Final 2B SP ‘12

1.  (30  POINTS)

A beam of randomly polarized light with intensity 41 W/m² is sent rightward through THREE polarizing sheets. The  randomly  polarized wave is shown below moving rightward  toward Sheet 1.  The polarizing direction of  Sheet 1 is  vertical. The polarizing direction of  Sheet 2 makes a  45⁰ - angle with the vertical . The polarizing direction of  Sheet 3 makes a  65⁰ – angle with the vertical .

                                
(a)  (10 points) What is the Intensity of the light after it passes through Sheet 1 before reaching  Sheet 2?
(b)  (10 points)  What is the Intensity of the light after it passes through Sheet 2 before reaching  Sheet 3?

(c)  (10 points)  What is the Intensity of the light after it passes through  Sheet 3?

SOLUTIONS answers in W/m*m.

(a)  I1 = Io/2 = 20.5

(b)  I2 = I1*cos²45 = 10.25

(c)  I3 = I2*cos²20 = 9.05

2. (30  POINTS)  A concave mirror has a radius of curvature R = 10.0 cm. When an upright object is  placed a distance d_o in front (on the left) of the mirror,  a real, inverted  image is produced on the same side (on the left)  with magnification m = -2. 

(a) (3 points)    What is the mirror’s focal length f ?  
(b) (12 points)  What is the object distance d_o ? 
(c) (11 points)  What is the image distance d_i ?
(d) (4 points) Use ray tracing to sketch the object and image of this problem. Sketch on the diagram below. Using ray tracing, sketch rays  from the tip of  object,  to the mirror and finally to the tip of inverted image  left of mirror. Draw two or three “strategic” rays from  object arrow tip that reflect off mirror and finally converge at arrow tip of real image.

SOLUTIONS answers in cm

(a)   f = R/2 = 5.00

(b)  1/f = 1/do   + 1/di  and di = 2*do leads to do = 7.5

(c)   di = 15

3. (36 points)  MAGNIFYING GLASS: A magnifying glass is formed from a single double convex (converging) lens with a focal length of f = +12.0 cm .  The lens, with focal points labeled by  F,  is shown in the schematic  (figure 1) next page.  It could be the model for a Bart passenger’s  reading glasses while she scans the afternoon’s news headlines.  The distance | d_i |  is the distance between one of the  lens’s and the virtual  IMAGE of the  news story  of the Sharks victory she wants to read about.  Note: do  f in this problem. Figure 1 applies to  more general situations like part (a) and also to a special condition like  part (b), (c) and (d) with certain adjustments of diagram. Figure 1 also shows the unaided eye viewing the object  a distance 25 cm away. Note that N = near point distance = 25 cm.
            
(a) (14 points) Suppose d_o = 10.0 cm.  Compute d_i  and find the linear magnification m = -d_i/d_o .

    
  Now change the problem for a special situation in parts (b), (c ) and (d). Assume d_o = f for these next parts:

(b) (8 points)  In this case, we know the lens forms  a virtual image at d_i =  -.  What is the angular magnification M = ϴ’/ϴ?  

   
(c) (8 points) Suppose the object is an insect 2.50 mm in length.
What is angle  ϴ  in the formula for the angular magnification M? 
 
(d) (6  points) What is   ϴ’ in the formula for the angular magnification M? 

SOLUTIONS answers in cm or no units

(a)  1/f = 1/do   + 1/di  and di = -60

(b)  M = 25/f = 2.08
(c)  0.25/25 = 0.01

(d)  M = (2.08)*(0.01)  = 0.0208

4. (36  points) TWO SLIT INTERFERENCE: A coherent electromagnetic wave  of wavelength 520 nm passes through two thin slits that are a distance d = 0.0400 mm apart.  The light  then falls on a screen a distance R = 70.0 cm away.

(a)  (11  points)   How far away from the center of the central bright fringe on the screen is the 3^rd  bright fringe (not counting the central bright fringe) 

(b)  (11 points)   How far away from the center of the central bright fringe on the screen is the 3^rd    dark fringe? Also,  please compute the difference between this answer and the answer you got in part (a).

(c)  (5 points)   How far away from the center of the central bright fringe on the screen is the 1^st dark fringe?

(d) (5 points)   From your answer to part (c), what is the width of the central  maximum? 

