REAL TEST 4  2B SP ‘12

1.  (22  POINTS)

A beam of vertically polarized light with intensity 41 W/m² is sent leftward through two polarizing sheets. The vertically polarized wave is shown below moving   toward Sheet 1.  The polarizing direction of  Sheet 1 makes an angle of 70⁰ with the vertical. The polarizing direction of  Sheet 2 makes an angle of  40⁰ with the horizontal.

(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?

(c)  ( 2 points) We assume the initial leftward moving electro-magnetic wave (before reaching sheet 1) is linearly polarized in the vertical direction with the electric field oscillating up and down parallel to the page. For the initial wave, suppose at a given location in space and moment in time, the electric field points vertically up.  What direction is the magnetic field vector at that same location and instant of time, IN or OUT  of the page. Explain your reasoning or any right hand rules  you use.  
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SOLUTIONS MANUAL:
(a) Use I₁ = I_o*cos²70.   =  4.8 W/m².  
(b) Use I₂ = I₁*cos²20 = 4.2 W/m²  since angle between the second axis and the first is 20 degrees. Check out it.  Work it  out using complimentary angles for example. The second angle (Sheet 2) makes an angle of 50 degrees (= 90 - 40) with the vertical and thus 70 - 50 = 20 for the second angle.   
(c) Vector B is into the the page. 

2.  (22 points) A converging lens has a focal length f = 25 cm. When an upright object is  placed a distance d_o in front (on the left) of the lens,  a real, inverted  image is produced on the other side (on the right)  with magnification m = -2.  FOR FULL CREDIT, SKETCH  THE RAY DIAGRAM FOR THE PROBLEM.

(a) (11 points) What is the object distance d_o ? 
(b) (11 points) What is the image distance d_i ? 
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SOLUTIONS MANUAL:
(a) Use  1/f = 1/d_i  + 1/d_o,  m = -2 = -di/do and substitution of the second equation into the first equations to get d_o = 37.5 cm
(b) After you get d_o , then get d_i using d_i = 2*d_o  =  75.0 cm  

3. (22 points)  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) below.  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.  Figure 1  below applies to  general situations and also to  part (a) with certain adjustments. Figure 1 also shows the unaided eye viewing the object  a distance 25 cm away. Note that N = near point distance = 25 cm.

(a) (8 points)  What is the angular magnification ϴ’/ϴ when this lens forms  a virtual image at - ?   What is d_o in this case?  

(b) (8 points) Suppose the object is an insect 2.00 mm in length. What is angle  θ when you view the insect with the unaided eye at the  near point?

(c) (6  points) What is   ϴ’ under the assumption of part (a)?
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SOLUTIONS MANUAL:
(a) M = (y/f)/(y/25) = 25/f = 2.08. 
(b) ϴ = y/25 = 0.008 radians, where y = 0.200 cm
(c) ϴ’ = y/f = 0.0167 radians, where y = 0.200 cm

4. (22  points) Coherent light of wavelength 525 nm passes through two thin slits that are a distance d = 0.0415 mm apart.  The light  then falls on a screen a distance R = 75.0 cm away. How far away from the central bright fringe on the screen is:

(a)  (11  points)   the 3^rd  bright fringe (not counting the central bright fringe.)

(b)  (11 points)  the fifth dark fringe? HINT: The numbering for 2-slit destructive interference is a little different. The solution to #5, TEST 3,  should be corrected since the 8th bright fringe corresponds to m’ = 7, not 8 . (i.e., m ‘ = 0 gives the first dark fringe beyond the central maximum for positive angles above  x-axis.)

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SOLUTIONS MANUAL:

 (a) d*sin theta  = d*tan theta = d*(y₃/R)
d*(y₃/R) = 3*lambda; solve for y₃ = 2.85 cm 

(b) d*sin theta  = d*tan theta = d*(y₅/R)
d*(y₅/R) = (4 + 1/2)*lambda; solve for y₅ = 4.27 cm.  Note we use the index "4"  for the fifth fringe as noted above. 

5.  (22  points)  Light of wavelength 656.3 nm passes through a single slit with width a = 0.450 mm. Assume the screen is R = 1.75 m from   slit.
(a)   (10  points)    What is the distance from the central maximum to the first minimum (zero) in intensity on screen?
(b)   (10  points)    What is the distance between the first minimum and the 3^rd minimum  in intensity on screen?
(c)   ( 2 points) From your answer to part (a),  what is the width of the central maximum? 
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SOLUTIONS MANUAL:

 (a) a*sin theta  = d*tan theta = d*(y₁/R)
d*(y₁/R) = 1*lambda; solve for y₁ = 2.55x10^-3 m . Note: R = 1.75 m

(b) d*sin theta  = d*tan theta = d*(y_n/R) =n*lambda  for any n !
d*(y_n+2/R) - d*(y_n/R) = (n+ 2)*lambda  - n*lambda
  d*(y_n+2/R) - d*(y_n/R) = 2*lambda =  5.11x10^-3 m  for any n.
Thus, the answer to (b) is  twice the answer to (a).  NOTE: n = 1 and n + 2 = 3.
(c) From symmetry about the axis,  the width of the central maximum is twice the answer to (a).  

6. EXTRA CREDIT. (12 points)  A muon is created 52.00 km above the Earth’s surface ( as measured in the Earth’s reference frame). The average lifetime of a muon, measured in its own rest frame, is 2.20  x10 ^{- 6} seconds.

In the frame of the muon, the Earth is moving toward the muon with speed 0.9860c. Thus, in the frame of the Earth, the muon is moving toward the Earth with speed 0.9860c.  SHOW ALL WORK.

(a) (2 points)  In the Earth’s frame, what is the lifetime of the muon?
(b) (2 points) In the Earth’s frame, how far  does the muon travel during its lifetime?

(c) (2 points) What is the  ratio of your  answer to part (b) to  the muon’s original height in the Earth’s frame?

(d) (2 points) In the muon’s frame, how much closer does the Earth get during the muon’s lifetime?  

(e) (2 points) In the muon’s frame, what is its  initial height above the Earth’s surface?

(f) (2 points) What is the ratio of your  answer to part (d) to  the muon’s initial  height in the muon’s  frame given in part ( e) ?

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 SOLUTIONS MANUAL:
(a) earth time = (2.20  x10 ^{- 6}) / (1 - v²/c²)^1/2 =  1.31x10^-5 s.  Longer than  muon frame  measurement. 
(b) earth measurement = v*(earth time) = 3903 m, where earth time is from part (a).
(c) Find  ratio  v*(earth time)/52 km  0.075 when we  convert to m.  
(d) distance closer (Earth frame) =  v*( 2.20  x10 ^{- 6} seconds)=  651 m.
(e) muon perceived height = 52 km*(1 - v²/c²)^1/2 = 8.67 km. Shorter than   Earth  frame  measurement.
(f) Find  ratio  v*( 2.20  x10 ^{- 6} seconds) / [52 km*(1 - v²/c²)^1/2 ] = 0.075.
NOTE: [v*( 2.20  x10 ^{- 6} seconds)/(1 - v²/c²)^1/2] /  52 km  = v*(earth time)/52 km = same answer as (c).