Advancing by pengxiang


									Advancing Physics AS

Chapter 1 - Imaging
Student notes

Specification              Imaging only

In the context of the digital revolution in communication, this section introduces elementary ideas about
image formation and digital imaging, and about the storage of digital information.

The material can be taught using up-to-date contexts such as medical scanning. There are opportunities to
address human and social concerns.

Candidates should demonstrate evidence of:

(a)   knowledge and understanding of phenomena, concepts and relationships by describing:

          the formation of a real image by a thin converging lens, understood as the lens changing the
          curvature of the incident wave-front;

           the storage of images in a computer as an array of numbers which may be manipulated to alter the

(b)   comprehension of the language and representations of physics by making appropriate use of the terms:

          pixel, bit, byte, focal length and power, magnification, resolution

          and by sketching and interpreting:

          diagrams of the passage of light through a converging lens

(c)   quantitative and mathematical skills, knowledge and understanding by making calculations and
      estimates involving:

          the amount of information in an image = no. of pixels  bits per pixel

          power of a converging lens, as change of curvature of wavefronts produced by the lens; use of

                              1 1 1
                                  (Cartesian convention); linear magnification;
                              v u f

                             restricted to thin converging lenses and real images

            v  f ;
          amounts and rates of transmission of information.

(d)   initiative and independence in learning by giving and explaining their own example of:

          an application of image formation;
Ch 1 Imaging         Learning outcomes

Section 1.1
Seeing invisible things – Ultrasound, electromagnetic spectrum, ‘seeing’ atoms

     Images can be formed with many kinds of signals, including ultrasound and all regions of
      the electromagnetic spectrum
     Images can be recorded electronically by microsensors; an example is the charge-coupled
      device (ccd); such images are composed of discrete picture elements – pixels
     Images on the atomic scale can be recorded by scanning methods; an example is the
      scanning tunnelling microscope (STM)
     The resolution of an image is the smallest distance over which a change can be seen

Section 1.2
Information in images – Image processing, amount of information, log scales

     Images can be stored as an array of pixels, each defined by a number; the amount of
      information stored = number of pixels x bits per pixel.
     Images can be smoothed by suitable averaging; images can be sharpened by identifying
      edges; images can be enhanced by increasing contrast or by false colouring.
     Quantities that cover a large range of values can usefully be displayed on a logarithmic
      ('times') scale. Prefixes for scientific units are chosen at multiples of 1000.
     1 bit of information contains 2 choices (0 or 1); 1 byte of information contains 8 choices
      (256 = 28 alternatives); information I provides N = 2I alternatives.

Section 1.3
With your own eyes – Eye and vision, real image from a thin lens

     The eye is like an intelligent video camera, sending out a stream of processed signals. It
      detects edges and movement.
     A converging lens adds a constant curvature to light falling on it. The curvature added is the
      power of the lens.
     Power of lens = 1 / f = 1 / v – 1 / u (Cartesian sign convention)
     Magnification = height of image / height of object
     Light takes the same time to travel on all paths from a point on the source to the image via
      the lens (or mirror).

Section 1.1: Seeing invisible things

Ultrasound imaging etc
Images are introduced with a discussion of an ultrasound scan of a 20 week old foetus. The image
is shown on p1 of Student Book, and is the first image in
Activity 10S Software based 'Looking at images'
Further information about this image can be found on File 10I Image 'The variety of uses of
scientific images'

Discussion of Activity 10S will focus on the following points to enable students to continue
confidently with the exercise themselves using other images:

1.        How was the image made?
         [Show image of foetus and the equipment used to produce it.
         Display Material 10S Computer screen 'Hospital ultrasound scanning equipment'
         Use magnification tool to show pixels. Can undo with ‘Edit Undo’.
         Discuss the image as an array of 256 x 256 pixels each having one of 256
         shades of grey (p 2).
         Discuss ‘Making an image with ultrasound’ pp 3-4 and diagram p 3.

        The delay times of the reflections tell the scanner where the denser tissue of the baby is.
         This can be illustrated with:
         Activity 30D Demonstration 'Distance measurement with ultrasound'
         together with a demonstration of an ultrasonic tape measure.

        The strengths of the reflections tell the scanner how dense the various tissues are.

