TIME OF COMPLETION_______________            NAME______SOLUTION_______________________

 

 

                                           DEPARTMENT OF NATURAL SCIENCES

 

PHYS 1112, Exam 3                                                                                                      Section 1

Version 1                                                                                                         December 1, 2006

Total Weight: 100 points

 

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:               10:30 a.m.

Stop:                11:20 a.m.

 

 

 

              PROBLEM

 

                 POINTS

 

                CREDIT

 

                     1-6

 

                     25

 

 

 

                      7

 

                     25

 

 

 

8

 

                     25

 

 

 

                      9

 

                     25

 

 

 

                     10

 

                     25

 

 

 

                 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.   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  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.

 

 

  1. What is the index of refraction of the transparent substance?

 

 

q1 = 58.0o

 

q2 = 90.0o - 58.0o = 32.0o

 

n1 sin q1 = n2 sin q2

                                                

n2 = n1 sin q1/ sin q2

 

n2 = (1.00) sin (58.0o)/ sin (32.0o) = 1.60

 

 

    b. What is the critical angle for total internal reflection in this substance?

 

sin qc = n2 / n1

                                                

sin qc = (1.00)/(1.60)

qc = 38.7o