10.0 cm+

125.0 cm

=2.5

25.0 cm+

125.0 cm

=3.5

25.0 cm

f = 7.14 cm

DIF: IIIB OBJ: 13-3.1 14. ANS: C DIF: II OBJ: 13-1.4 15. ANS: B DIF: I OBJ: 13-1.3 16. ANS: B DIF: I OBJ: 13-1.1

ID: A

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17. ANS: CGiven

p = 29 cm

f = 9.50 cm

Solution

Rearrange the mirror equation,1p

+1q

=1f

,and solve for q.

1q

=1f−

1p

=1

9.50 cm−

129 cm

=3.0529 cm

−1

29 cm=

2.0529 cm

p = 14 cm

DIF: IIIB OBJ: 13-3.1 18. ANS: B DIF: II OBJ: 13-1.3 19. ANS: A DIF: I OBJ: 13-1.3 20. ANS: B DIF: I OBJ: 13-2.1 21. ANS: C DIF: II OBJ: 13-2.1 22. ANS: A

Given

f = −20.0 cm

p = 30.0 cm

Solution

Rearrange the mirror equation,1p

+1q

=1f

,and solve for q.

1q

=1f−

1p

=1

−20.0 cm−

130.0 cm

= −1.5

30.0 cm−

130.0 cm

= −2.5

30.0 cm

q = −12 cm

DIF: IIIB OBJ: 13-3.1

ID: A

5

23. ANS: AGiven

f = 4.2 × 1014 Hz = 4.2 × 1014 s −1

c = 3.00 × 108 m/s

Solution

Rearrange the wave speed equation,c = fλ, to isolateλ, and calculate.

λ =cf

=3.00 × 108 m/sÊËÁÁÁÁ

ˆ¯˜̃̃˜

4.2 × 1014 s −1ÊËÁÁÁÁ

ˆ¯˜̃̃˜

= 7.1 × 10−7 m

DIF: IIIA OBJ: 13-1.2 24. ANS: D

Given

p = 40.0 cm

q = −15.0 cm

Solution

Use the mirror equation,1p

+1q

=1f

,and solve for f.

1f

=1p

+1q

=1

40.0 cm+

1−15.0 cm

=3.00

120. cm−

8.00120. cm

= −5.00

120. cm

f = −24.0 cm

DIF: IIIB OBJ: 13-3.1 25. ANS: B DIF: I OBJ: 13-3.1