         Check calculations on diagram panel on p 3.

         Use the diagram panel on p 4 to revise v = f and f = 1/T in the context of the ultrasound

           What you should know about waves
           The amplitude of a wave (symbol A) is the maximum displacement of the medium. This is
           measured from the undisturbed position. This is measured in metres (m) etc.

           The wavelength (symbol ) of a transverse wave is the distance between two
           neighbouring crests or troughs. The wavelength of a longitudinal wave is the distance
           between two neighbouring compressions or expansions (rarefactions). This is measured
           in metres (m) etc.

           The wave speed (symbol v) is the speed of the wave profile such as a crest or
           compression depending on the type of wave. This is measured in metres per second
           (m/s) etc.

           The time for one complete cycle is known as the period (symbol T). This is measured in
           seconds (s). If the period is 0.5 s, one complete wave is made every 0.5 s, so two
           complete waves are made every 1.0 s. The number of waves made every second is
           called the frequency (symbol f) of the wave. Frequency used to be measured in cycles
           per second but is now measured in hertz (Hz).

           f = 1/T and T = 1/f

           wave speed (m/s) = frequency (Hz) x wavelength (m)
           or v = f 
2.      What is the scale of the image? [actual size/size on image]

3.      What does the image tell us, and how?

4.      In what way is what you see 'invisible'?

5.      What wavelength of radiation, if any, is involved?
        [Why must the frequency of the ultrasound pulses be so high? See p 3.]

6.      What is the resolution of the image (the smallest size of thing that can be
        distinguished)? This is the size of one pixel in a digital image. The actual
        size of the object being imaged must be known for this. See p 2.

Students work through other examples from Activity 10S Software based 'Looking at images' in
ICT suite. Students should address each of the points 1-6 above. Further information about each
image can be found on:

File 10I       Image 'The variety of uses of scientific images'
File 20I        Image 'Astronomical images: Problems of noise and resolution'
File 30I        Image 'X-ray images in medical physics'
File 40I        Image 'Ultrasound images in medical physics'
File 50I        Image 'Magnetic resonance imaging
File 60I        Image 'Gamma rays detecting abnormalities'
File 70I        Image 'Satellite images of towns in Europe'

The ideas behind the ultrasound scanning can be revisited using:
Activity 40D Demonstration 'How fast sound moves in a solid'

Demonstrate digital image capture at work using a Web Cam intended for video conferencing. The
image is displayed directly on the computer screen.
Activity 20D Demonstration 'Electronic image capture'

Images from the Universe
Discuss the operation of the charge-coupled device (ccd), an example of a microsensor, for
converting light into digitally stored images. A digital image with 1 million pixels would have been
made with a ccd with a 1000  1000 array of elements.

’Images from the Universe’ are taken with many wavelengths form the electromagnetic spectrum
not just with visible light.

The electromagnetic spectrum (see p 7 for a diagrammatic representation)

Type of        Approximate       Approximate       Sources of         Detectors        Notes (uses,
radiation      wavelength in     frequency in      radiation                           dangers etc.)
               metres            hertz
Radio          104 to 10-3       104 to 1011       Accelerating       Aerial, tuning   Broadcasting
                                                   charges            circuit          (television and
                                                                                       radio), mobile

Microwaves     10-1 to 10-3      109 to 1011       Accelerating       Horn and         Microwave oven,
(sub-set of                                        charges            dish             radar
radio                                              (magnetron)
Infra-red       10-4 to 10-6    1012 to 1014       Hot objects      Phototransist   Radiant heaters,
                                                                    or etc.         night vision,
                                                                                    burglar alarms,
                                                                                    remote controls,
                                                                                    thermography, i.r.
                                                                                    lasers used with
                                                                                    optical fibres and
                                                                                    CD players

Visible light   (710-7) to     6 1014            Hot objects,     Eyes,           Vision
                (4 10-7)                          excited atoms    photographic
Ultra-violet    10-7 to 10-9    1015 to 1017       Very hot         Photographic    Fluorescence,
                                                   objects,         film,           sun lamps, killing
                                                   fluorescent      fluorescent     germs, astronomy
                                                   lights           screen
                                                   (converted to
                                                   visible light)
X-rays          10-9            1017 upwards       Collision of     Photographic    Medical
                downwards                          fast electrons   film,           diagnosis,
                                                   with target as   ionisation      analysis of matter
                                                   in X-ray tubes   effects
Gamma           10-11           1019 upwards       Radioactive      Photographic    Studying nuclear
rays            downwards                          atoms            film,           structure, cancer
                                                                    ionisation      treatment, food
                                                                    effects         sterilisation,

‘Spy in the sky’ (p6) suggests some uses for satellite imaging. Note the references to resolution
and to the idea of false colour.

