TIME OF COMPLETION______       NAME__SOLUTION___________________________

 

 

                                           DEPARTMENT OF NATURAL SCIENCES

 

PHYS 1112, Exam 3                                                                                                      Section 1

Version 1                                                                                                         December 6, 2004

Total Weight: 100 points

 

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:               6:00 p.m.

Stop:                7:15 p.m

 

 

 

              PROBLEM

 

                 POINTS

 

                CREDIT

 

                     1-6

 

                     30

 

 

 

                      7

 

                     20

 

 

 

8

 

                     15

 

 

 

                      9

 

                     20

 

 

 

                     10

 

                     15

 

 

 

                 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 of the following statements is correct for a series RLC circuit?  

 

    1. The voltage across the capacitor leads the voltage across the inductor by 90o.

 

    1. The voltage across the inductor leads the voltage across the capacitor by 90o.

(6)

    1. The voltage across the inductor leads the voltage across the resistor by 180o.

 

    1. The voltage across the inductor leads the voltage across the capacitor by 180o.

 

 

 

 

  1. 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?

 

    1. 1.30.

 

    1. 2.40.

(6)

    1. 2.60.

 

    1. 0.400.

 

 

 

 

 

 

  1. 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?

 

    1. 78o.

 

    1. 88o.

(6)

    1. 68o.

 

    1. 49o.

 

 

 

  1. A laser beam (l = 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?

 

    1. 3.47 x 10-3.

 

    1. 3.47 x 10-2.

(6)

    1. 3.47 x 10-4.

 

    1. 3.47 x 10-6.

 

 

 

 

 

 

  1. A monochromatic (single frequency, single wavelength) light ray in air (n = 1.00)

enters a glass prism (n = 1.50). In the glass prism

 

    1. Both the frequency and the wavelength are the same as in air.

 

    1. The frequency is the same, but the wavelength is greater than in air.

(6)

    1. The frequency is the same, but the wavelength is smaller than in air.

 

    1. The wavelength is the same, but the frequency is greater than in air.

 

 

 

 

 

  1. Which ray diagram is correct?

 

    1. A.

 

    1. B.

(6)

    1.  C.

 

    1. D.  

 

 

 

 

 

  1. A series ac circuit contains a 50.0-W resistor, a 15.0-mF capacitor, a 0.200-H inductor, and a 60.0-Hz generator that has an rms output of 90.0 V.  

 

    1. Find the impedance of the circuit.

 

XL = 2 p f L = 2 p (60.0 Hz) (0.200 H) = 75.4 W

 

XC = 1/2 p f C = 1/(2 p (60.0 Hz) (15.0 x 10-6 F)) = 177 W

 

Z = (R2 + (XL – XC)2)1/2 = ((50.0 W)2 + (75.4 W  – 177 W)2)1/2   = 113 W

 

    1. Find the maximum current that flows in the resistor.

 

DVmax = 21/2 DVrms

 

DVmax = 21/2 (90.0 V) = 127 V 

 

Imax = Vmax /Z

 

Imax = (127 V)/(113 W) = 1.13 A

 

 

    1. Find the maximum potential difference across the capacitor.

 

DVmax = Imax XC = (1.13 A)(177 W) = 200 V

 

    1. Does the voltage lead the current in the circuit or the current leads the voltage? EXPLAIN.

 

f = tan-1 ((XL – XC)/R)

 

f = tan-1 ((75.4 W – 177 W)/ (50.0 W)) = -63.8o

 

f < 0, hence voltage lags behind current. 

 

 

  1. 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)

 

 

 

  1. A small kitten finds itself 150 cm in front of a diverging lens.

 

    1. 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

 

    1. Determine the magnification.

 

M = -q/p = - (-50.0 cm)/(150 cm) = 1/3

 

    1. Is the image real or virtual?

 

Virtual (q < 0)

 

    1. Is the image upright or inverted?

 

Upright (M > 0)

 

    1. Draw a neat ray diagram of the situation.  

 

 

  1. 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(q1) = n2 sin(q2) 

 

sin(q2) = n1 sin(q1)/ n2  

 

sin(q2) = (1.00) sin(35.0o)/(1.571) = 0.365 

 

q1 = 21.41o

                                                                                                       

Violet:

 

n1 sin(q1) = n2 sin(q2) 

 

sin(q2) = n1 sin(q1)/ n2  

 

sin(q2) = (1.00) sin(35.0o)/(1.594) = 0.360 

 

q1 = 21.09o

 

d = 21.41o - 21.09o = 0.32o