TIME OF COMPLETION______
NAME__SOLUTION___________________________
DEPARTMENT OF NATURAL SCIENCES
PHYS 1112, Exam 3
Version 1
Total Weight: 100 points
Section 1
December 6, 2004
1.
Check your examination for completeness prior to starting. There are a total of ten (10)
problems on eight (8) pages.
2.
Authorized references include your calculator with calculator handbook, and the
Reference Data Pamphlet (provided by your instructor).
3.
You will have 75 minutes to complete the examination.
4.
The total weight of the examination is 100 points.
5.
There are six (6) multiple choice and four (4) calculation problems. Work five (5)
multiple choice and all calculation problems. Show all work; partial credit will be given for
correct work shown.
6.
If you have any questions during the examination, see your instructor who will
be located in the classroom.
7.
Start:
Stop:
6:00 p.m.
7:15 p.m
PROBLEM
POINTS
1-6
30
7
20
8
15
9
20
10
15
TOTAL
100
PERCENTAGE
CREDIT
CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN
MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK
FOR PARTIAL CREDIT.
1. Which of the following statements is correct for a series RLC circuit?
a. The voltage across the capacitor leads the voltage across the inductor by 90o.
b. The voltage across the inductor leads the voltage across the capacitor by 90o.
(6)
c. The voltage across the inductor leads the voltage across the resistor by 180o.
d. The voltage across the inductor leads the voltage across the capacitor by 180o.
2. Light is refracted through a diamond. If the angle of incidence is 30.0o, and the angle of
refraction is 12.0o, what is the index of refraction?
a. 1.30.
b. 2.40.
(6)
c. 2.60.
d. 0.400.
3. A layer of ethyl alcohol (n = 1.361) is on top of water (n = 1.333). At what angle relative
to the normal to the interface of the two liquids is light totally reflected?
a. 78o.
b. 88o.
(6)
c. 68o.
d. 49o.
4. A laser beam ( = 694 nm) is incident on two slits 0.100 mm apart. Approximately how
far apart (in m) will the bright interference fringes be on the screen 5.00 m from the
double slits?
a. 3.47 x 10-3.
b. 3.47 x 10-2.
(6)
c. 3.47 x 10-4.
d. 3.47 x 10-6.
5. A monochromatic (single frequency, single wavelength) light ray in air (n = 1.00)
enters a glass prism (n = 1.50). In the glass prism
a. Both the frequency and the wavelength are the same as in air.
b. The frequency is the same, but the wavelength is greater than in air.
(6)
c. The frequency is the same, but the wavelength is smaller than in air.
d. The wavelength is the same, but the frequency is greater than in air.
6. Which ray diagram is correct?
a. A.
b. B.
(6)
c.
C.
d. D.
7. A series ac circuit contains a 50.0- resistor, a 15.0-F capacitor, a 0.200-H inductor,
and a 60.0-Hz generator that has an rms output of 90.0 V.
a. Find the impedance of the circuit.
XL = 2  f L = 2  (60.0 Hz) (0.200 H) = 75.4 
XC = 1/2  f C = 1/(2  (60.0 Hz) (15.0 x 10-6 F)) = 177 
Z = (R2 + (XL – XC)2)1/2 = ((50.0 )2 + (75.4  – 177 )2)1/2 
b. Find the maximum current that flows in the resistor.
Vmax = 21/2 Vrms
Vmax = 21/2 (90.0 V) = 127 V
Imax = Vmax /Z
Imax = (127 V)/(113) = 1.13 A
c. Find the maximum potential difference across the capacitor.
Vmax = Imax XC = (1.13 A)(177) = 200 V
d. Does the voltage lead the current in the circuit or the current leads the voltage?
EXPLAIN.
 tan-1 ((XL – XC)/R)
 tan-1 ((75.4  – 177 )/ (50.0 )) = -63.8o
 < 0, hence voltage lags behind current.
8. A basketball player of height 2.20 m is standing 3.00 m in front of a convex spherical
mirror of radius of curvature of 4.00 m.
a. What is the focal length of the mirror?
f = R/2 = - 2.00 m (mirror is convex)
b. Locate the image.
p = 3.00 m
1/p + 1/q = 1/f
1/q = 1/f – 1/p = 1/(-2.00 m) – 1/(3.00 m) = -5/(6.00 m)
q = -1.20 m
c. Determine the size of the image.
M = -q/p = - (-1.20 m)/(3.00 m) = 0.400
M = h’/h
h’ = M h = (0.400)(2.20 m) = 0.880 m
d. Is the image real or virtual?
Virtual (q < 0)
e. Is the image upright or inverted?
Upright (M > 0)
9. A small kitten finds itself 150 cm in front of a diverging lens.
a. If the absolute value of the image distance is 50.0 cm, find the focal length of the
lens.
p = 150 cm
q = -50.0 cm (lens is diverging , image is virtual)
1/f = 1/p + 1/q
f = -75.0 cm
b. Determine the magnification.
M = -q/p = - (-50.0 cm)/(150 cm) = 1/3
c. Is the image real or virtual?
Virtual (q < 0)
d. Is the image upright or inverted?
Upright (M > 0)
e. Draw a neat ray diagram of the situation.
10. A certain sample of a flint glass has an index of refraction of 1.571 for red light (656 nm)
and 1.594 for violet light (434 nm). If white light is incident from air at an incidence
angle of 35.0o, what is the angular separation of the red and violet rays in the refracted
beam?
Red:
n1 sin(1) = n2 sin(2)
sin(2) = n1 sin(1)/ n2
sin(2) = (1.00) sin(35.0o)/(1.571) = 0.365
1 = 21.41o
Violet:
n1 sin(1) = n2 sin(2)
sin(2) = n1 sin(1)/ n2
sin(2) = (1.00) sin(35.0o)/(1.594) = 0.360
1 = 21.09o
 = 21.41o21.09o= 0.32o
TIME
OF
COMPLETION_______________
NAME______SOLUTION_______________________
DEPARTMENT OF NATURAL SCIENCES
PHYS 1112, Exam 3
Version 1
Total Weight: 100 points
Section 1
December 1, 2006
1.
Check your examination for completeness prior to starting. There are a total of ten (10)
problems on seven (7) pages.
2.
Authorized references include your calculator with calculator handbook, and the
Reference Data Pamphlet (provided by your instructor).
3.
You will have 50 minutes to complete the examination.
4.
The total weight of the examination is 100 points.
5.
There are six (6) multiple choice and four (4) calculation problems. Work five (5)
multiple choice and three (3) calculation problems. Show all work; partial credit will be given
for correct work shown.
6.
If you have any questions during the examination, see your instructor who will
be located in the classroom.
7.
Start:
Stop:
10:30 a.m.
11:20 a.m.
PROBLEM
POINTS
1-6
25
7
25
8
25
9
25
10
25
TOTAL
100
PERCENTAGE
CREDIT
CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN
MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK
FOR PARTIAL CREDIT.
1. If the radius of curvature of the concave mirror is r, the focal length is
a. 2r.
b. r.
(5)
c. r/2.
d. Cannot be determined from the information given.
2. Light arriving at a concave mirror on a path parallel to the axis is reflected
a. Back parallel to the axis.
b. Back on itself.
(5)
c. Through the focal point.
d. Through the center of curvature.
3. A negative magnification for a mirror means
a. The image is inverted, and the mirror is concave.
b. The image is inverted, and the mirror is convex.
(5)
c. The image is inverted, and the mirror may be concave or convex.
d. The image is upright, and the mirror is convex.
e. The image is upright, and the mirror may be concave or convex.
4. The index of refraction of diamond is 2.42. This means that a given frequency of light travels
a. 2.42 times faster in air than it does in diamond.
b. 2.42 times faster in diamond than it does in air.
(5)
c. 2.42 times faster in vacuum than it does in diamond.
d. 2.42 times faster in diamond than it does in vacuum.
5. Lenses that are thicker at the center
a. Spread out light rays.
b. Bend light rays to a point beyond the lens.
(5)
c. Have no effect on light rays.
d. Reflect light rays back.
6. The critical angle for a beam of light passing from water into air is 48.8°. This means that all
light rays with an angle of incidence greater than this angle will be
a. Absorbed.
b. Totally reflected.
(5)
c. Partially reflected and partially transmitted.
d. Totally transmitted.
