Physics 1230 “Light and Color”: Exam #3 Your Last Name____________________________________________________

Your First & Middle Name __________________________________________

General  information:  This  exam  will  be  worth  100  points.    There  are  15  multiple  choice  questions  worth  3  points  each  (part  1  of  the  exam)  and  problems  worth  a  total  of  55  points  (part  2  of  the  exam).  There  will  be  no  partial  credit  for  the  multiple  choice  problems.  Your  answers  to  the  problems  should  be  on  the  same  pages  as  the  exam  assignment  in  the  provided  space  (you  can  use  the  reverse  sides  of  the  pages  if  you  need  more  space  and  for  calculations).      

Rules  for  the  exam:  The  exam  will  be  held  during  a  regular  class  period.  All  exam  solutions  need  to  be  turned  in  by  5PM.  You  can  use  a  calculator  and  a  single  8.5  x  11  sheet  of  paper  with  equations  or  notes.  The  paper  must  have  your  name  and  student  number  on  the  top  of  both  sides;  the  rest  of  the  sheet  may  be  in  any  format  that  you  choose.  If  you  are  eligible  for  special  considerations  because  of  a  disability,  you  must  bring  a  letter  from  the  CU  Office  of  Disability  Services.  The  letter  will  be  in  effect  for  the  entire  term,  and  you  need  submit  it  only  once  before  the  first  exam.  I  will  make  special  arrangements  on  a  case-­‐by-­‐case  basis.  You  may  be  excused  from  an  exam  because  of  a  medical  problem;  please  give  me  a  note  from  a  doctor  in  this  case  if  possible.  I  will  deal  with  these  situations  on  a  case-­‐by-­‐case  basis.  The  exam  solutions  will  be  posted  at  http://www.colorado.edu/physics/phys1230/phys1230_fa11/Exams.htm  soon  after  the  exam.  The  exam  scores  will  be  posted  at  https://culearn.colorado.edu      

Useful information: Relationship  between  frequency  and  wavelength  of  light:      λ ·∙ν  =  c    λ = wavelength, ν = frequency, c = speed of light (3 x 108 m/s in empty space).

 Chromaticity  diagram:                                                  "wiring"  of  chromatic  channels      Polarized  vs.  unpolarized  light  

       

Visible  spectrum  of  light:      Exam  Assignment,  Part  1  (multiple  choice  problems,  45  points  total):  

1.1. Combinations of only red and blue

() produce colors that are complementary to yellow

() produce colors that are complementary to orange

*() produce colors that are complementary to green

() produce colors that are complementary to magenta

() produce colors that are complementary to cyan

1.2. The saturation of a color specifies

() its wavelength

() its brightness as perceived by the eye

() the amount of black ink needed to print the color

() the maximum value of the plot of its intensity as a function of wavelength

*() the width of the plot of its intensity as a function of wavelength

1.3. The hue of a combination of two monochromatic lights with wavelengths of 590 nm and 550 nm is

() red

*() yellow

() orange

() green

() blue

1.4.  To print text that is to appear red when viewed in white light, you would use

() an ink that absorbs only magenta

() an ink that absorbs only yellow

() an ink that absorbs only green

() an ink that absorbs only blue

*() an ink that absorbs only cyan

1.5. A filter that passes only monochromatic red light at 650 nm is placed on top of a filter that passes only monochromatic green light at 530 nm. When 460 nm blue light strikes one side of the pair of filters, the light that comes out of the other side (after passing through both filters) is

() red

() green

() blue

() yellow

*() black – no light comes out the other side

1.6. Light of 350 nm wavelength is combined with light of 750 nm wavelength. For a human with normal vision, this combination appears

*() black

() red

() blue

() magenta

() white

1.7. If you see two monochromatic lights which have exactly the same output powers and whose wavelengths are 410 nm and 420 nm, then

(a) the intensities look the same but the 420 nm light looks greener

(b) the intensities and the hues both look the same

(c) the 410 nm light appears brighter and the hues look the same

*(d) the 420 nm light appears brighter and the hues look the same

(e) the 420 nm light looks brighter and greener

In answering this and following question, you may find the following two figures helpful. They are taken from the textbook. The first is figure 9.11 on page 245 and the second is figure 10.5 on page 273. The figure on the left shows the response of the three types of cones in the eye and the figure on the right shows how colors are combined:

1.8. The hue of a combination of two monochromatic lights with colors of 650 nm and 570 nm is

(a) red

*(b) orange

(c) yellow

(d) green

(e) blue

1.9. In addition to the 3 subtractive primaries, 4 color printing also uses black ink because the 3 subtractive primaries