(e)  (4 points) In this problem, in what range is the electromagnetic
wave  ? Circle one:

(i) x-ray range   (ii) visible range  (iii) ultra–violet range (iv) in the range of radio waves

SOLUTIONS answers in cm

(a)   y3 = 3*lambda*R/d = 2.73 cm

(b)   y3 = 2.5*lambda*R/d = 2.28 cm; difference = 0.455 cm  
(c)   y1 = 0.5*lambda*R/d = 0.455 cm  

(d)   2*0.455 cm = 0.91 cm

(e) ii visible

5.    (18  points)  THIN FILM INTERFERENCE:

(a) (12 points) What is the smallest,  non-zero thickness T  of a soap bubble film that appears black (indicating destructive interference) when viewed by reflected light with a wavelength λ of 470 nm  in air ? The index of refraction of the film is 1.36.  There is air on both sides of the film. Assume the index of refraction of air is 1. See diagram below of a section of the film being irradiated by the light. 

(b) (6 points)  For the same bubble as part (a), what is the smallest,  non-zero thickness T  of a soap bubble film that appears BRIGHT (indicating constructive interference)

SOLUTIONS answers in nm and cm

(a)  T = lambda/(2*n) = 172.8 nm

(b)  T = lambda/(4*n) = 86.4 cm

6.  (26  points)  SINGLE  SLIT DIFFRACTION: A coherent electromagnetic wave  of wavelength 656.3 nm passes through a single slit with width a = 0.460 mm. Assume the screen is 1.82 m from the  slit.
(a)   (8  points)    What is the distance from the center of the central maximum to the first minimum (zero) in intensity on screen?
(b)   (8  points)    What is the distance between the first minimum and the 5^th  minimum  in intensity on screen?
(c)   (6 points) From your answer to part (a),  what is the width of the central maximum? 
(d)  (4 points) In this problem, in what range is the electromagnetic wave  ? Circle one:

(i) x-ray range   (ii) visible range  (iii) ultra–violet range (iv) in the range of radio waves

SOLUTIONS answers mm

(a) y1 = 1*lambda*R/a = 2.596 mm

(b) y5 -  y1 = 4*lambda*R/a  = 10.3867 mm
(c) (2)* lambda*R/a = 5.193 mm  

(d) ii visible

7. (30 points)  Cosmic ray particles are created 57.00 km above the Earth’s surface (as measured in the Earth’s reference frame). The average lifetime of a particle, measured in its own rest frame, is 2.00  x10 ^{- 6} seconds.

In the frame of the particle,  the Earth is moving toward the particle  with speed 0.9820c. Thus, in the frame of the Earth, the particle is moving toward the Earth with speed 0.9820c.  SHOW ALL WORK.
(a) (5 points)  In the Earth’s frame, what is the lifetime of the particle ?

(b) (5 points)  In the Earth’s frame, how far  does the particle travel during its lifetime?

(c) (5 points)  What is the  ratio of your  answer to part (b) to  the particle’s original height in the Earth’s frame?

(d) (5 points)  In the particle’s frame, how much closer does the Earth get during the particle’s lifetime?  

(e) (5 points)  In the particle’s frame, what is its  initial height above the Earth’s surface?

(f) (5 points)  What is the ratio of your  answer to part (d) to  the particle’s initial  height in the particle’s  frame given in part ( e) ?

SOLUTIONS answers use sqrt(1 – v²/c²) =  sqrt(1 – 0.9820²) =alpha =
0.188881  
(a) 2.00x10 ^-6/alpha = 1.06x10^-5 s
(b)  d = 0.9820*c*1.06x10^-5 s  = 3119.2 m
(c)   d/57.00 km = 0.0547.

(d)  d’ = = 0.9820*c*2.00x10^-6 s = 589.2 m 

(e)   57*alpha = 10.7665 km

(f)    d’/10.7665 km = 0.0547

8. (10 POINTS) Below is a energy level scheme of a hypothetical one-electron element Mathematicum. The potential energy is taken to be zero for an electron at an infinite distance from the nucleus.

(a) (2 points) How much energy does it take to ionize  an electron from the  ground  state level (n = 1) ?

(b) (6 points) What will be the final electron state if a photon of energy of 10 eV strikes a Mathematicum atom initially  in its ground  state level (n_i = 1)? In other words, find the  value n_f  of the final state if the atom absorbs a 10-eV photon.
(c) (2 points) What will happen if a photon of energy of 7 eV strikes a Mathematicum atom in its ground  state level? Explain why.

(d) EXTRA CREDIT: (12 points) Photons emitted in the Mathemticum transitions  n = 3 to n = 1 will eject photoelectrons from a certain metal.  The work function of the metal  is 4.00 eV . The work function is the minimum photon energy needed in order for an electron to escape the metal. Use conservation  of energy to answer the following question: What is the  kinetic energy of the ejected electrons?  

SOLUTIONS answers in eV.

(a) 0 –(-20 eV) = 20 eV.

(b)  n = 2

(c)   photon not absorbed

(d)   KE = 15 eV – 4 eV = 11 eV.