7.
A stamp collector uses a converging lens with focal length 24 cm to view a stamp 18 cm in
front of the lens.
a.. Where is the image located?
do = 18.0 cm
f = 24.0 cm
1/di + 1/do = 1/f
1/di = 1/f - 1/do = 1/(24.0 cm) - 1/(18.0 cm)
di = -72.0 cm
b. What is the magnification?
M = - di /do = - (-72.0 cm)/(18.0 cm) = 4.00
c. Is the image upright or inverted, real or virtual?
Upright (M > 0), Virtual (di < 0)
d. Draw the ray diagram to confirm your answers.
8.
A diverging lens with a focal length of 14 cm is placed 12 cm to the right of a converging
lens with a focal length of 18 cm. An object is placed 33 cm to the left of the converging
lens.
a. Where will the final image be located?
1st lens:
do1 = 33.0 cm
f1 = 18.0 cm
1/di1 + 1/do1 = 1/f1
1/di1 = 1/f1 - 1/do1 = 1/(18.0 cm) - 1/(33.0 cm)
di1 = 39.5 cm
2nd lens:
do2 = - (39.5 cm – 12.0 cm) = - 27.5 cm
1/di2 = 1/f2 - 1/do2 = 1/(-14.0 cm) - 1/(-27.5 cm)
di2 = - 28.5 cm (28.5 cm in front of the second lens)
b. What is the total magnification of the system?
M1 = - di1 /do1 = - (-39.5 cm)/(33.0 cm) = - 1.20
M2 = - di2 /do2 = - (-28.5 cm)/(-27.5 cm) = - 1.04
M = M1M2 = 1.24
c. Is the image upright or inverted, enlarged or reduced in size?
Upright (M > 0), Enlarged (|M| > 1)
9.
A 4.5-cm-tall object is placed 28 cm in front of a spherical mirror. It is desired to produce a
virtual image that is upright and 3.5 cm tall.
a. What type of mirror should be used?
Convex
b. Where is the image located?
M = hi /ho = (3.50 cm)/(4.5 cm) = 0.778
M = - di /do
di = - M do = - (0.778)(28.0 cm) = -21.8 cm
c. What is the focal length of the mirror?
do = 28.0 cm
di = -21.8 cm
1/di + 1/do = 1/f
1/f = 1/(28.0 cm) + 1/(-21.8 cm)
f = -98.3 cm
d. What is the radius of curvature of the mirror?
R = 2 f = -197 cm
10. Light is incident from air on a transparent substance at an angle of 58.0o with the normal.
The reflected and refracted rays are observed to be mutually perpendicular.
a. What is the index of refraction of the transparent substance?
1 = 58.0o

2 = 90.0o - 58.0o = 32.0o
n1 sin1 = n2 sin2
n2 = n1 sin1/ sin2
n2 = (1.00) sin58.0o/ sin32.0o
b. What is the critical angle for total internal reflection in this substance?
sinc = n2 / n1
sinc = (1.00)/(1.60)
c = 38.7o
TIME
OF
COMPLETION_______________
NAME___SOLUTION__________________________
DEPARTMENT OF NATURAL SCIENCES
PHYS 1112, Exam 3
Version 1
Total Weight: 100 points
Section 1
December 3, 2007
1.
Check your examination for completeness prior to starting. There are a total of ten (10)
problems on seven (7) pages.
2.
Authorized references include your calculator with calculator handbook, and the
Reference Data Pamphlet (provided by your instructor).
3.
You will have 50 minutes to complete the examination.
4.
The total weight of the examination is 100 points.
5.
There are six (6) multiple choice and four (4) calculation problems. Work five (5)
multiple choice problems and three (3) calculation problems. Show all work; partial credit will
be given for correct work shown.
6.
If you have any questions during the examination, see your instructor who will
be located in the classroom.
7.
Start:
Stop:
11:30 a.m.
12:20 p.m.
PROBLEM
POINTS
1-6
25
7
25
8
25
9
25
10
25
TOTAL
100
PERCENTAGE
CREDIT
CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN
MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK
FOR PARTIAL CREDIT.
1. A resistor is connected to an AC supply. On this circuit, the current
a. Leads the voltage by 90o.
b. Lags the voltage by 90o.
c. Is in phase with the voltage.
d. Leads the voltage by 180o.
2. The inductive reactance in an ac circuit changes by what factor when the frequency is
tripled?
a. 1/3.
b. 1/9.
c. 3.
d. 9.
3. Which one of the following is the correct order of the electromagnetic spectrum from low
to high frequencies?
a. Radio waves, infrared, microwaves, UV, visible, X-rays, gamma rays.
b. Radio waves, UV, X-rays, microwaves, infrared, visible, gamma rays.
c. Radio waves, microwaves, infrared, visible, UV, X-rays, gamma rays.
d. Radio waves, microwaves, visible, X-rays, infrared, UV, gamma rays.
4. The frequency of a microwave sygnal is 9.76 GHz. What is its wavelength in meters?
a. 3.07 x 10-2 m.
b. 2.07 x 10-2 m.
c. 1.07 x 10-2 m.
d. 5.07 x 10-2 m.
5. Which of the following is a correct statement?
a. A concave mirror always produces a real image.
b. A convex mirror always produces a virtual image.
c. A convex mirror always produces a real image.
d. A concave mirror always produces a virtual image.
6. Which of the following is correct for a convex mirror?
a. f = (1/4)R.
b. f = (1/2)R.
c. f = - (1/2)R.
d. f = 2R.
7. A 400-mH inductor with 200  resistance is connected in series to a 5.00 F capacitor
and a 60-Hz, 30-V source. Calculate
a. The impedance of the circuit.
XL = 2  f L = 2  (60.0 Hz) (400 x 10-3 H) = 151
XC = 1/(2  f C) = 1/(2  (60.0 Hz) (5.00 x 10-6 F)) = 530 
Z = (R2 + (XL – XC)2)1/2 = ((200 )2 + (151  – 530 )2)1/2 
b. The rms current that flows in the inductor.
Irms = Vrms /Z
Irms = (30.0 V)/(429) = 0.0700 A
c. The phase angle.
 tan-1 ((XL – XC)/R)
 tan-1 ((151  – 530 )/ (200 )) = -62.2o
d. Does the voltage in the inductor lead the voltage in the capacitor or lags behind?
Voltage in the inductor makes 180o angle with the voltage in the capacitor.
e. What is the resonant frequency of this circuit?
XL = XC
2  f0 L = 1/(2  f0 C)
f0 = 1/(2 Sqrt (LC)= 110 Hz
8. An incident ray makes an angle of 60o with the normal to the boundary between water (n =
1.33) and glass (n = 1.52). What is the angle between the reflected and refracted rays?
n1 sin(1) = n2 sin(2)
sin(2) = (1.33) sin(60.0o)/(1.52) = 0.758
2 = 49.2o
 = 180o – (49.2o + 60.0o) = 70.8o
9.Santa checks himself for soot, using his reflection in a shiny silvered Christmas tree ornament
25.0 cm away. The diameter of the ornament is 7.20 cm. Standard reference works state that
he is a “right jolly old elf”, so we estimate his height to be 1.60 m.
a. Where is the image of Santa formed by the ornament?
do= 25.0 cm
f = - 1.80 cm
1/di + 1/do= 1/f
1/di = 1/f - 1/do = 1/(-1.80 cm) - 1/(25.0 cm) = -0.596 cm-1
di = -1.68 cm
b. How tall is the image?
M = - di /do = -(-1.68 cm)/(25.0 cm) = 0.0671
hi= M ho= 0.107 m
c. Is the image upright or inverted?
Upright (M > 0)
d. Is the image real or virtual?
Virtual (di < 0)
TIME
OF
COMPLETION_______________
NAME____SOLUTION_________________________
DEPARTMENT OF NATURAL SCIENCES
PHYS 1112, Exam 3
Version 1
Total Weight: 100 points
Section 1
April 25, 2006
1.
Check your examination for completeness prior to starting. There are a total of ten (10)
problems on seven (7) pages.
2.
Authorized references include your calculator with calculator handbook, and the
Reference Data Pamphlet (provided by your instructor).
3.
You will have 75 minutes to complete the examination.
4.
The total weight of the examination is 100 points.
5.
There are six (6) multiple choice and four (4) calculation problems. Work all problems.
