TIME OF COMPLETION_______________            NAME_____SOLUTION________________________

 

 

                                           DEPARTMENT OF NATURAL SCIENCES

 

PHYS 1112, Exam 3                                                                                                             Section 1

Version 1                                                                                                                 December 2, 2002

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.

 

1.                  Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet  (provided by your instructor).

 

1.                  You will have 80 minutes to complete the examination.

 

1.                  The total weight of the examination is 100 points.

 

1.                  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.

 

1.                  If you have any questions during the examination, see your instructor who will

be located in the classroom.

 

7.         Start:                11:30 a.m.

Stop:                12:50 p.m

 

 

 

               PROBLEM

 

                 POINTS

 

                 CREDIT

 

                     1-6

 

                      30

 

 

 

                       7

 

                      15

 

 

 

8

 

                      20

 

 

 

                       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.                  *         Two parallel slits are illuminated with monochromatic light, and an interference pattern is observed on a screen. If the distance between the slits were increased, would the distance between the bright fringes:

 

    1. Increase.

 

    1. Decrease.    y = l L m / d, d goes up, y goes down.

(5)

    1. Remain the same.

 

    1. The effect will depend on wavelength of the light.

 

 

 

  1. A girl is standing in front of a concave mirror. Consider two rays of light, one from her nose and one from her mouth, that are parallel as they are traveling toward the mirror. These rays will come together:

 

    1. At the focal point.

 

    1. At the center of curvature.

(5)

    1. At the image point.

 

    1. Behind the mirror if she is too close to the mirror.

 

 

  1. Dispersion occurs when:

 

    1. Some materials bend light more than other materials.

 

    1.  A material slows down some wavelengths more than others.

(5)

    1. A material changes some frequencies more than others.

 

    1. Light has different speeds in different materials.

 

 

 

 

  1. When the reflection of the object is seen in plane mirror, the image is:

 

    1. Real and upright.

 

    1. Real and inverted.

(5)

    1. Virtual and upright.

 

    1. Virtual and inverted.

 

 

  1. A light beam is incident upon a still water surface. What is the maximum possible value for the angle of refraction? (n water = 1.333).

 

    1. 76.2o.

(5)

    1. 67.5o.

 

    1. 54.4o.

 

d.      48.6o.  sin(q1) = 1.333 sin(q2)    sin(q2) = sin(q1)/1.333,  the maximum value corresponds to sin(q1) = 1.00,  q2  = arcsin(1.00/1.333)

 

 

 

6. Which of the following best describes the image for a thin concave lens that forms whenever the magnitude of the object distance is less than that of the lens’ focal length?

 

a.       Inverted, enlarged, and virtual.

 

b.      Upright, enlarged, and virtual.

(5)

c.       Upright, diminished, and virtual.  Just draw a ray diagram!!!

 

d.      Inverted, diminished, and real.

 

 

 

 

 

 

 

 

 

7. Two glass prisms are placed together as shown below. If a beam of light strikes the face of one of the prisms at normal incidence, at what angle does it emerge from the other prism?

 

 

 

For the boundary  between prism 1 and prism 2 : 

(15)

                                             1.60sin(q1) = 1.40 sin(q2), where  q1 = 45.0o.      

 

                                             (1.60/1.40) sin(q1) = sin(q2),   q2 = 53.9o.      

 

 

                                                                 For the boundary between prism 2 and air: 

 

                                              1.40sin(q1) = 1.00 sin(q2), where  q1 = 8.91o.      

 

                                             (1.40/1.00) sin(q1) = sin(q2),   q2 = 12.5o.      

 

 

8. An object is 15.0 cm from a converging lens whose focal length is 10.0 cm. On the opposite side of this lens, at a distance of 60.0 cm, is a converging lens with a focal length of 20.0 cm.

 

    1. Where is the final image formed?

 

1/p1 + 1/q1 = 1/f1

 

1/q1 = - 1/p1 + 1/f1

 

1/q1 = - 1/(15.0 cm) + 1/(10.0 cm) = 0.0333 cm-1

 

q1 = 30.0 cm                   p2 = 60.0 cm – 30.0 cm = 30.0 cm

 

1/p2 + 1/q2 = 1/f2

 

1/q2 = - 1/p2 + 1/f2

 

1/q2 = - 1/(30.0 cm) + 1/(20.0 cm) = 0.0167 cm-1

 

q2 = 59.9 cm     

 

    1. If the object has a height of 4.00 cm, how tall is the image?

 

M = M1 M2

 

M1 = - q1/p1 = - (30.0 cm)/(15.0 cm) = -2.00

 

M2 = - q2/p2 = - (60.0 cm)/(30.0 cm) = -2.00

 

M = (-2.00)(-2.00) = 4.00

 

M = h’/h               h’ = h M = (4.00 cm)(4.00) = 16.0 cm

 

 

    1. Is the image upright or inverted?

 

Upright (M is positive).

 

  1. In a double-slit experiment using monochromatic light, a screen is placed 1.25 m away from the slits, which have a separation distance of 0.0250 mm. The third-order bright fringe is 6.60 cm from the center of the central maximum.

 

    1.  Find the wavelength of the light.

 

 

        ymax = l L m /d

      

        ymax d/ L m = l

      

        l = (6.60 x 10-2 m) (2.50 x 10-5 m) / (1.25 m)(3) = 4.40 x 10-7 m = 440 nm

      

 

    1. What is the distance between the second-order bright maximum and the center of the central maximum?

 

 

         y2 = l L 2 /d = (4.40 x 10-7 m)(1.25) (2)/ (2.50 x 10-5 m) = 4.40 cm

 

 

  1. A hummingbird is hovering 50.0 cm in front of a spherical garden mirror of diameter 20.0 cm. Hmmm.

 

    1. Determine the location of the image of our feathered friend.

 

 

f = - 5.00 cm   p = 50.0 cm

 

1/p + 1/q = 1/f

 

1/q = - 1/p + 1/f = -1/(50.0 cm) + 1/(-5.00 cm) = -0.220 cm-1

 

q = -4.55 cm

 

    1. Determine the magnification.

 

M = -q/p = -(-4.55)/(50.0 cm) = 0.091.

 

    1. Draw a ray diagram to verify your answers.

 

 

 

 

 

    1. Is the image real or virtual, upright or inverted, reduced or magnified?

 

Virtual (q is negative), upright (M is positive), and reduced (M is less than one).