1. (53 points) The  upright object shown below is to the left of a  converging lens  with a focal length of 20 cm ( not shown). The object produces a real image  which is one half the size of the object.     
 

SOLUTIONS

Obviously the image is TO THE  RIGHT OF THE OBJECT BECAUSE THE IMAGE IS REAL

 (a)  1/f = 1/d_o  + 1/d_i  where we need d_o AND  d_i ! BUT magnification m = -1/2; remember  when the image is real the value of m is negative in the case of a thin converging lens. Get ready for this concept on the final exam  IN COMPARISON to magnifying glass below, a converging lens producing a virtual image behind (to the left of)  the object !

Now,  -1/2 = -d_i/d_o.  Thus,  solve for d_i = do/2.   Thus 1/0.20 = 3/do so that do = 0.60 m.
(b)  m = -1/2 means image is inverted.

(c) d_i   =  do/2   =  0.30 m  via  substitution in  above equations.

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(a) (30 points)   How far from the lens is the object?
(b) ( 3  points)  Is the image upright or inverted (upside down)?   Explain.
(c) (13 points) How far from the lens is the image?

 (d) (4  points)   On the diagram above sketch the lens on the right of the object.  Draw the image by ray tracing as we did in class. Ray trace by drawing two rays from the tip of the arrow, one parallel to the axis and the other going though the center of the lens. Label the object distance d_o and image distance d_i.

2. (25 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.

SOLUTIONS

The first  polarizers' axes makes a  70 degree angle  with vertical and the second polarizer makes angle 50 degrees with the vertical using basic geometry/trig. Review your trig !!.  Thus the polarizers make an angle of 20 degrees between each other, word is bond.  Let I_o be the intensity of the light before going through the  first polarizer. Assume this light is polarized along a direction making an angle 70 degrees with the first polarizer axis. The intensity after passing through polarizer 1 is I₁ = I_o*cos² 70 .  The intensity is reduced again  after transmitting through polarizer  2. Since polarizer 2's  axis  makes a 20 degree angle with polarizer 1's axis,  we have
I₂ = I₁*cos² 20.  That is: I₂ =  I_o*cos²70*cos² 20  = 4.2 W/m² given initial intensity of  41 W/m² .
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(a)  (12 points) What is the Intensity of the light after it passes through Sheet 1 before reaching  Sheet .
(b)  (13 points)  What is the Intensity of  the light after it passes through   Sheet 2?

3.  (38 points) Two  parallel  plates of glass are separated by an air gap of thickness t.  Light of wavelength 625 nm falls  perpendicular  to the plates.  

 

SOLUTIONS

(a) 2*t = lambda/2--- note the role reversal switching the definition of constructive and destructive interference. Thus,  t = 156 nm.
(b)  2*t = lambda--- note the role reversal---giving t = 313 nm.
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(a) (19 points)  What  minimum thickness t will produce  constructive  interference at the detector?

(b) (19 points)  What  minimum non zero thickness t will produce  destructive  interference at the detector?

4. (37 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 image of the news story  of the Sharks victory she wants to read about.  The figure 1  below applies directly   to part (a).  Note that N = near point distance = 25 cm. A modified diagram would have to be drawn for part (b) in figure 2’s space.
SOLUTIONS

(a)  See figure 25.16.  Note:  d_i = -N, where is the near point distance .

M = 1/f + 1 = 3.1

(b) 1/f = 1/d_o  + 1/d_i     or 1/d_o   = 1/f - 1/d_i  = 1/f - (-1/N) = 1/f + 1/N so that  d_o= N*f/(N + f) = 0.081 m. You can also do it an easier way but please know the relationship between methods which contains important hidden information M = N/d_o  so that  d_o = N/M = 0.25/3.1 = 0.081 m, same  result.  Get ready for this concept on the final exam  IN COMPARISON to thin lens example above involving  a converging lens producing a real image to right of (behind)  object !
(c) M = N/f = 2.1.
 d_o = f  and the sketch is like that of figure 25-17. PLEASE COMPARE WITH figure 25-16, where the image is at a distance from lens less than infinity . 
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(a) (12 points) What is the angular magnification θ’/θ when this lens forms a virtual  image at the person’s near point, assumed to be 25 cm?
(b) (15 points) In part (a), what is the distance d_o  between the object and lens? 
(c) (10 points)  What is the angular magnification θ’/θ when this lens forms  a virtual image at - ? Please sketch the modified ray tracing diagram of  lens and rays  using   figure 2’s axis on next page. What is d_o in this case?