Show all work; partial credit will be given for correct work shown.
6.
If you have any questions during the examination, see your instructor who will
be located in the classroom.
7.
Start:
Stop:
10:30 a.m.
11:45 a.m.
PROBLEM
POINTS
1-6
30
CREDIT
7
20
8
15
9
15
10
20
TOTAL
100
PERCENTAGE
CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN
MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK
FOR PARTIAL CREDIT.
1. When a beam of light (wavelength = 590 nm), originally traveling in air, enters a piece of
glass (index of refraction 1.50), its frequency
a. Increases by a factor of 1.50.
b. Is reduced to 2/3 its original value.
(5)
c. Is unaffected.
d. None of the given answers.
2. Light arriving at a concave mirror on a path parallel to the axis is reflected
a. Back parallel to the axis.
b. Back on itself.
(5)
c. Through the focal point.
d. Through the center of curvature.
3. An object is situated between a concave mirror's surface and its focal point. The image
formed in this case is
a. Real and inverted.
b. Real and erect.
(5)
c. Virtual and erect.
d. Virtual and inverted.
4. Light enters air from water. The angle of refraction will be
a. Greater than the angle of incidence.
b. Equal to the angle of incidence.
(5)
c. Less than the angle of incidence.
d. Cannot be determined.
5. Lenses that are thicker at the center
a. Spread out light rays.
b. Bend light rays to a point beyond the lens.
(5)
c. Have no effect on light rays.
d. Reflect light rays back.
6. At what frequency does a 10-μF capacitor have a reactance of 1200 Ω?
a. 13 Hz.
b. 42 Hz.
(5)
c. 60 Hz.
d. 83 Hz.
7. Two 28.0-cm-focal-length converging lenses are placed 16.5 cm apart. An object is placed
36.0 cm in front of one lens.
a. Where will the final image formed by the second lens be located?
1st Lens:
do1= 36.0 cm
f1 = 28.0 cm
1/di1 + 1/do1= 1/f1
1/di1 = 1/f1 - 1/do1
1/ di1 = 1/(28.0 cm) - 1/(36.0 cm) = 0.00792 cm-1
di1 = 126 cm
M1 = - di1 /do1 = - 3.50
2nd Lens:
do2= 16.5 cm - 126 cm = -109.5 cm
f2 = 28.0 cm
1/di2 + 1/do2= 1/f2
1/di2 = 1/f2 - 1/do2
1/ di2 = 1/(28.0 cm) - 1/(-109.5 cm) = 0.0448 cm-1
di2 = 22.3 cm
M2 = - di2 /do2 = 0.204
b. What is the total magnification?
M = M1 M2 = (-3.50)(0.204) = -0.713
c. Is the image upright or inverted, real or virtual, enlarged or reduced in size?
Real (di2 >0 ), inverted (M < 0), and reduced in size (|M| < 1).
8. An aquarium filled with water has flat glass sides whose index of refraction is 1.52. A beam
of light from outside the aquarium strikes the glass at a 43.5° angle to the perpendicular.
What is the angle of this light ray when it enters
a. the glass, and then
n1 sin(1) = n2 sin(2)
sin(2) = n1 sin(1) / n2 = (1.00) sin(43.5°)/(1.52) = 0.453
2 = 26.9°
b. the water?
n2 sin(2) = n3 sin(3)
sin(3) = n2 sin(2) / n3 = (1.52) sin(26.9°)/(1.33) = 0.518
3 = 31.2°
c. What would be the refracted angle if the ray entered the water directly?
n1 sin(1) = n3 sin(3)
sin(3) = n1 sin(1) / n3 = (1.00) sin(43.5°)/(1.33) = 0.518
3 = 31.2°
9. A 135-mH inductor with 2.0- resistance is connected in series to a 20-  F capacitor and a 60Hz, 45-V source. Calculate
a. The impedance of the circuit.
XL = 2  f L = 2 (60.0 Hz) (135 x 10-3 H) = 50.9 
XC = 1/(2  f C) = 1/(2 (60.0 Hz) (20.0 x 10-6 F)) = 133 
Z = (R2 + (XL – XC)2)1/2 = 81.8 
b. The rms current.
Irms = Vrms/Z = (45.0 V)/(81.8 ) = 0.550 A
c. The phase angle.
tan  = (XL – XC)/R = (50.9133 )/ (2.00 
 = -88.6o
d. Does the current lead the voltage or the voltage leads the current in this circuit. EXPLAIN.
Current leads the voltage as indicated by the negative phase angle.
10. A 20.0-cm tall Teddy bear stands 120.0 cm from the vertex of a concave mirror having a
radius of curvature of 60.0 cm.
a. How far is the image from the mirror?
do= 120 cm
f1 = R/2 = 30.0 cm
1/di + 1/do= 1/f
1/di = 1/f - 1/do
1/ di = 1/(30.0 cm) - 1/(120 cm)
di = 40.0 cm
b. What is the height of the Teddy image?
M = - di /do = - (40.0 cm)/(120 cm) = -1/3
M = hi /ho
hi= ho M = (20.0 cm)(-1/3) = - 6.67 cm
c. Draw a ray diagram to check your answers.
c. Is the image real or virtual, upright or inverted, enlarged or reduced?
Real (image is formed in front of the mirror), inverted (M < 0), reduced (|M| < 1).
TIME
OF
COMPLETION_______________
NAME___SOLUTION__________________________
DEPARTMENT OF NATURAL SCIENCES
PHYS 1112, Exam 3
Version 1
Total Weight: 100 points
Section 1
April 24, 2007
1.
Check your examination for completeness prior to starting. There are a total of ten (10)
problems on eight (8 ) pages.
2.
Authorized references include your calculator with calculator handbook, and the
Reference Data Pamphlet (provided by your instructor).
3.
You will have 75 minutes to complete the examination.
4.
The total weight of the examination is 100 points.
5.
There are six (6) multiple choice and four (4) calculation problems. Work all problems.
Show all work; partial credit will be given for correct work shown.
6.
If you have any questions during the examination, see your instructor who will
be located in the classroom.
7.
Start:
Stop:
10:30 a.m.
11:45 a.m.
PROBLEM
POINTS
1-6
20
7
20
8
20
9
20
10
20
TOTAL
100
PERCENTAGE
CREDIT
CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN
MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK
FOR PARTIAL CREDIT.
1. Light arriving at a concave mirror on a path through the center of curvature is reflected
a. Back parallel to the axis.
b. Back on itself.
(4)
c. Through the focal point.
d. Midway between the focal point and the center of curvature.
2. An object is placed at a concave mirror's center of curvature. The image produced by the
mirror is located
a. Out beyond the center of curvature.
b. At the center of curvature.
(4)
c. Between the center of curvature and the focal point.
d. At the focal point.
3. Concave spherical mirrors produce images which
a. Are always smaller than the actual object.
b. Are always larger than the actual object.
(4)
c. Are always the same size as the actual object.
d. Could be smaller than, larger than, or the same size as the actual object, depending on
the placement of the object.
4. Light enters air from water. The angle of refraction will be
a. Greater than the angle of incidence.
(4)
b. Equal to the angle of incidence.
c. Less than the angle of incidence.
5. Lenses that are thicker at the center
a. Spread out light rays.
b. Bend light rays to a point beyond the lens.
(4)
c. Have no effect on light rays.
d. Reflect light rays back.
6. Radio, microwaves, infrared radiation, and visible light are
a. Portions of the electromagnetic spectrum.
b. Portions of the electromagnetic spectrum, except radio waves which are not.
(4)
c.
Portions of the electromagnetic spectrum, except radio and micro waves which are
not.
d.
Portions of the electromagnetic spectrum, except infrared radiation which is not.
7. An object is 16.0 cm to the left of the lens. The lens forms an image 36.0 cm to the right of the
lens.
a. What is the focal length of the lens?
do= 16.0 cm
di= 36.0 cm
1/di + 1/do= 1/f
1/f = 1/(16.0 cm) + 1/(36.0 cm) = 0.0903 cm-1
f = 11.1 cm
b. Is the lens converging or diverging?
Converging (f > 0)
c. If the object is 8.00 cm tall, how tall is the image?
M = - di /do = -2.25
hi= M ho= -18.0 mm
d. Is it upright or inverted?
Inverted (M < 0)
e. Draw a ray diagram.