‘Seeing’ atoms
Discuss ‘Seeing’ atoms’ and ‘Images for the mind’s eye’ pp 10-11. This includes brief detail of the
scanning tunnelling microscope [See A-Z for further details].

Section 1.2 Information in images

Image processing
Start by considering the possibilities of some image processing software with a demonstration of
the key elements of Activity 50S Software based 'Image processing: The surface of Mercury'.
The following exercise helps to explain how to smooth sharp edges, remove noise and fond edges.

References:     Advancing Physics AS pp 16-17
                Activities 50S, 60S, 70S and 80S

Various operations:
(a) Replacing each pixel by the mean of its value and those of its neighbours.
(b) Replacing each pixel by the median of its value and those of its neighbours. When a series of
    numbers is place in ascending order, the number in the middle is the median.

(c) Subtracting the N, S, E and W neighbours from four times the value of each pixel.
Calculate only the values for the squares inside the darkened box to avoid edge effects.

1      Smoothing edges by calculating mean values [operation (a)].


       0   0   0   1   1   1
       0   0   0   1   1   1
       0   0   0   1   1   1
       0   0   0   1   1   1
       0   0   0   1   1   1        becomes

       0   0   0   1   1   1

       Using the mean rounds off sharp corners.

3      (i) Removing noise by calculating mean values [operation (a)].

       1   1   1   1   1   1
       1   2   1   1   1   1
       1   1   1   1   1   1
       1   1   1   0   1   1
       1   0   1   1   1   1        becomes

       1   1   1   1   1   1

       (ii) Removing noise by calculating median values [operation (b)].

       1   1   1   1   1   1
       1   2   1   1   1   1
       1   1   1   1   1   1
       1   1   1   0   1   1
       1   0   1   1   1   1
       1   1   1   1   1   1

       Taking mean and median values both reduce noise. Which process is the most effective?

4      Finding edges by subtracting the values of the neighbours [operation (c)].

       (i) An image with an edge.

       1   1   1   2   2   2
       1   1   1   2   2   2
       1   1   1   2   2   2        becomes
       1   1   1   2   2   2
       1   1   1   2   2   2
       1   1   1   2   2   2

          (ii) An image with a uniform brightness gradient.

          1   2   3   4   5   6
          1   2   3   4   5   6
          1   2   3   4   5   6
          1   2   3   4   5   6        becomes
          1   2   3   4   5   6
          1   2   3   4   5   6

          Areas of uniform brightness are removed, as are areas where there is a uniform brightness
          gradient. Regions where the gradient of the brightness changes abruptly are enhanced.

Students should go on to try enhancing a variety of images. The essential points to establish are
smoothing, noise reduction, contrast and edge enhancement.

Activity 50S Software based 'Image processing: The surface of Mercury'
(All students start with this)

Students try at least one example of image enhancement:
Activity 60S 'Image enhancing: Volcanoes on Io' or
Activity 70S 'Medical uses of x-ray images' or
Activity 80S 'Medical uses of ultrasound images

The key points to stress are that images are made of pixels, that a pixel is defined by a number,
and that modifying images means doing arithmetic on pixels. The most important general
modification to emphasise is averaging which is a good way of removing random or rapid
fluctuations in all sorts of experiments.

[Students can get some striking and sometimes artistic effects with image processing software
such as PhotoShop.]

Amount of information and log scales
Introduce powers of 10 with ‘Powers of Ten’ video (9 minutes).

Decimal and binary number systems

Decimal: 102 101 100
          2 4 7
The number 247 (two hundred and forty seven) is made up as follows:
(2100) + (410) + (71)

Binary: 22 21 20
        1 0 1
The number 101 in binary represents the number (14) + (02) + (11) = 5 in decimal.

