Physics 1230 “Light and Color”: Exam #2* Your Last Name____________________________________________________

Your First & Middle Names__________________________________________

General  information:  This  exam  will  be  worth  100  points.    There  are  11  multiple  choice  questions  worth  3  points  each  (part  1  of  the  exam)  and  problems  worth  a  total  of  67  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  the  course  web  page  soon  after  the  exam.  The  exam  scores  will  be  posted  at  https://culearn.colorado.edu      

Equations: Lens equation io xxf111

+=, where f is focal length of a lens (positive for convex

converging lens), xo is distance from center of lens to object, xi is distance from center of lens to image (positive if on opposite side of lens from object)

 Magnification  equation:    o

i

o

i

xx

SS

M ==,  where    si  is  the  size  of  image  (perpendicular  to  axis),  s0  =  size  of  the  

object  (in  direction  perpendicular  to  the  axis);  xo  is  distance  from  lens  to  object,  xi  is  distance  to  image.  

Law  of  refraction:    Light  bending  at  an  interface  of  two  media  is  determined  by  the  law  of  refraction  (the  so-­‐called  Snell's  law):    ni  sinθi  =  nt  sinθt  ,  where    θi  =  angle  between  incident  ray  and  normal,  θt  =  angle  between  transmitted  ray  and  normal,  ni  and  nt  are  the  indices  of  refraction  in  the  medium  containing  the  incident  ray  and  in  the  medium  containing  the  transmitted  ray.    

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).

Critical angle & total internal reflection:

n1 and n2 are the indices of refraction in the medium containing the incident ray and in the medium containing the transmitted ray, respectively.

Bending power of a lens: P = 1/f P = power in diopters, f = focal length of the lens in meters

Combined power of two thin lenses in contact: P1 + P2 = P P1 = power of lens 1; P2 = power of lens 2; P = power of combined lenses f-number (f/stop) = (focal length of lens)/(diameter of lens opening at that f/stop)

Metric system distances: 100 cm = 1 m, 10 mm = 1 cm, 25.4 mm = 1 inch

Exam  Assignment,  Part  1*  (multiple  choice  problems,  3  points  each,  33  points  total):  1.1. Your  exposure  meter  shows  that  f/5.6  at  1/100s  is  a  correct  setting  for  your  photograph.  Which  of  the  following  settings  are  also  correct?  A.  f/11  @  1/25  B.  f/11  @  1/50  C.  f/4  @  1/50  D.  f/4  @  1/200  E.  f/2.8  @  1/200  F.  f/2.8  @  1/400  G.  f/2.8  @  1/50  Possible  answers:    

1.  A,  C,  E  

2.  B,  D,  F  

3.  A,  D,  F  

4.  B,  C,  G  

5.  A,  D,  G   1.2. The  following  exposure  combinations  are  equivalent:    (1)  f/2  @  1/200s  (2)  f/5.6  @  1/25s  Which  setting  would  you  prefer  for  photographing  (a)  a  dog  running  by  (b)  flowers  in  a  garden,  some  near,  some  relatively  far  away  

a.  (a)  =  1  (b)  =  2  

b.  (a)  =  2  (b)  =  1  

c.  Both  1  

d.  Both  2  

e.  None  of  these   1.3. Suppose  you  have  an  f/5.6  lens  on  your  camera.  If  the  lens  has  a  56-­‐mm  focal  length,  what  is  its  effective  diameter?    

a.  0.1  mm  

b.  1  mm  

c.  5.6  mm  

d.  10  mm  

e.  560  mm  

1.4.  Which  of  the  following  statements  is  wrong  and  is  NOT  describe  a  difference  between  the  eye  and  a  simple  camera?    

a.  The  eye  focuses  by  changing  the  focal  length  of  the  lens;  a  simple  camera  changes  the  distance  between  the  lens  and  the  image  plane  (film).  

b.  The  eye  has  a  blind  spot;  a  camera  does  not.  

c.  The  size  of  the  aperture  in  a  camera  affects  the  depth  of  field  of  the  image;  the  size  of  the  pupil  does  not.  