8. An object 9.00 mm tall is placed 12.0 cm to the left of the vertex of a convex spherical
mirror whose radius of curvature has a magnitude of 20.0 cm.
a. Determine the position of the image.
do= 12.0 cm
f = - 10.0 cm
1/di + 1/do= 1/f
1/di = 1/f - 1/do
1/ di = 1/(-10.0 cm) - 1/(12.0 cm) = 0.183 cm-1
di = -5.45 cm
b. How tall is the image?
M = - di /do = 0.455
hi= M ho= 4.09 mm
c. Is the image real or virtual?
Virtual (di <0)
d. Is the image upright or inverted?
Upright (M > 0)
9. An object 10.00 cm high is placed 12.0 cm to the left of the converging lens of focal
length 8.00 cm. A second converging lens of focal length of 6.00 cm is placed 36.0 cm
to the right of the first lens on the same principal axis.
a. Where is the image produced by the combination located?
1st Lens:
do1= 12.0 cm
f1 = 8.00 cm
1/di1 + 1/do1= 1/f1
1/di1 = 1/f1 - 1/do1
1/ di1 = 1/(8.00 cm) - 1/(12.0 cm) = 0.0417 cm-1
di1 = 24.0 cm
M1 = - di1 /do1 = - 2.00
2nd Lens:
do2= 36.0 cm – 24.0 cm = 12.0 cm
f2 = 6.00 cm
1/di2 + 1/do2= 1/f2
1/di2 = 1/f2 - 1/do2
1/ di2 = 1/(6.00 cm) - 1/(12.0 cm) = 0.0833 cm-1
di2 = 12.0 cm
M2 = - di2 /do2 = -1.00
b. How high is it?
M = M1 M2 = (-2.00)(-1.00) = 2.00
hi= M ho= 20.0 cm
c. Is the image upright or inverted?
Upright (M > 0)
10. You have a 200- resistor, a 0.400-H inductor, and a 6.00-F capacitor. They are connected
in series with an ac source of peak voltage of 30.0 V operating at 100 Hz frequency.
a. What is the impedance of the circuit?
XL = 2  f L = 2 (100 Hz) (0.400 H) = 251 
XC = 1/(2  f C) = 1/(2 (100 Hz) (6.00 x 10-6 F)) = 265 
Z = (R2 + (XL – XC)2)1/2 = 200 
b. What is the rms current flowing in the resistor?

Vrms = Vmax/(2)1/2 = 21.2 V
Irms = Vrms/Z = (21.2 V)/(200 ) = 0.106 A
c. What is the rms potential drop across the resistor?
VR, rms = Irms R = (0.106 A)(200 ) = 21.2 V
d. What is the rms power dissipated by the circuit?
Prms = Irms VR, rms = (0.106 A)(21.2 V) = 2.25 W
e. At this frequency, will the voltage lead or lag the current?
tan  = (XL – XC)/R = (251 265 )/(200 
 = -4.00o Voltage lags the current.
TIME
OF
COMPLETION_______________
NAME_________SOLUTION____________________
DEPARTMENT OF NATURAL SCIENCES
PHYS 1112, Exam 3
Version 1
Total Weight: 100 points
Section 1
April 17, 2008
1.
Check your examination for completeness prior to starting. There are a total of ten (10)
problems on eight (8) pages.
2.
Authorized references include your calculator with calculator handbook, and the
Reference Data Pamphlet (provided by your instructor).
3.
You will have 75 minutes to complete the examination.
4.
The total weight of the examination is 100 points.
5.
There are six (6) multiple choice and four (4) calculation problems. Work five (5)
multiple choice problems and four (4) calculation problems. Show all work; partial credit will be
given for correct work shown.
6.
If you have any questions during the examination, see your instructor who will
be located in the classroom.
7.
Start:
Stop:
10:30 a.m.
11:45 a.m.
k
PROBLEM
POINTS
1-6
20
7
20
8
20
CREDIT
9
20
10
20
TOTAL
100
PERCENTAGE
CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN
MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK
FOR PARTIAL CREDIT.
1. Which one of the following is the correct number for the magnification of the plane mirror?
A) 1.0.
B) 0.5.
(4)
C) 1.5.
D) 2.0.
2. The perpendicular distance from the center of curvature to a spherical mirror is called
A) The center of curvature.
B) The radius of curvature.
(4)
C) The focal length.
D) Twice the radius of curvature.
3. A beam of light that is parallel to the principal axis is incident on a convex mirror. What
happens to the reflected beam of light?
A) It also is parallel to the principal axis.
B) It is perpendicular to the principal axis.
(4)
C) It appears to be coming from the focal point on the other side of the mirror.
D) It appears to be coming from the center of the curvature of the mirror.
4. A ray of light in air enters a glass surface at right angles to the surface. What is the angle of
refraction?
A) 0o.
B) 90o.
(4)
C) 180o.
D) 270o.
5. The speed of light in a certain medium is 2.20 x 108 m/s. What is the index of refraction of
this medium?
A) 0.73.
B) 1.00.
(4)
C) 1.36.
D) 1.60.
6. Which one of the following properties of light is evidence of the wave nature of light?
A) Reflection of light.
B) Refraction of light.
(4)
C) Interference of light.
D) None of the above.
7. An object with a height of 6.50 cm is placed 8.00 cm in front of a lens. An upright image
of the object is formed at a distance of 3.00 cm in front of the lens.
a. What is the focal length of the lens?
do= 8.00 cm
di= -3.00 cm
1/di + 1/do= 1/f
1/ f = 1/(8.00 cm) + 1/(-3.00 cm) = -5/(24.0 cm)
f = -4.80 cm
b. Is the lens converging or diverging?
Diverging (f < 0)
c. What is the height of the image?
M = - di /do = - (-3.00 cm)/(8.00 cm) = 0.375
M = hi /ho
hi= ho M = (6.50 cm)(0.375) = 2.44 cm
d. Is the image real or virtual?
Virtual (formed in front of the lens).
8. An insect 3.75 mm tall is placed 22.5 cm to the left of a thin plano-convex lens. The left
surface of the lens is flat, the right surface has a radius of curvature of magnitude 13.0
cm, and the index of refraction of the lens material is 1.70.
a. Is this lens converging or diverging?
Converging.
b. What is the focal length of the lens?
1/f = (n-1)(1/R1 – 1/R2)
R1 = ∞
R1 = -13.0 cm
1/f = (1.70 - 1)(1/∞ – 1/(-13.0 cm)) = 0.0539 cm-1
f = 18.6 cm
c. What is the location of the image this lens forms of the insect?
do= 22.5 cm
1/di + 1/do= 1/f
1/di = 1/f - 1/do
1/ di = 1/(18.6 cm) - 1/(22.5 cm)
di = 106 cm
d. Is the image real or virtual? Upright or inverted? Enlarged or reduced in size?
Real (di >0), inverted (M < 0), enlarged (|M| >1)
e. Draw a ray diagram to confirm your answers.
9. A concave mirror forms an image, on a wall 3.00 m in front of the mirror, of the filament
of a headlight lamp positioned 10.0 cm from the mirror.
a. What is the focal length of the mirror?
do= 10.0 cm
f = 300 cm
1/di + 1/do= 1/f
1/ f = 1/(10.0 cm) + 1/(300 cm)
f = 9.68 cm
b. What is the radius of curvature of the mirror?
R=2f
R = 2 (9.68 cm) = 19.4 cm
c. What is the height of the image if the height of the object is 5.00 mm?
M = - di /do = - (300 cm)/(10.0 cm) = -30.0
M = hi /ho
hi= ho M = (5.00 mm)(-30.0) = -150 cm = -15.0 cm
d. Is the image real or virtual?
Real (image is formed in front of the mirror)
e. Is the image upright ore inverted?
Inverted (M < 0)
10. The light beam makes an angle of 20.0o with the normal NN’ in the linseed oil. Determine the
angles  and ’. (The index of refraction for linseed oil is 1.48, the index of refraction for water
is 1.33.)
