TIME OF COMPLETION_______________            NAME____SOLUTION_________________________

 

 

                                           DEPARTMENT OF NATURAL SCIENCES

 

PHYS 1112, Exam 3                                                                                                             Section 1

Version 1                                                                                                                 December 1, 2003

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 (5) multiple choice problems 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:                1:30 p.m.

Stop:                2:45 p.m

 

 

               PROBLEM

 

                 POINTS

 

                 CREDIT

 

                     1-6

 

                      30

 

 

 

                       7

 

                      20

 

 

 

8

 

                      20

 

 

 

                       9

 

                      15

 

 

 

                      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. A lens made of glass (n = 1.50) has a convex first surface with a radius of curvature of 2.00 m and a second concave surface with a radius of curvature of 1.00 m. Its focal length is
    1. 3.00 m.

                                                                                                1/f = (n-1)(1/R1 – 1/R2)

    1. 0.500 m.                        

(6)                                                                                                  1/f = (1.50 - 1)(1/(2.00 m)- 1/(1.00 m)) = -1/4

    1. – 0.250 m.

 

    1. – 4.00 m.

 

 

 

 

  1. Suppose that you make a calculation of the magnification of the image of a real object located somewhere in front of a convex spherical mirror and it turns out to be –1.50. You conclude that

 

    1. The calculation was wrong.

                                                                                   q is always negative for a convex mirror, which means that M can be only positive for a real object             

    1. The image is smaller than the object.

(6)

    1. The object was upside down.

 

    1. The image is larger than the object.

 

 

  1. A chicken is standing 1.00 m in front of a vertical plane mirror. A woman is standing 5.00 m from the mirror, behind and in line with the bird. How far from her will she see the image of the chicken?

 

    1. 5.00 m.

 

    1. 1.00 m.

(6)

    1. 6.00 m.

 

    1. 4.00 m.

 

  1. When light is incident on an interface, the angle(s) of

 

    1. The reflected and the refracted beams depend on the wavelength.

 

    1. The reflected and the refracted beams are independent on frequency.

(6)

    1. The refracted beam is independent on frequency.

 

    1. The reflected beam is independent on frequency.

 

 

 

  1. Nowadays it is possible to directly measure, using electronic techniques, the frequencies of electromagnetic oscillations ranging up to 500 MHz. That corresponds to a wavelength of

 

    1. 0.600 m.

 

    1. 0.600 cm.

(6)

    1. 6.00 m.

 

    1. 6.00 cm.

 

 

 

  1. In comparison to ultraviolet, visible light has

 

    1. Wavelengths that are shorter.

 

    1. Frequencies that are lower.

(6)

    1. Wavelengths that are equal.

 

    1. Frequencies that are equal.

 

 

 

 

  1. An arrow 2.00 cm long is located 75.0 cm from a lens, which has a focal length of +30.0 cm.

 

  1. If the arrow is perpendicular to the principal axis of the lens, what is its lateral magnification, defined as h/h0?

 

f = + 30.0 cm

 

p = 75.0 cm

 

1/p + 1/q = 1/f

 

1/q = 1/f – 1/p

 

1/q = 1/(30.0 cm) – 1/(75.0 cm)

 

q = 50.0 cm

 

M = -q/p = -0.667

 

  1. Suppose, instead, that the arrow lies along the principal axis, extending from 74.0 cm to 76.0 cm from the lens. What is the longitudinal magnification of the arrow, defined as L/L0?

 

p1 = 74.0 cm

 

1/p1 + 1/q1 = 1/f

 

1/q1 = 1/f – 1/p1

 

1/q1 = 1/(30.0 cm) – 1/(74.0 cm)

 

q1 = 50.5 cm

 

 

p2 = 76.0 cm

 

1/p2 + 1/q2 = 1/f

 

1/q2 = 1/f – 1/p2

 

1/q2 = 1/(30.0 cm) – 1/(76.0 cm)

 

q1 = 49.6 cm

 

L = 50.5 cm – 49.6 cm  = 0.889 cm

 

L0 = 2.00 cm

 

L/L0 = (0.889 cm)/(2.00 cm) = 0.445

 

 

  1. Draw a ray diagram to represent the situation in which the arrow is perpendicular to the principal axis.

 

 

  1. A person stands 20.0 m from a thin positive lens having a focal length of 0.500 m. Behind this lens a distance of 2.00 m, and centered on the same axis, is another positive lens having a focal length of 1.00 m.

 

    1. Where will the image of the person appear?

 

p1 = 20.0 m

 

1/p1 + 1/q1 = 1/f1

 

1/q1 = 1/f1 – 1/p1

 

1/q1 = 1/(0.500 m) – 1/(20.0 m)

 

q1 = 0.513 m

 

p2 = 1.49 m

 

1/p2 + 1/q2 = 1/f2

 

1/q2 = 1/f2 – 1/p2

 

1/q2 = 1/(1.00 m) – 1/(1.49 m)

 

q2 = 3.05 m

 

    1. What is the lateral magnification of this system of lenses?

 

M1  = -q1/p1 = - (0.513 m)/(20.0 m) = -0.026

 

M2  = -q2/p2 = - (3.05 m)/(1.49 m) = - 2.05

 

M = M1  M2 = 0.053

 

 

    1. Describe the image. (Real or virtual, enlarged or reduced in size, upright or inverted?)

 

Real (q2 >0)

 

Reduced (M < 1)

 

Upright (M >0)

 

  1. A display lamp having a bright 5.00-cm long vertical filament is positioned 30.0 cm from a concave mirror that projects the bulb’s image onto a wall 9.00 m from the vertex.

 

    1. What is the radius of curvature of a mirror?

 

q = 900 cm

 

p = 30.0 cm

 

1/p + 1/q = 1/f

 

1/f = 1/q + 1/p

 

1/f = 1/(30.0 cm) + 1/(900 cm)

 

q = 29.0 cm

 

R = 2 f = 58.0 cm

 

 

    1. How big is the image?

 

 

M = -q/p =  - (900 cm)/(30.0 cm) = -30.0

 

M = h’/h

 

h’ = M h

 

h’ = (-30.0)(5.00 cm) = -150 cm

 

 

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

 

 

Real (q > 0), Inverted (M < 1)

 

 

  1. Pictured below is a semicircular piece of clear test material surrounded by air.

 

 

    1. Determine the index of refraction of the sample.

 

 

n1 sin (q1) = n2 sin (q2)

 

1.00 sin (30.0o) = n2 sin (19.0o)

 

n2 = sin (30.0o) / sin (19.0o) = 1.54

 

    1. What is the speed of light in the material?

 

N = c/Vn

 

Vn = c/n = (3.00 x 108 m/s)/ 1.54 = 1.98 x 108 m/s

 

 

    1. Explain why the emerging ray is unbent.

 

Ray is traveling in radial direction, which means normal incidence => the angle is not refracted.