(a) used together produce magenta

(b) produce a color that depends on the order they were printed

*(c) do not produce a deep enough black when used together

(d) used together produce a dark green

(e) used together produce a dark blue

1.10. The description of the hue of a color as perceived by the brain is based on

(a) total intensity, red-blue difference and yellow-green difference

*(b) total intensity, red-green difference and yellow-blue difference

(c) total intensity, red-yellow difference and green-blue difference

(d) intensities of yellow, cyan and magenta

(e) intensities of red, green and blue

1.11.  Combinations of only red and blue

(a) produce colors that are complementary to yellow

(b) produce colors that are complementary to orange

(c) produce light beams that have no well defined color

(d) produce light beams that have very low intensity

*(e) produce colors that have no corresponding wavelength

1.12.  The curve shows the intensity distribution of a light bulb. What do we perceive the color to be?

a) Saturated reddish blue

b) Red

c) Unsaturated magenta

d) Unsaturated blue

*e) Unsaturated green

1.13. Refer to the chromaticity diagram on the first page. What colors can you get by additively mixing 580 nm yellow with 480 nm blue of different intensities

a) Off white

b) Any color

c) Any of the colors with a yellowish or bluish look

*d) Any of the colors along the line joining the two wavelengths

e) Green

Intensity  

 

700  

 

600  

 

500  

 

Wavelength  (nm)  

 

400  

 

1.14.  Light coming directly from the Sun through the Earth atmosphere is

(a) linearly polarized

*(b) unpolarized

(c) circularly polarized

(d) with linear polarizaqtion state alternating between two orthogonal directions (parallel and perpendicular to horizont)

(e) scattered

1.15.  The Sky is blue because of

(a) Rayleigh scattering of green and red light coming from the Sun

(b) additive mixing of colors

(c) subtractive mixing of colors

(d) a combination of absorption and total internal reflection;

* (e) scattering of blue light in the Earth atmosphere.

Exam  Assignment,  Part  2  (55  points  total) 2.1. Problems on color mixing and properties of colors (15  points).    (a)  On  the  Chromaticity  diagram  below,  show  how  you  can  find  complementary  colors.  Explain  how  one  finds  

complementary  colors  for  red  and  green  using  the  diagram  (5  points).  

 The  complement  to  any  wavelength  color  on  the  edge  of  the  chromaticity  diagram  is  obtained  by  drawing  a  straight  line  from  that  color  through  white  to  the  other  edge  of  the  diagram.  The  lines  extended  to  the  opposite  edges  of  the  diagram  show  that  the  complement  to  700  nm  red  is  490  nm  cyan  and  the  complement  to  green  is  magenta  -­‐  a  non-­‐wavelength  color.  

 (b)  Describe  the  color  shown  by  a  filled  circle  on  the  chromaticity  diagram  below  so  that  people  worldwide  can  

understand  what  color  you  have  in  mind  (using  both  words  and  numbers)  and  find  the  dominant  hue  of  this  color  using  the  diagram  (5  points).    

 Since  the  chromaticity  diagram  is  a  world  standard,  we  can  describe  this  color  (greenish  yellow)  using  its  chromaticity  diagram  coordinates:  x  =  0.4;  y  =  0.5.  To  find  the  dominant  hue  of  the  color  indicated  by  the  black  dot,  we  draw  a  straight  line  from  white  through  the  point  to  get  dominant  wavelength,  and  hence,  hue  (568  nm  greenish  yellow).  This  method  works  because  additive  mixture  of  white  with  a  fully-­‐saturated  (wavelength)  color  gives  the  desaturated  color  of  the  original  point.  

 (c).  Describe  how  common  color  printers  work;  explain  what  the  dots  of  different  size  on  the  image  below  are  intended  to  represent  and  how  they  are  related  to  the  color  in  the  squares  on  the  right:  

The  printers  commonly  use  four  inks  (cyan,  magenta,  yellow,  and  black  -­‐  CMYK).  This  image  shows  3  examples  of  color  halftoning  with  CMYK  separations.  From  left  to  right  we  see  what  is  printed  by  the  cyan,  magenta,  yellow,  and  black  inks.  Because  of  color  mixing,  the  combined  halftone  pattern  in  the  5th  column  will  be  perceived  by  the  human  eye  as  shown  in  the  6th  column  on  the  right  side.        2.2. Problems on color mixing and properties of colors (15  points  total).    (a)  Using  the  cone  sensitivity  graph  on  1st  page,  determine  what  color  will  be  perceived  by  a  human  eye  with  normal  color  vision  in  the  case  of  light  intensity  distribution  shown  below.  Explain  how  you  determine  this  (5  points).    

 The  eye  will  see  this  light  being  of  white  color.  Here  the  eye  is  viewing  a  mixture  of  blue  and  yellow  (e.g.,  460  nm  blue  of  intensity  1  and  575  nm  yellow  of  intensity  1.66).  The  blue  light  excites  mainly  s-­‐cones  but  also  a  bit  of  i-­‐cones  and  a  bit  of  L-­‐cones.  The  yellow  light  excites  i-­‐cones  and  (slightly  more)  L-­‐cones  but  no  s-­‐cones.  The  result  is  an  equal  response  of  s-­‐cones,  i-­‐cones  and  L-­‐cones,  which  will  be  perceived  as  white  color.    (b)  Using  the  chromatic  channel  diagram  and  cone  sensitivity  curves  below,  describe  the  work  of  chromatic  channels    