Air-Linseed oil Boundary
n1 sin(1) = n2 sin(2)
(1.00) sin() = (1.48) sin()
sin() = (1.00) sin(20.0o)/(1.48) = 0.506
 = 30.4o
Linseed oil – Water Boundary
n1 sin(1) = n2 sin(2)
(1.48) sin() = (1.33) sin(')
sin(') = (1.48) sin(20.0o)/(1.33) = 0.381
' = 22.4o
TIME
OF
COMPLETION_______________
NAME___SOLUTION__________________________
DEPARTMENT OF NATURAL SCIENCES
PHYS 1112, Exam 3
Version 1
Total Weight: 100 points
Section 1
July 31, 2006
1.
Check your examination for completeness prior to starting. There are a total of ten (10)
problems on eight (8) pages.
2.
Authorized references include your calculator with calculator handbook, and the
Reference Data Pamphlet (provided by your instructor).
3.
You will have 75 minutes to complete the examination.
4.
The total weight of the examination is 100 points.
5.
There are six (6) multiple choice and four (4) calculation problems. Work all problems.
Show all work; partial credit will be given for correct work shown.
6.
If you have any questions during the examination, see your instructor who will
be located in the classroom.
7.
Start:
Stop:
2:30 p.m.
3:50 p.m.
PROBLEM
POINTS
1-6
20
7
20
8
20
9
20
10
20
TOTAL
100
PERCENTAGE
CREDIT
CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN
MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK
FOR PARTIAL CREDIT.
1. Plane mirrors produce images which
a. Are always smaller than the actual object.
b. Are always larger than the actual object.
(4)
c. Are always the same size as the actual object.
d. Could be smaller, larger, or the same size as the actual object, depending on the
placement of the object.
2. Is it possible to see a virtual image?
a. No, since the rays that seem to emanate from a virtual image do not in fact emanate
from the image.
b. No, since virtual images do not really exist.
(4)
c. Yes, since almost everything we see is virtual because most things do not themselves
give off light, but only reflect light coming from some other source.
d. Yes, but only indirectly in the sense that if the virtual image is formed on a sheet of
photographic film, one could later look at the picture formed.
3. Light arriving at a concave mirror on a path through the focal point is reflected
a. Back parallel to the axis.
b. Back on itself.
(4)
c. Through the focal point.
d. Through the center of curvature.
4. A pure inductor is connected to an AC power supply. In this circuit, the current
a. Leads the voltage by 90°.
b. Lags the voltage by 90°.
(4)
c. Is in phase with the voltage.
d. None of the given answers.
5. The capacitive reactance in an ac circuit changes by what factor when the frequency is
tripled?
a. 1/3.
b. 1/9.
(4)
c. 3.
d. 9.
6. All electromagnetic waves travel through a vacuum at
a. The same speed.
b. Speeds that are proportional to their frequency.
(4)
c. Speeds that are inversely proportional to their frequency.
d. None of the given answers.
7. A 2.00-cm-high insect is 1.20 m from a 135-mm-focal-length lens.
a. Where is the image?
do= 120 cm
f = 13.5 cm
1/di + 1/do= 1/f
1/di = 1/f - 1/do
1/ di = 1/(13.5 cm) - 1/(120 cm)
di = 15.2 cm
c.
How high is it?
M = - di /do = - (15.2 cm)/(120 cm) = -0.127
M = hi /ho
hi= ho M = (2.00 cm)(-0.127) = - 0.253 cm
d. Is the image real or virtual?
Real (di > 0)
e. Is the image upright or inverted?
Inverted (M < 0)
f. Is the image reduced in size or magnified?
Reduced in size (|M| < 1)
(Continued on the next page)
g. Draw a ray diagram to confirm your answers.
8. Two 28.0-cm-focal-length converging lenses are placed 16.5 cm apart. An object is
placed 36.0 cm in front of one lens.
a. Where will the final image formed by the second lens be located?
1st Lens:
do1= 36.0 cm
f1 = 28.0 cm
1/di1 + 1/do1= 1/f1
1/di1 = 1/f1 - 1/do1
1/ di1 = 1/(28.0 cm) - 1/(36.0 cm) = 0.00792 cm-1
di1 = 126 cm
M1 = - di1 /do1 = - 3.50
2nd Lens:
do2= 16.5 cm - 126 cm = -109.5 cm
f2 = 28.0 cm
1/di2 + 1/do2= 1/f2
1/di2 = 1/f2 - 1/do2
1/ di2 = 1/(28.0 cm) - 1/(-109.5 cm) = 0.0448 cm-1
di2 = 22.3 cm
M2 = - di2 /do2 = 0.204
b. What is the total magnification?
M = M1 M2 = (-3.50)(0.204) = -0.713
9. What is the rms current in an RC circuit if R  8.80 k, C  1.80  F, and the rms applied voltage
is 120 V at 60.0 Hz?
XL = 0
XC = 1/(2  f C) = 1/(2 (60.0 Hz) (1.80 x 10-6 F)) = 1470 
Z = (R2 + (XL – XC)2)1/2 = 8922 
Irms = Vrms/Z = (120 V)/(8922 ) = 0.0134 A
b. What is the phase angle between voltage and current?
tan  = (XL – XC)/R = (01470 )/ (8800 
 = -9.48o
c.What are the voltmeter readings across R and C?

Vrms, C = XC Irms
Vrms, C = (1470 )(0.0134 A) = 19.7 V
Vrms, R = R Irms
Vrms, R = (8800 )(0.0134 A) = 118 V
d. Does the current lead the voltage in the circuit or lags behind the voltage?
Current leads the voltage as indicated by the negative phase angle.
e.What is the rms power dissipated in the resistor?
P = Irms Vrms cos  = (0.0134 A)(120 V) cos(-9.48o) = 1.59 W
10. A certain sample of a flint glass has an index of refraction of 1.571 for red light (656 nm) and
1.594 for violet light (434 nm). If white light is incident from air at an incidence angle of 35.0o,
what is the angular separation of the red and violet rays in the refracted beam?
Red:
n1 sin(1) = n2 sin(2)
sin(2) = n1 sin(1)/ n2
sin(2) = (1.00) sin(35.0o)/(1.571) = 0.365
1 = 21.41o
Violet:
n1 sin(1) = n2 sin(2)
sin(2) = n1 sin(1)/ n2
sin(2) = (1.00) sin(35.0o)/(1.594) = 0.360
1 = 21.09o
 = 21.41o21.09o= 0.32o
TIME
OF
COMPLETION_______________
NAME_____SOLUTION________________________
DEPARTMENT OF NATURAL SCIENCES
PHYS 1112, Exam 3
Version 1
Total Weight: 100 points
Section 1
July 26, 2007
1.
Check your examination for completeness prior to starting. There are a total of ten (10)
problems on eight (8) pages.
2.
Authorized references include your calculator with calculator handbook, and the
Reference Data Pamphlet (provided by your instructor).
3.
You will have 75 minutes to complete the examination.
4.
The total weight of the examination is 100 points.
5.
There are six (6) multiple choice and four (4) calculation problems. Work all problems.
Show all work; partial credit will be given for correct work shown.
6.
If you have any questions during the examination, see your instructor who will
be located in the classroom.
7.
Start:
Stop:
1:30 p.m.
2:45 p.m.
PROBLEM
POINTS
1-6
20
7
20
8
20
9
20
10
20
TOTAL
100
PERCENTAGE
CREDIT
CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN
MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK
FOR PARTIAL CREDIT.
1. If you stand in front of a convex mirror, at the same distance from it as its focal length,
a. You won’t see your image because there is none.
b. You won't see your image because it's focused at a different distance.
(4)
c. You will see your image and you will appear smaller.
d. You will see your image and you will appear larger.
e. You will see your image at your same height.
2. A negative magnification for a mirror means
a. The image is inverted, and the mirror is concave.
b. The image is inverted, and the mirror is convex.
(4)
c. The image is inverted, and the mirror may be concave or convex.
d. The image is upright, and the mirror is convex.
e. The image is upright, and the mirror may be concave or convex.
3. A ray of light, which is traveling in air, is incident on a glass plate at a 45° angle. The angle
of refraction in the glass
a. Is less than 45°.
b. Is greater than 45°.
(4)
c. Is equal to 45°.
d. Could be any of the above; it all depends on the index of refraction of glass.
4. Which one of the following is not an electromagnetic wave?
a. UV.
b. Infrared.
(4)
c. Radio waves.
d. Sound waves.
e. Gamma rays.
5. Resonance in a series RLC circuit occurs when
a. XL is greater than XC.
b. XL is less than XC.
