TIME OF COMPLETION_______________            NAME___SOLUTION__________________________

 

 

                                           DEPARTMENT OF NATURAL SCIENCES

 

PHYS 1112, Exam 3                                                                                                             Section 1

Version 1                                                                                                                       April 22, 2004

Total Weight: 100 points

 

1.         Check your examination for completeness prior to starting.  There are a total of ten (10) problems on nine (9) 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:                1:30 p.m.

Stop:                2:45 p.m

 

 

 

               PROBLEM

 

                 POINTS

 

                 CREDIT

 

                     1-6

 

                      30

 

 

 

                       7

 

                      20

 

 

 

8

 

                      20

 

 

 

                       9

 

                      20

 

 

 

                      10

 

                      10

 

 

 

                 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. A light ray in air enters and passes through a block of glass. What can be stated with regard to its speed after it emerges from the block?

 

    1. Speed is less than when in glass.

 

    1. Speed is less than before it entered glass.

(5)

    1. Speed is the same as that in glass.

 

    1. Speed is the same as that before it entered the glass.

 

 

 

  1. Which of the following describes what will happen to a light ray incident on a glass-to-air boundary at greater than the critical angle?

 

    1. Total internal reflection.

 

    1. Total transmission.

(5)

    1. Partial reflection, partial transmission.

 

    1. Partial reflection, total transmission.

 

 

 

  1. Find the frequency of X-rays of wavelength 10-10 m.

 

    1. 3.00 x 1018 Hz.                           l f = c            f = c / l                         

(5)

    1. 3.00 x 1010 MHz.

 

    1. 6.00 x 109 Hz.

 

    1. 3.00 x 108 Hz.

 

 

 

 

 

  1. When the reflection of an object is seen in a plane mirror, the distance from the mirror to the image depends on:

 

    1. The wavelength of light used for viewing.

 

    1. The distance from the object to the mirror.  q = p

(5)

    1. The distance from the observer to the mirror.

 

    1. The size of the object.

 

 

 

  1. Which of the following best describes the image formed by a thin concave lens whenever the magnitude of the object distance is less than the lens’ focal length?

 

    1. Inverted, enlarged, and real.

 

    1. Upright, enlarged, and virtual.

(5)

    1. Upright, diminished, and virtual.

 

    1. Inverted, diminished, and real.

 

 

 

 

  1. Ansel places an object 30.0 cm from a thin converging lens. If a real image is formed 10.0 cm from the lens, what is the lens’ focal length?

 

    1. 30.0 cm.

 

    1. 15.0 cm.

(5)

    1. 10.0 cm.

 

    1. 7.50 cm.        1/f = 1/p + 1/q

 

 

 

 

 

 

  1. Two positive thin lenses with focal lengths of 25.0 cm and 40.0 cm are separated by 130 cm. A 5.00-cm tall chess piece is placed 50.0 cm in front of the first lens.

 

a.       Where will its image appear?

 

p1 = 50.0 cm

f1 = 25.0 cm

 

1/p1 + 1/q1 = 1/f1

1/q1 = 1/f1 – 1/p1

1/q1= 1/(25.0 cm) – 1/(50.0 cm)

q1 = 50.0 cm

 

p2 =  130 cm - q1 = 80.0 cm

f2 = 40.0 cm

 

1/p2 + 1/q2 = 1/f2

1/q2 = 1/f2 – 1/p2

1/q2= 1/(40.0 cm) – 1/(80.0 cm)

 

q2 = 80.0 cm (80.0 cm behind the second lens)

 

b.      How tall is the image?

 

M1 = -q1/p1

M1 = - (50.0 cm)/(50.0 cm) = -1.00

 

M2 = -q2/p2

M2 = - (80.0 cm)/(80.0 cm) = -1.00

 

M = M1 M2 = 1.00

 

h’ = h M = (5.00 cm) (1.00) = 5.00 cm

 

c.       Is the image upright or inverted, real or virtual?

 

Real (q > 0), upright (M >0)

 

8. A 15.0-cm tall Teddy bear stands 60.0 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?

 

 

f =  R/2 = 30.0 cm

p = 60.0 cm

 

1/p + 1/q = 1/f

1/q = 1/f – 1/p

1/q = 1/(30.0 cm) – 1/(60.0 cm)

 

q =  60.0 cm

 

b.      What is the height of Teddy’s image?

 

 

M = -q/p

M = - (60.0 cm)/(60.0 cm) = -1.00

 

h’ = h M = (15.0 cm) (-1.00) = -15.0 cm

 

 

c.       Draw a ray diagram to check your answers.

 

 

d.      Is the image real or virtual, upright or inverted, enlarged or reduced?

 

Real (q > 0), inverted (M <0), and same size (|M| = 1).

 

 

 

9.  A convex (positive) lens having a 60.0-cm focal length is placed 100 cm from the frog.

 

a.       Where the image of the frog is located?

 

 

f = 60.0 cm

p = 100 cm

 

1/p + 1/q = 1/f

1/q = 1/f – 1/p

1/q = 1/(60.0 cm) – 1/(100 cm)

 

q =  150 cm

 

b.      Find the lateral magnification.

M = -q/p

 

M = - (150 cm) / (100 cm) = -1.50

 

 

c.       Describe the image.

 

Real (q > 0), inverted (M < 0), and enlarged (|M| > 1).

 

 

d.      Draw a ray diagram .

 

 

10. 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.

 

    1. What is the index of refraction of the plastic?

 

 

n1 sin (q1) = n2 sin (q2)

 

1.00 sin (62.7o) = n2 sin (48.1o)

 

n2  = sin (62.7o) / sin (48.1o) = 1.19

 

    1. Find the speed of light in the plastic.

 

n2  = c / v

 

v = c/n2

 

v = (3.00 x 108 m/s) / 1.19 = 2.51 x 108 m/s