  The  3  kinds  of  cones  are  related  to  r-­‐g  and  y-­‐b  by  the  way  they  are  connected  to  neural  cells  (ganglion  cells).  Cones  of  each  kind  are  attached  to  3  different  neural  cells  which  control  the  two  chromatic  channels,  y-­‐b  and  r-­‐g,  and  the  white  vs.  black  channel  called  the  achromatic  channel  (lightness).  When  light  falls  on  the  L-­‐cones  they  tell  all  3  neural  cells  to  increase  the  electrical  signal  they  send  to  the  brain.  When  light  falls  on  the  i-­‐cones  they  tell  the  r-­‐g  channel  cell  to  decrease  (inhibit)  its  signal  but  tell  the  other  cells  to  increase  their  signal.  When  light  falls  on  the  s-­‐cones  they  tell  the  y-­‐b  channel  cell  to  decrease  (inhibit)  its  signal  but  tell  the  other  cells  to  increase  their  signal.  The  neural  cell  for  the    y-­‐b  chromatic  channel  has  its  signal  inhibited  when  (blue)  light  excites  the  s-­‐cone,    INTERPRETED  AS  BLUE,  and  enhanced  when  light  excites  the  i-­‐  &  L-­‐  cones,  INTERPRETED  AS  YELLOW.  The  neural  cell  for  the  r-­‐g  chromatic  channel  has  its  signal  inhibited  when  (green)  light  falls  on  the  i-­‐cone  INTERPRETED  AS  GREEN  and  enhanced  when  light  excites  the  s  and  L  cone,  INTERPRETED  AS  MAGENTA  (Psychological  red).  The  neural  cell  for  the  achromatic  channel  has  its  signal  enhanced  when  light  excites  any  of  the  cones      (c)  Light  falling  on  the  center/surround  of  the  receptive  fields  of  the  chromatic  channels  of  a  human  eye  with  normal  is  shown  below.  On  each  of  the  5  cases,  mark  what  color  will  be  perceived/interpreted  and  explain  how  double-­‐opponent  chromatic  channels  work.      

Red Green Green Yellow Yellow R in center and G in the surround both enhance signal (R-G channel ganglion cell signals red) R in surround inhibits the signal (R-G channel ganglion cell signals green) R in surround and G in the center both inhibit signal (R-G channel ganglion cell signals green) Y in center and B in the surround both enhance signal (Y-B channel ganglion cell signals yellow) Y in center enhances signal (Y-B channel ganglion cell signals yellow) 2.3. Questions about light polarization & scattering (10 points). (a). Describe why Sun looks yellow but sunset looks red. (5 points for this problem). The propert of Rayleigh scattering is that the shorter the wavelength, the more light of given wavelength is scattered. For this reason, blue light is scattered much more than red light. One can think of white light from the Sun being a mixture of red, green and blue light. Blue light is scattered the most and does not come to our eyes directly from the Sun. For this reason sun looks yellow (red + green that are left). However, when light travels through even more atmosphere (in the case of Sunset), the light with the next shortest wavelengths (green) is scattered too, so that the Sunset looks red.

(b). Assume that light propagates in z-direction and electric field of the propagating electromagnetic wave is shown by blue arrows below. What is the linear polarization direction? Is it possible to extinct this light using a linear polarizer (5 points  for  this  problem).

The linear polarization of this light wave is along the y-direction. Yes, this light can be extinct by an x-polarized linear polarizer.

2.4. Problems involving material of chapters 9, 10, 11, and 13 (15 points total). (a) Mark the Intensity vs. wavelength distribution corresponding to a monochromatic laser source. In a dark room in which you have only one of these light sources, light of what source will not pass through a very good blue color filter? Explain why. (5 points for this problem).

   

Laser

A blue color filter adsorbs all visible light except for blue light. The intensity vs. wavelength distribution for the red laser (figure on the left above) is such that there is essentially no light emitted within the spectral range of transmission of a blue color filter. Because of this, this laser beam will be blocked by the blue color filter. In the case of the two other light sources, some light will go through the blue color filter.

 (b).  Describe  how  one  obtains  green  color  when  printing  photos  using  a  conventional  color  printer  with  yellow,  cyan,  magenta,  and  black  cartridges.  (5  points  for  this  problem).  

Predominantly  by  using  cyan  and  yellow  inks  (with  possible  addition  of  black  if  dark  green).  The  inks  are  printed  in  the  form  of  dots  separated  by  small  distances  of  the  order  of  10-­‐100  microns  that  cannot  be  distinguished  by  human  eye  without  optical  instruments  like  microscopes.  

 (c).  Describe  how  one  obtains  yellow  color  in  a  conventional  display  having  only  red,  blue  and  green  pixels.  (5  points  for  this  problem).

Because of additive mixing of colors used in displays, red and green colors in displays can combine to yield yellow color. Since the pixels are of size smaller than what a human eye can resolve without the use of optical instruments such as microscopes, the red and green light from the corresponding indistinguishable pixels mixes to produce yellow color.