The two values used in the binary system are the digits 1 and 0. A binary digit is called a ‘bit’.
Electronically the idea of bits may be represented by a lamp that is ‘ON’ or ‘OFF’ or by a voltage
that is ‘HIGH’ or ‘LOW’.


(a) Convert the following 4-bit binary numbers into decimal numbers:

   1111                        1010                           0010                            0111

(b) Convert the following 8-bit binary numbers into decimal numbers:

   00010000                    11111111                  10101010                       01010101

(c) Write the following decimal numbers as 8-bit binary numbers:

   25                          100                            255                             180

The next section uses logarithmic thinking. We use logarithms every time we check the size of a
computer file in bytes (I byte = 8 bits). It will be shown that the space needed to store a file is a
logarithmic measure of the amount of information in it.

A message which can only be 'yes' or 'no' (or 'with sugar' or 'without sugar') has only two
alternatives. We need one bit (zero or one) to store it.

One pixel of a grey scale image can have 256 different alternative values. But 256 bits are not
needed to store it. Eight are enough, because there are 256 different numbers, which can be
represented by eight binary digits, or bits.

The decimal value of the binary number 11111111 is (127) + (126) + (125) + (124) + (123) +
(122) + (121) + (120) which is (128 + 64 + 32 + 16 + 8 + 4 + 2 + 1) which is 255. From 0 to 255
there are 256 numbers.
That is, the number N of alternatives which can be represented by information I bits is just N = 2I.

When I = 7, N = 27 = 128 and when I = 8, N = 28 = 256.
Thus if you add one bit of information you double the number of alternatives.

        Display Material 50O OHT 'Bits and bytes'                             Display Material 60O OHT ''Plus' and
                                                                              'times' scales of information'

Bits and bytes                                                               ‘Plus’ and ‘times’ scales of information
One byte stores 256 alternatives
                                                                             Number of alternatives = 2 number of bits
                                                    Decimal   Number of
    8 bits = 1 byte       4 bits       2 bits 1 bit value     alternatives
                                                                             If 8 bits of information are used:
       0    0     0   0      0     0     0    0       0
       0    0     0   0     0      0     0    1       1                                  number of alternatives = 28 = 256
       0    0     0   0     0      0     1    0       2        21 =2
                                                                             In general, if the amount of information is I bits:
       0    0     0   0     0      0     1    1       3
       0    0     0   0     0      1     0    0       4        22 =4
                                                                                         number of alternatives = 2I
       0    0     0   0     0      1     0    1       5
       0    0     0   0     0      1     1    0       6                      If the number of bits increases by one, the number of alternatives doubles.
                                                                             Informaton is measured on a ‘plus’ scale; number of alternatives on a ‘times’
       0    0     0   0     0      1     1    1       7                      scale:
       0    0     0   0     1      0     0    0       8        23 =8

       0    0     0   0     1      1     1    1       15
                                                                               ‘PLUS’ SCALE                                        ‘TIMES’ SCALE
       0    0     0   1     0      0     0    0       16       24 =16          (LINEAR)                                            (LOGARITHMIC)

       0    0     0   1     1      1     1    1       31
                                                                               Amount of information                               Number of alternatives
       0    0     1   0     0      0     0    0       32       25 =32
                                                                               increase by equal additions                         increase by equal multiples

                  1   1     1      1     1    1                                amount = I                                          number of alternatives = 2I
       0    0                                         63
       0    1     0   0     0      0     0    0       64       26 =64          amount = log2 N                                     number of alternatives= N

       0    1     1   1     1      1     1    1       127
      1     0     0   0     0      0     0    0       128      27 =128

      1     1     1   1     1      1     1    1       255                    The OHT ‘Bits and Bytes’ shows that the number of alternative values
                                                                             which can be represented grows rapidly as the amount of memory used
1      0    0     0   0     0      0     0    0       256      28 =256       increases.

     The discussion leads to work on logarithmic scales, in the form of ladders of quantities where
     adding one rung of the ladder multiplies the quantity by some amount. The link between adding
     one kind of quantity and multiplying another is the heart of the matter.

     Consider the meaning of logarithms numerically using base 10 and base 2.
     log10 1000 = 3 because 103 = 1000.
     log2 8 = 3 because 23 = 8.