d.  A  camera  usually  uses  a  shutter  to  allow  a  specific  amount  of  light  to  enter;  the  eye  has  continuous  imaging.   1.5. A  light  object  looks  even  lighter  to  you  when  it  is  surrounded  by  a  dark  background.  This  is  due  to:    

a.  dispersion  

b.  accomodation  

c.  binocular  disparity  

d.  polaroid  sunglasses  

e.  simultaneous  lightness  contrast     1.6. Give  two  reasons  why  eye  movements  are  beneficial  (select  two  out  of  6  possible  choices  below):    

a.  reduction  of  afterimages  and  saturation  effects  

b.  reduction  of  latency  due  to  improved  ganglion  response  

c.  increased  clarity  by  keeping  edges  in  view  

d.  reduction  of  lateral  inhibition  

e.  depth  perception  

f.  increased  depth  of  field  due  to  edge  enhancement  

1.7. If a person has an eye with a near point of 50 cm and a far point of 60 cm, then

(a) The person has myopia and needs eye glasses with a positive lens

(b) The person has myopia and needs eye glasses with a negative lens

(c) The person has hyperopia and needs eye glasses with a positive lens

(d) The person has hyperopia and needs eye glasses with a negative lens

(e) The person needs bifocals (bifocals are eye glasses with lenses which have two focal lengths: one for viewing objects at a distance and a second for viewing objects up close)

1.8. If a person has an eye with a near point of 200 cm and a far point of infinity, then

(a) The person has myopia and needs eye glasses with a positive lens

(b) The person has myopia and needs eye glasses with a negative lens

(c) The person needs bifocals

(d) The person has hyperopia and needs eye glasses with a positive lens

(e) The person has hyperopia and needs eye glasses with a negative lens

1.9. In order to increase the depth of field when taking a picture, you should

() Use a larger f/number and a longer exposure time

() Use a smaller f/number and a longer exposure time

() Use a larger f/number and a shorter exposure time

() Use a smaller f/number and a shorter exposure time

() Use a faster shutter speed and a film with a higher ASA rating

1.10. A brief image is projected on a screen for 0.01 secs. You perceive an afterimage after a delay time shown in the figure. The afterimage lasts for a persistence time, as shown. When should the next image be projected on the screen to give the illusion of smooth motion (a movie)?

() During the projection interval of the brief image

() During the delay interval

() During the persistence time of the after-image

() After the persistence time of the after-image

1.11. Bending (refraction) of light rays which enter our eyes is done by the

a) The aqueous and vitreous humors, retina, and sclera.

b) The eyelens and the retina.

c) The cornea and the eyelens.

d) The sclera and the eyelens.

e) The retina.

Exam  Assignment,  Part  2*  (67  points  total) 2.1. Photography (questions  worth  15  points). (a) Compare  a  simple  lens  camera  to  a  pinhole  camera  (i.e.,  what  are  the  differences/similarities?).  What  are  the  advantages  of  even  a  simple  lens  camera  over  a  pinhole  camera?  What  is  the  main  difference  between  a  simple  film  camera  and  a  digital  camera?    (8  points)   The  simple  lens  camera  has  a  lens  and  an  adjustable  aperture  (stop),  whereas  the  pinhole  camera  only  has  a  very  small  opening.  The  simple  camera  has  the  advantage  of  the  large  aperture  lens  to  capture  more  light  than  the  pinhole  opening.  This  makes  it  possible  to  have  lower  exposure  times  and  to  take  pictures  of  fast  events.  (It  also  has  greater  flexibility  with  focusing/depth-­‐of-­‐field.)  The  film  and  digital  cameras  are  similar,  with  the  exception  that  the  film  is  replaced  by  a  CCD  device  that  captures  the  image.  

   (b) List the components of a simple commercially available lens camera and describe their functions. (7 points) In a simple camera, there is a lens to focus the picture onto the CCD or film, CCD or film to record the image, a light-tight case (to keep the CCD/film from being exposed), and some kind of shutter to prevent the CCD/film from being exposed after and prior to taking the picture. 2.2. Eye and human vision (questions worth 15 points). (a) Describe the process of eye accommodation. Why is accommodation important for human vision? (7 points) Accommodation is the process that the eye uses to focus on objects at different distances. It is achieved by changing the focal length of the eye-lens. The eye-lens is normally focused on a very distant object when the muscles are relaxed. As an object moves closer to the eye, the muscles that do this increase the strength (lens power) of the eye-lens and therefore shorten its focal length until the object is in focus on the retina.