(4)
c. XL is equal to XC.
d. (XL - XC) is equal to R2.
6. If you triple the rms voltage in an ac circuit, what happens to the maximum voltage?
a. It increases by a factor of 3½ .
b. It decreases by a factor of 3½.
(4)
c. It increases by a factor of 3.
d. It decreases by a factor of 3.
7. A series RC circuit has R  8.80 k, C  1.80  F, and the rms applied voltage is 120 V at
60.0 Hz.
a. What is the impedance of the circuit?
XL = 0
XC = 1/(2  f C) = 1/(2  (60.0 Hz) (1.80 x 10-6 F)) = 1474 
Z = (R2 + (XL – XC)2)1/2 = ((8800 )2 + (0  – 1474 )2)1/2 
b. What is the rms current flowing in the circuit?
Irms = Vrms /Z
Irms = (120 V)/(8923) = 0.0134 A
c. What is the phase angle between voltage and current?
 tan-1 ((XL – XC)/R)
 tan-1 ((0  – 1474 )/ (8800 )) = -9.51o
 < 0, hence voltage lags behind current.
d. What is the average power dissipated in the circuit?
P = Irms Vrmscos0.0134 A)(120 V) cos(-9.51o ) = 1.59 W
e. What are the voltmeter readings across R and C?

VC, rms = Irms XC = (0.0134 A)(1474) = 19.8 V

VR, rms = Irms R = (0.0134 A)(8800) = 118 V
8. A horse is 2.25 m tall and it stands with its face 15.0 m from the thin lens with the focal
length of +3.00 m.
a. Determine the location of the equine nose.
do= 15.0 m
f = + 3.00 m
1/di + 1/do= 1/f
1/di = 1/f - 1/do = 1/(3.00 m) - 1/(15.0 m) = 4/(15.0 m)
f = 15/4 m = 3.75 m
b. What is the magnification?
M = - di /do = -(3.75 m)/(15.0 m) = - 0.250
c. How tall is the image?
hi= M ho= -0.563 m
d. Is the image real or virtual?
Real (di >0)
e. Is the image upright or inverted?
Inverted (M < 0)
f. Draw a ray diagram to check your answers.
9. We wish to place an object 45.0 cm in front of a spherical mirror and have its image appear on
a screen 90.0 cm in front of the mirror.
d. Should we use a concave or a convex mirror?
Concave (Convex mirror will form an image in the back of the lens)
e. What must be the radius of curvature of the mirror?
do= 45.0 cm
di= 90.0 cm
1/di + 1/do= 1/f
1/f = 1/(45.0 cm) + 1/(90.0 cm) = 1/(30.0 cm)
R = 60.0 cm
f. What is the lateral magnification?
M = - di /do = - (90.0 cm)/(45.0 cm) = -2.00
g. Is the image real or virtual? Upright or inverted? Enlarged or reduced in size?
Real (di >0 ), Inverted (M < 0), Enlarged (|M| > 1)
h. Draw a ray diagram to check your answers.
10.A light ray is incident at an angle of 25o on the surface between air and heavy flint glass (n =
1.65). What angle does the refracted ray make with the normal to the surface when the ray is
incident from
a. The air side?
n1 sin(1) = n2 sin(2)
sin(2) = (1.00) sin(25.0o)/(1.65) = 0.256
1 = 14.8o
b. From the glass side?
n1 sin(1) = n2 sin(2)
sin(2) = (1.65) sin(25.0o)/(1.00) = 0.697
1 = 44.2o
TIME
OF
COMPLETION_______________
NAME____SOLUTION_________________________
DEPARTMENT OF NATURAL SCIENCES
PHYS 1112, Exam 3
Version 1
Total Weight: 100 points
Section 1
July 17, 2008
1.
Check your examination for completeness prior to starting. There are a total of ten (10)
problems on seven (7) pages.
2.
Authorized references include your calculator with calculator handbook, and the
Reference Data Pamphlet (provided by your instructor).
3.
You will have 75 minutes to complete the examination.
4.
The total weight of the examination is 100 points.
5.
There are six (6) multiple choice and four (4) calculation problems. Work five (5)
multiple choice problems and four (4) calculation problems. Show all work; partial credit will be
given for correct work shown.
6.
If you have any questions during the examination, see your instructor who will
be located in the classroom.
7.
Start:
Stop:
1:30 p.m.
2:45 p.m.
PROBLEM
POINTS
1-6
20
7
20
8
20
9
20
10
20
TOTAL
100
PERCENTAGE
CREDIT
CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN
MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK
FOR PARTIAL CREDIT.
1. The inductive reactance of a 20.0-mH inductor at a certain frequency is 120 . What is the
angular frequency in rad/s?
(A) 2000 rad/s.
(B) 4000 rad/s.
(4)
(C) 500 rad/s.
(D) 6000 rad/s.
(E) 7000 rad/s.
2. In a RLC circuit, the values of the inductance and capacitance are both doubled. In
comparison with the resonance frequency of the original circuit, the new frequency will be
(A) The same as before.
(B) Reduced to one-half the original value.
(4)
(C) Reduced to one-quarter the original value.
(D) Increased by a factor of two.
(E) Increased by a factor of four.
3. The angle of incidence is the angle between
(A) The incident ray and the reflected ray.
(B) The incident ray and the normal drawn at the point where this ray meets the mirror.
(4)
(C) The reflected ray and the normal drawn at the point where this ray meets the mirror .
(D) The incident ray and the surface of the mirror.
(E) The reflected ray and the surface of the mirror.
4. The index of refraction, n is given by
(A) n = c v .
(B) n = cv2.
(4)
(C) n = v/c.
(D) n = c/v.
(E) None .of the other answers is correct
5. A ray of light strikes the mirror perpendicular to the mirror. What is the angle of reflection?
(A) 0o.
(B) 90o.
(4)
(C) 180o.
(D) 360o.
(E) 270o.
6. Lucite has an index of refraction of 1.50. What is its critical angle of incidence?
(A) 1.16o.
(B) 15.0o.
(4)
(C) 41.8o.
(D) 65.2o.
(E) 87.4o.
7. An ac generator with a maximum emf of 20.0 V is connected in series with a 20.0-F
capacitor and an 80.0- resistor, There is no inductance in the circuit. Find
a. the power factor,
XL = 0
XC = 1/( C) = 1/((400 rad/s) (20.0 x 10-6 F)) = 125 
tan  = (XL – XC)/R = (0125 )/ (80.0 
 = -57.4o
cos  = 0.539
b. the rms current,
Z = (R2 + (XL – XC)2)1/2 = 148 
Irms = Vrms/Z = [(20 V)/sqrt(2)]/(148 ) = 0.0955 A
c. the average power if the angular frequency of the generator is 400 rad/s.
P = Irms Vrms cos  = (0.0955 A)(14.1 V) cos(-57.4o) = 0.728 W
8. A slab of glass with an index of refraction of 1.50 is submerged in water with an index of
refraction of 1.33. Light in the water is incident on the glass. Find the angle of refraction if the
angle of incidence is
a. 60.0o.
n1 sin(1) = n2 sin(2)
sin(2) = n1 sin(1)/ n2
sin(2) = (1.33) sin(60.0o)/(1.50) = 0.768
1 = 50.2o
b. 45.0o.
n1 sin(1) = n2 sin(2)
sin(2) = n1 sin(1)/ n2
sin(2) = (1.33) sin(45.0o)/(1.50) = 0.627
1 = 38.8o
c. 30.0o
n1 sin(1) = n2 sin(2)
sin(2) = n1 sin(1)/ n2
sin(2) = (1.33) sin(30.0o)/(1.50) = 0.443
1 = 26.3o
9. A series ac circuit contains a 100 - resistor, a 30.0-F capacitor, a 0.100-H inductor,
and a 50.0-Hz generator that has an rms output of 45.0 V.
a. Find the impedance of the circuit.
XL = 2  f L = 2 (50.0 Hz) (0.100 H) = 31.4 
XC = 1/(2  f C) = 1/(2 (50.0 Hz) (30.0 x 10-6 F)) = 106 
Z = (R2 + (XL – XC)2)1/2 = 125 
b. Find the maximum current that flows in the resistor.
Imax = Vmax/Z = (45.0 V) sqrt(2)/(125 ) = 0.510 A
c. Find the maximum potential difference across the capacitor.