     (a) Write down log10 1000000

     (b) What is the number whose logarithm to the base 10 is 5?

     (c) Write down log2 32

(d) What is the number whose logarithm to the base 2 is 10?
Display Material 70O OHT 'Comparing                                                     Display Material 80O OHT 'Logarithmic
logarithms of base 2 and 10'                                                            ('times') ladder of distance'

   Comparing logarithms base 2 and base 10                                               Logarithmic ladder of distance

   Logarithms and bases

   logarithms of                                                                         A ladder of distances in multiples of metres
                                           numbers                      logarithms of
     numbers to
                     10                1024 1000                        numbers to
          base 2                                                 3      base 10

                     9                 512

                                                                                         1021 galaxy                                        Equal multiple scales
                     8                 256

                                                                                         1018                             1 Em (exa)        A logarithmic scale is one on
                                                                                                    nearest stars                           which equal spaces correspond
   log2 (128) = 7    7                 128 100                   2    log10 (100) = 2
                                                                                         1015                             1 Pm (peta)       to equal multiples.

                     6                 64                                                1012                             1 Tm (tera)       In this distance scale each
                                                                                                    distance to the Sun                     upward step multiplies the
                                                                                         109                              1 Gm (giga)       distance by 1000.
                     5                 32                                                           the Earth
                                                                                         106                              1 Mm (mega)       Each downward step divides
                                                                                                                                            the distance by 1000.
                     4                 16                            log10 (16) = 1.2
   log2 (10) = 3.4                                                                       103        small town            1 km (kilo)
   (approx)                                   10                 1                                                                          Each upward step adds 3 to
                                                                                         100        human body            1m                the logarithm (base 10) of the
                     3                 8                                                                                                    distance.
                                                                                         10–3       width of a hair       1 mm (milli)
                                                                                                                                            Each downward step subtracts
                     2                 4
                                                                                         10–6       microchip element     1 m (micro)      3 from the logarithm of the
                     1                 2                                                 10–9       molecule              1 nm (nano)
                                                                                              –12                         1 pm (pico)
                                                                                         10         atomic nucleus
                     0                 1      1                  0                                                                              × 1000
                                                                                         10–15                            1 fm (femto)

                                                                                         10–18                            1 am (atto)              1
                                                                                                                                                × 1000
   210 = 1024             If base index = number                     103 = 1000

   10 = log2 (1024)       Then index = logarithm base (number)       3 = log10 (1000)

 Display Material 90O OHT 'Logarithmic                                                     Display Material 100O          OHT
 ('times') ladder of time'                                                                 'Logarithmic ('times') ladder of mass'

                                                                                         Logarithmic ladder of mass

                                                                                         A ladder of masses in multiples of grams


                                                                                                                                               Equal multiple scales
                                                                                         1018                                1 Eg (exa)
                                                                                                                                               A logarithmic scale is one on
                                                                                         1015                                1 Pg (peta)       which equal spaces correspond
                                                                                                                                               to equal multiples.
                                                                                         1012                                1 Tg (tera)
                                                                                                                                               In this mass scale each
                                                                                         109                                 1 Gg (giga)       upward step multiplies the
                                                                                                                                               mass by 1000.
                                                                                         106        a car                    1 Mg (mega)
                                                                                                                                               Each downward step divides
                                                                                         103                                 1 kg (kilo)       the mass by 1000.

                                                                                         100                                 1g                Each upward step adds 3 to
                                                                                                                                               the logarithm (base 10) of the
                                                                                         10–3 a mosquito                     1 mg (milli)      mass.

                                                                                         10–6                                1 g (micro)      Each downward step subtracts
                                                                                                                                               3 from the logarithm of the
                                                                                         10   –9
                                                                                                                             1 ng (nano)       mass.

                                                                                         10–12                               1 pg (pico)

                                                                                         10–15                               1 fg (femto)
                                                                                                                                                     × 1000
                                                                                         10                                  1 ag (atto)

                                                                                                                                                     ×    1

Logarithmic ladder of time

A ladder of times in multiples of seconds


                                                   Equal multiple scales
1018 age of Universe               1 Es (exa)
                                                   A logarithmic scale is one on
1015                               1 Ps (peta)     which equal spaces correspond
                                                   to equal multiples.
1012                               1 Ts (tera)
                                                   In this time scale each
10   9
                                   1 Gs (giga)     upward step multiplies the
           one year                                time by 1000.
106                                1 Ms (mega)
                                                   Each downward step divides
103                                1 ks (kilo)     the time by 1000.