(b) Describe where (within your eye) the image is formed when you see the person/object in colors and during the day and what changes when you see the same person/object during the night or in a dark room (not much light) (8 points). The image of a person/object you see is formed on the retina. Looking at someone means their image is on your fovea, the small region near the center of the retina. Fovea is used for sharp detailed viewing and has the most cones (needed for the precise color vision) but practically no rods (used for low light, less precise viewing). Outside in the sun you are mainly seeing by using your cones and they enable you to see in full color (this is called photoptic vision). In the case of night or dark-room conditions, you switch to rods, which are more sensitive but do not allow you to see colors (this is called scotopic vision) and are located mostly outside of the fovea. 2.3. Vision correction (questions worth 10 points).

Assuming that the lens power for a normal (unaccommodated) eye is P = 60 Diopters, find what prescription eyeglasses or contact lenses are needed to correct vision of eyes with lens power of 63 Diopters. Describe the type of contact lenses that need to be used for vision correction. Give the focal lens distance for these lenses. (10 points  for  this  problem).

The combined lens power of two lenses is the sum of their lens powers: Pcorrected vision=Peye + Peyeglasses . From this, we find Peyeglasses= Pcorrected vision- Peye=60-63=-3D. Therefore, the lens power of the eyeglasses or contact lenses should be -3 Diopters. One needs to use concave (diverging) lenses to correct for this type of vision problem. The focal length distance of the lens can be found using the expression that relates this focal length to the lens power P: P=1/f or f=1/P (where P is in Diopters). One finds f=1/P= -1/3=-0.333 m = -33.3 cm.

2.4. Optical instruments.

Describe the difference between a telescope and an optical microscope. Discuss why these optical instruments can be useful, what are their main components, and what is called “intermediate image” formed in these instruments. Diagrams of simple microscope and telescope are encouraged but not required to answer this question (10 points).

Microscope: Telescope:

Both telescope and optical microscope have an objective and an eyepiece and the simplest versions of them can be built from two lenses (or using also curved mirrors in the case of reflective telescopes). In both of the instruments, the objective forms an intermediate image of an object that serves as a source of new rays and is viewed using the second lens (eyepiece). A telescope is an instrument designed for the observation of remote objects. The name "telescope" was derived from the Greek tele = 'far' and skopein = 'to see’. In the telescope, the intermediate image (red arrow shown below, much closer than the object) serves as a source of new rays and is viewed using the second lens (eyepiece). The simple astronomical telescope consists of two lenses as shown in the following figure. The first lens, called the objective, has a focal length that is as long as possible.

In its simplest form, the microscope consists of two positive lenses, as shown in the diagram below. The first lens has a short focal length, and the object is placed in front of this lens just beyond the focal distance. Since the lens is a positive lens, it forms a real, inverted image behind the lens. The second lens (eyepiece) uses this real intermediate image as its object and acts as a standard simple magnifier.

2.5. Image processing by the human eye and understanding illusions (17 points)

(a). Describe how ganglion cells respond to rays of light that enter different parts of the receptive fields of ganglion cells (use a schematic drawing). (5 points)

The ganglion cell emits electrical signals at a medium (ambient) level without any stimulation. Depending on where the light from an image falls on the receptive field it can either further excite or inhibit the electrical signal from the cell to the brain. Light falling on the center of the receptive field causes a stronger signal from the ganglion cell. Light falling on the sides (surround) of the receptive field inhibits the ambient signal (reduces its strength). Light falling partly on the center & partly on the surround partly stimulates & partly inhibits the ambient signal from the ganglion cell in proportion to the number of receptors it falls on.

(b) Describe & explain the nature of different illusions that you see on the left and right images (5 points):

These optical illusions are based on lateral inhibition. Hermann grid (left image above): when the image is on the receptive field #1 in-between 4 black squares, there is more light falling on the surround than in the position between only two black (receptive field #2).Because of this, there is more suppression of light intensity and thus the illusion of a dark spot at the first location (#1).

Stripes on the right Image above: This effect is due to edge enhancement - at an edge between light and dark, the dark is made darker due to lateral inhibition of the cells that are seeing the lighter region. Although each vertical band has equal color across the width, we perceive that the grayscale has a gradient in each region.

(c) Looking on the painting below, describe how George Seurat used the properties of image processing by human eye (7 points):

French artist George Seurat used edge enhancement by lateral inhibition to make figures stand out sharply. This is done using darker/lighter stripes just before the edges, as on the example shown below.