Vmax, C = XC Imax
Vmax, C = (106 )(0.510 A) = 54.1 V
d. Does the voltage lead the current in the circuit or the current leads the voltage?
EXPLAIN.
tan  = (XL – XC)/R = (31.4 106 )/ (100 
 = -36.7o
Current leads the voltage since the phase angle is negative.
10. The current in a 0.800-H solenoid increases from 20.0 mA to 160 mA in 7.00 s.
a. What is the average emf induced in the solenoid during that time interval?
 = -L t
 = (0.800 H)(0.160 A – 0.020 A)/(7.00 s) = 0.0160 V
b. How much energy is stored in the solenoid at the moment the current reaches 160mA value?
U = ½ L I2 = ½ (0.800 H)(0.160 A)2 = 0.0102 J
TIME
OF
COMPLETION_______________
NAME___SOLUTION__________________________
DEPARTMENT OF NATURAL SCIENCES
PHYS 1112, Exam 3
Version 1
Total Weight: 100 points
Section 1
December 2, 2009
1.
Check your examination for completeness prior to starting. There are a total of nine (9)
problems on six (6) pages.
2.
Authorized references include your calculator with calculator handbook, and the
Reference Data Pamphlet (provided by your instructor).
3.
You will have 50 minutes to complete the examination.
4.
The total weight of the examination is 100 points.
5.
There are six (6) multiple choice and three (3) calculation problems. Work five (5)
multiple choice and three (3) calculation problems. Show all work; partial credit will be given for
correct work shown.
6.
If you have any questions during the examination, see your instructor who will
be located in the classroom.
7.
Start:
Stop:
12:00 p.m.
12:50 p.m
PROBLEM
POINTS
1-6
25
7
35
8
25
9
25
TOTAL
100
PERCENTAGE
CREDIT
CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN
MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK
FOR PARTIAL CREDIT.
5. An electric current is induced in a conducting loop by all but one of these processes.
Which one does NOT produce an induced current?
a. Rotating the loop inside a constant magnetic field.
b. Placing the loop so that its area is perpendicular to the increasing magnetic field.
(5)
c. Moving the loop parallel to the uniform magnetic field.
d. Expanding the area of the loop while it is perpendicular to a uniform magnetic
field.
6. A horizontal rod (oriented in the east-west direction) is moved northward at constant
velocity through a magnetic field that points straight down. Make a statement concerning
the potential induced across the rod.
a. The west end of the rod is at higher potential than the east end.
b. The east end of the rod is at higher potential than the west end.
(5)
c. The top surface of the rod is at higher potential than the bottom surface.
d. The bottom surface of the rod is at higher potential than the top surface.
7. What is the rms current value for an ac current with the maximum value of 10.0 A?
a. 28.2 A.
b. 3.10 A.
(5)
c. .
d. .
e.
8. A series RLC circuit is connected to an as generator. When the generator frequency
varies ( but the maximum emf is constant), the average power is:
a. A minimum when |XL – XC | = R.
b. A minimum when XL = XC.
(5)
c. A maximum when |XL – XC | = R.
d. A maximum when XL = XC.
9. Which of the following statements is correct for a series RLC circuit?
a. The voltage across the capacitor leads the voltage across the resistor by 45o.
b. The voltage across the inductor leads the voltage across the resistor by 180o.
(5)
c. The voltage across the resistor leads the voltage across the capacitor by 90o.
d. The voltage across the inductor leads the voltage across the capacitor by 90o.
10. The current in the long wire is decreasing. What is the direction of the current induced in
the conducting loop below the wire?
a. Counterclockwise.
b. Clockwise.
(5)
c. Clockwise or counterclockwise depending on the size of the loop.
d. No current is induced.
7. A square, single-turn wire loop 1.00 cm on the side is placed inside a solenoid that has a
radius of 3.00 cm and is 20.0 cm long. The solenoid has 100 turns. The current in the
solenoid is 30.0 A flowing counterclockwise.
c. Find the magnetic flux through the square loop.
B = 0 n I = (4  x 10-7 T-m/A)(100 turns/0.200 m)(30.0 A) = 0.0188 T
i = B A cos (0o) = (0.0188 T)((0.0100 m)2) = 1.885 x 10-6 Wb
d. The square loop is turned through 90o angle so that its plane is now oriented parallel to
the axis of the solenoid. What is the magnetic flux through the solenoid now?
f = B A cos (90o) = 0
e. If it took 22.0 ms to complete the rotation, what is the average emf induced in the square
loop?
 = (f -i)/t = (0 – 1.885 x 10-6 Wb)/(0.022 s) = 8.57 x 10-5 V
8. A series RLC circuit contains a 1.00 F capacitor, a 5.00 mH coil, a 100 Ohms resistor,
and a generator producing a maximum voltage of 100 V at a frequency of 5000 Hz.
a. Calculate the impedance of the circuit.
XL = 2  f L = 2  (5000 Hz) (0.005 H) = 157 
XC = 1/2  f C = 1/(2  (5000 Hz) (1.00 x 10-6 F)) = 31.8 
Z = (R2 + (XL – XC)2)1/2 = ((100 )2 + (157  – 31.8 )2)1/2 
b. Calculate the maximum current in the circuit.
Imax = Vmax /Z
Imax = (100 V)/(160) = 0.625 A
c. Calculate the maximum voltage drop across the resistor, the capacitor, and the
inductor.
VC,max = Imax XC = (0.625 A)(31.8) = 20.0 V
VL,max = Imax XL = (0.625 A)(157) = 98.1 V
VR,max = Imax R = (0.625 A)(100) = 62.5 V
d. Does the current in this circuit lead the voltage or does the voltage lead the
current? EXPLAIN your answer.
 tan-1 ((XL – XC)/R)
 tan-1 ((157  – 31.8 )/ (100 )) = 51.3o
Voltage leads the current in this circuit (phase angle is positive).
9.A solenoid is made of 300 turns of wire wrapped around a hollow cylinder of radius
1.20 cm and length 6.00 cm.
a. What is the self-inductance of the solenoid?
L = 0AN2/l = (4  x 10-7 Tm/A)(300)2(0.0120 m)2/(0.0600 m) = 8.53 x10-4 H
b. If the current in the solenoid is decreasing at a rate of 35.0 A/s, what is the emf induced in
the solenoid??
|| = L I/t = (8.53 x 10-4 H)(35.0 A/s) = 0.0298 V
TIME
OF
COMPLETION_______________
NAME___SOLUTION__________________________
DEPARTMENT OF NATURAL SCIENCES
PHYS 1112, Exam 3
Version 1
Total Weight: 100 points
Section 1
December 1, 2010
1.
Check your examination for completeness prior to starting. There are a total of nine (9)
problems on six (6) pages.
2.
Authorized references include your calculator with calculator handbook, and the
Reference Data Pamphlet (provided by your instructor).
3.
You will have 50 minutes to complete the examination.
4.
The total weight of the examination is 100 points.
5.
There are six (6) multiple choice and three (3) calculation problems. Work five (5)
multiple choice problems and three (3) calculation problems. Show all work; partial credit will
be given for correct work shown.
6.
If you have any questions during the examination, see your instructor who will
be located in the classroom.
7.
Start:
Stop:
11:00 a.m.
11:50 a.m.
PROBLEM
POINTS
1-6
25
7
25
8
25
9
25
TOTAL
100
CREDIT
PERCENTAGE
CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN
MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK
FOR PARTIAL CREDIT.
1. A layer of water (n = 1.333) floats on a container of carbon tetrachloride (n = 1.42). What is
the critical angle of the interface?
a. 88o.
b. 78o.
(5)
c. 66o.
d. 58o.
e. 43o.
2. A circular coil lies flat on a horizontal table. A bar magnet is held above its center with
its north pole pointing down. The stationary magnet induces (when viewed from above)
e. No current in the coil.
f. A clockwise current in the coil.
(5)
g. A counterclockwise current in the coil.
h. A current whose direction cannot be determined from the information given.
3. A horizontal rod (oriented in the east-west direction) is moved northward at constant
velocity through a magnetic field that points straight down. Make a statement concerning
the potential induced across the rod.
a. The west end of the rod is at higher potential than the east end.
b. The east end of the rod is at higher potential than the west end.