100                                1s              Each upward step adds 3 to
                                                   the logarithm (base 10) of the
10–3 flap of a fly’s wing          1 ms (milli)    time.

     –6                                            Each downward step subtracts
10                                 1 s (micro)
                                                   3 from the logarithm of the
10–9 light crosses a room 1 ns (nano)              time.

10–12                              1 ps (pico)
     –15                                                 × 1000
10                                 1 fs (femto)

10–18                              1 as (atto)

Display Material 110O          OHT
'Logarithmic ('times') ladder of speed'

Logarithmic ladder of speed

A ladder of speeds in multiples of metres per second


1021          No speeds                           Equal multiple scales
             greater than
1018        speed of light     1 Em s–1 (exa)
                                                  A logarithmic scale is one on
1015                           1 Pm s–1 (peta)    which equal spaces correspond
                                                  to equal multiples.
1012                           1 Tm s–1 (tera)
                                                  In this speed scale each
109                            1 Gm s–1 (giga)    upward step multiplies the
                                                  speed by 1000.
           light speed
106                            1 Mm s–1 (mega)
                                                  Each downward step divides
     3                                            the speed by 1000.
10                             1 km s–1 (kilo)
100        walking speed       1 m s–1            Each upward step adds 3 to
                                                  the logarithm (base 10) of the
     –3                                           speed.
10                             1 mm s–1 (milli)
10         continental drift   1 m s–1 (micro) Each downward step subtracts
                                                3 from the logarithm of the
10                             1 nm s–1 (nano) speed.
10–12                          1 pm s–1 (pico)
10–15                          1 fm s–1 (femto)
                                                    × 1000

10                             1 am s–1 (atto)
                                                    ×    1

Computer files
Remember: the number N of alternatives which can be represented by information I bits is just N =
2I and I = log2 N.

The space needed to store a file is the number of bits (or more realistically – bytes) which is a
logarithmic measure of the amount of information in it i.e. the logarithm of the number of
alternatives. The number of alternatives will be a very large number when referring to a computer
file; the logarithm of that number is much smaller and more manageable.

A ‘bit’ is a small amount of information. It is common to work in bytes, where 1 byte is 8 bits. In
word-processing, one byte is used for one character (such as a letter). A large file will contain
many bytes of information so it is better to work in kilobytes. One kilobyte is 2 10 or 1024 bytes but
we normally call this 1000 bytes for simplicity. One megabyte is 220 or 1048576 bytes but we
normally call this 1000000 bytes for simplicity.

(a) Show that a 10 kilobyte word-processed file may contains less than 2000 words if the author
    writes an average of 6 characters per word.

(b) Estimate the size of a computer file in kbytes if it contains a 4000 word essay.

Digital images
In the context of digital images, a ccd image is a 2 dimensional
array of pixels. The larger the number of pixels in an image, the better the

Each pixel of a grey scale image may require 1 byte (8 bits) of information
(28 = 256 variations of grey) whereas each pixel of a colour image may
require 3 bytes (3  8 = 24 bits), one for each of the primary colours. A high
resolution colour image with 1 million pixels will therefore require 3  106
bytes of information (approximately 3 megabytes).

The amount of information stored = number of pixels  bits per pixel.

Colour images require a lot of bytes of memory if stored on a computer. Downloading large files
takes a long time if the information can only be transferred at, say 56000 bits per second.

Time for download = number of bits to be transferred  transfer rate in bits per second

(a) The image of the Moon completely fills a ccd containing 1000 000 elements. The diameter of
    the Moon is 3.5  106m. Calculate the resolution of the image.

(b) Calculate the number of bits of information required for each pixel if the image is coloured and
    there are 3 bytes to a pixel.

(c) Calculate how many different colours are available.

(d) Calculate the number of bits in the image file and show that it will take over five minutes to
   download this file if a modem operating at 56 000 bits per second is used.