(5)
c. The top surface of the rod is at higher potential than the bottom surface.
d. The bottom surface of the rod is at higher potential than the top surface.
4. Which of the following is a correct statement?
e. A concave mirror always produces a real image.
f. A convex mirror always produces a virtual image.
g. A convex mirror always produces a real image.
h. A concave mirror always produces a virtual image.
5. Which of the following is correct for a convex mirror?
e. f = (1/4)R.
f. f = (1/2)R.
g. f = - (1/2)R.
h. f = 2R.
6. . A ray of light strikes the mirror perpendicular to the mirror. What is the angle of
reflection?
PHYS 1112 Exam 3, Version 1
Fall 2004
67
a.
0o.
b. 90o.
c. 180o.
d. 270o.
8. A spherical concave shaving mirror has a radius of curvature of 25.0 cm.
a. Where is the image of a person’s face when it is 10.0 cm to the left from the mirror?
f = R/2 = 12.5 cm
p = 10.0 cm
1/p + 1/q = 1/f
1/q = 1/f – 1/p = 1/(12.5 cm) – 1/(10.0 cm) = -1/(50.0 cm)
q = -50.0 cm
b. What is the magnification?
M = -q/p = - (-50.0 cm)/(10.0c m) = 5.00
c. Is the image real or virtual? Upright or inverted? Enlarged or reduced in size?
Virtual (q < 0)
PHYS 1112 Exam 3, Version 1
Fall 2004
68
Upright (M > 0)
Enlarged (|M| > 1)
8. Bar ab moves without friction on conducting rails as shown below. There is a uniform
magnetic field with magnitude of 0.600 T directed into the plane of the page. You want to
make an exercise machine out of this apparatus, in which the person exercises by pushing
the bar back and forth with an average speed of 4.00 m/s. What should be the circuit
resistance if the person moving the bar is to do work at an average rate of 200 W?
 = B V L = (0.600 T)(4.00 m/s)(3.00 m) = 7.20 V
P =  /R
R =  /P = 0.259 
9. Light traveling in air is incident on the surface of a block of plastic at an angle of 62.7o to the
normal and is bent so that it makes a 48.1o angle with the normal in the plastic.
a. What is the index of refraction of the plastic?
n1 sin (1) = n2 sin (2)
n2 = n1 sin (1)/sin (2) = 1.19
b. Find the speed of light in the plastic.
n = c/v
v = c/n=(3.00 x 108 m/s)/1.19 = 2.51 x 108 m/s
PHYS 1112 Exam 3, Version 1
Fall 2004
69
TIME
OF
COMPLETION_______________
NAME__SOLUTION___________________________
DEPARTMENT OF NATURAL SCIENCES
PHYS 1112, Exam 3
Version 1
Total Weight: 100 points
Section 1
April 27, 2011
1.
Check your examination for completeness prior to starting. There are a total of nine (9)
problems on six (6) pages.
2.
Authorized references include your calculator with calculator handbook, and the
Reference Data Pamphlet (provided by your instructor).
3.
You will have 50 minutes to complete the examination.
4.
The total weight of the examination is 100 points.
5.
There are six (6) multiple choice and three (3) calculation problems. Work all calculation
problems and 5 (five) multiple choice. Show all work; partial credit will be given for correct
work shown.
6.
If you have any questions during the examination, see your instructor who will
be located in the classroom.
7.
Start:
Stop:
2:00 p.m.
2:50 p.m.
PROBLEM
POINTS
1-6
25
7
25
8
25
9
25
TOTAL
100
PHYS 1112 Exam 3, Version 1
Fall 2004
70
CREDIT
PERCENTAGE
CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN
MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK
FOR PARTIAL CREDIT.
1. A proton in a magnetic field does not experience a force. Which of the following
statements is correct with respect to the situation?
I. The proton may have a velocity of zero m/s.
II. The proton may be moving parallel to the direction of the magnetic field.
(A) Statement I only.
(B) Statement II only.
(5)
(C) Neither statement I nor statement II.
(D) Both statement I and statement II.
2. A resistor, a capacitor, and an inductor are connected in series across an AC source. Which of
the following statements is false? (Select all that apply.)
(A)The instantaneous voltage across the capacitor lags the current by 90°.
(B)The instantaneous voltage across the inductor leads the current by 90°.
(5)
(C)The instantaneous voltage across the resistor is in phase with the current.
(D)The voltages across the resistor, capacitor, and inductor are not in phase.
(E)The rms voltage across the combination of the three elements equals the
algebraic sum of the rms voltages across each element separately.
PHYS 1112 Exam 3, Version 1
Fall 2004
71
3. A rectangular conducting loop is placed near a long wire carrying a current I as shown
in the figure below. If I decreases in time, what can be said of the current induced in the
loop?
(A)The direction of the current depends on the size of the loop.
(B)The current is clockwise
(5)
(C)The current is counterclockwise
(D)The current is zero
(E)Nothing can be said about the current in the loop without more information.
4. The bar shown in the figure below moves on rails to the right with a velocity , and a
uniform, constant magnetic field is directed out of the page. Which of the following
statements are correct?
(A)The induced current in the loop is zero.
(B)The induced current in the loop is clockwise.
(5)
(C)The induced current in the loop is counterclockwise.
(D) The induced current in the loop is to the left.
PHYS 1112 Exam 3, Version 1
Fall 2004
72
(E) No force is required to keep the bar moving at constant speed.
5. Three particles travel through a region of space where the magnetic field is out of the
page, as shown below. The electric charge of each of the three particles is, respectively,
(A) 1 is neutral, 2 is negative, and 3 is positive.
(B) 1 is neutral, 2 is positive, and 3 is negative.
(5)
(C) 1 is positive, 2 is neutral, and 3 is negative.
(D)1 is negative, 2 is neutral, and 3 is positive.
(E) It is impossible to find based on the given information.
6. At a particular instant, a proton moves eastward at speed V in a uniform magnetic field that is
directed straight downward. The magnetic force that acts on it is
(A) Zero.
(B) Directed upward.
(5)
(C) Directed to the north.
(D) Directed to the south.
(E) Directed downward.
PHYS 1112 Exam 3, Version 1
Fall 2004
73
7. A solenoid of radius 2.70 cm has 370 turns and a length of 21.0 cm.
(a) Find its inductance.
L
0 N 2 A
l

(4 10 7 Tm / A)(370) 2  (0.0270 m) 2
(0.210 m)
 1.88 10 3 H
(b) Find the rate at which current must change through it to produce an emf of 72.0 µV.
|  | L
I
t
I |  |

 38.3 10 3 A / s
t
L
8. A horizontal power line of length 62.7 m carries a current of 2.08 kA northward as shown
in the figure below. The Earth's magnetic field at this location has a magnitude of 5.40
10-5 T. The field at this location is directed toward the north at an angle 65° below the
power line.
(a) Find the magnitude of the magnetic force on the power line.
F  IlB sin( )  (2.08 103 A)(62.7 m)(5.40 10 5 T ) sin(65.0o )  6.38 N
PHYS 1112 Exam 3, Version 1
Fall 2004
74
(b) Find the direction of the magnetic force on the power line.
out of the page
into the page
counterclockwise
clockwise
9. An RLC circuit consists of a 130 resistor, a 21.0 µF capacitor, and a 400 mH inductor,
connected in series with a 120 V, 60.0 Hz power supply.
(a)What is the impedance of the circuit?
X L  2fL  2 (60 Hz)(0.400H )  151 
X C  1 /( 2fC )  1 /( 2 (60 Hz)(21.0 10 6 F ))  126 
Z  R 2  ( X L  X C ) 2  132 
(b) Find the rms current in the inductor.
I rms 
Vrms 120 V

 0.909 A
Z
132 
(c) Find the rms voltage across the capacitor.
VC , rms  X C I rms  (126 )(0.909 A)  115 V
(d) Find the phase angle.
tan( ) 
X L  XC
 0.192
R
  10.9o
(e) Does the current lead the voltage in this circuit, or lag behind?
Current lags behind since the phase angle is positive.
PHYS 1112 Exam 3, Version 1
Fall 2004
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Fall 2004
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Fall 2004
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PHYS 1112 Exam 3, Version 1
Fall 2004
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PHYS 1112 Exam 3, Version 1
Fall 2004
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PHYS 1112 Exam 3, Version 1
Fall 2004
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