[In practice, digital cameras use data compression methods to reduce the size of the image file. A
3 Mbyte file may be reduced to 1 Mbyte.]

Section 1.3 With your own eyes
Eye and illusion
                                                            Cross-section of the eye

                                                              vitreous humour

                                      iris (controls
                                      light into eye)
                                                                                       retina (light-sensitive cells)

                         (refracts light
                         and protects

                                                                                                     fovea (where
                                                                                                     retinal cells are
                         eye pupil                                                                   densest)

                                                                                                        blind spot

                         aqueous humour

                                           ciliary muscle
                                           (controls lens
                                                               eye lens (focuses
                                                               light onto retina)

Page 19 gives a simple description of the eye [See A-Z for more detail]. The eye is like an
intelligent video camera, sending out a stream of processed signals. The retina consists of two
types of light-sensitive cells. Rods predominate except in and near the fovea which is at the centre
of the retina. Rods respond to different intensities and they are not colour sensitive which is why
objects away from the centre of the field of vision are seen in shades of grey, not in colour. Several
rods are joined to each nerve fibre so less detail is seen in images away from the centre of the field

of vision. Cones are sensitive to red or green or blue light. These types of cells are found in and
near the fovea. They are most densely packed in the fovea so an image formed on this part of the
retina is seen in greatest detail.

Many important features can be seen on the model available.

Activity 150D      Demonstration 'Models of the eye'
The model shows how the cornea does most of the focussing; the lens inside the eye (not
available in this demonstration) is needed to focus at different distances (accommodation). The
use of correcting lenses for coping with short and long sight can be demonstrated. A separate 2
dimensional model with adjustable power lens can be shown to illustrate accommodation.

Remember that most of the focusing (curvature of wave fronts added) is done at the cornea; the
eye lens adds an adjustable further small amount of curvature.

It is helpful to compare focussing on a Web Cam with focussing of the eye.

Activity 140E Experiment 'The intelligent eye'
These experiments show how the eye and brain work together and also enable the width of the
fovea to be estimated.

The rods and cones are cross-connected with inhibiting links, so as to enhance contrasts at edges.
Hence the eye detects edges and movement very effectively.
Activity 130D Demonstration 'Grey step: Edge enhancement in the retina'

The Hermann grid can be used to test the function of inhibiting links (see p20).

Display Material 120O         OHT 'The Hermann grid'

Lenses and real images
Only real images formed by converging lenses are discussed. Real images are those that can be
formed on a screen.

See the formation of a real image in space using:
Activity 160D 'Image in mid-air'

Display Material 160O                                   Display Material 170O
OHT 'Rays and waves'                                    OHT 'Rays and waves focused'

                                                                   Wave and ray points of view:

                                                                   Wave point of view:
                                                                   The lens adds curvature to the wave entering it
Ray point of view                Wave point of view


     Ray point of view:                 Wave point of view:
     Light travels out in               Light spreads out in
                                                                                                    focal length f
     straight lines from a              spherical wave fronts
     small source                       from a small source

                                                                   Ray point of view:
                                                                   The lens bends the rays, bringing them to a focus


   Ray point of view:                Wave point of view:
   Light in a parallel beam or       Wave fronts in a parallel
   from a very distant source        beam or from a very distant
   has rays (approximately)          source are straight (not
   parallel to one another           curved) and parallel

Activity 200D Demonstration 'Focusing water ripples'

This introduces idea of lens adding curvature.

The image-lens distance when the object is at infinity is known as the focal length, f. The waves
coming from infinity have no curvature and the lens adds curvature to the waves passing through
it. The radius of the spherical wavefronts after waves from infinity have passed through the lens is
f. The curvature of a sphere of radius r is 1/r, so the lens has added curvature 1/f to the wavefronts.

It can be seen that a more powerful lens (shorter focal length) adds more curvature.
The power of a lens is a measure of the curvature it adds.

power (dioptres) P = 1/f where f is in metres.

A camera lens with a focal length of 50 mm, which is 0.050 m, has a power of 1/0.050m = 20
dioptre. When placed together the powers add so a 20 dioptre and 30 dioptre lens placed together
with have a total power of 50 dioptre.

Calculate the focal length of a 50 dipotre lens.

If you think that’s short, try calculating the focal length of a WebCam lens with a power of 278

It is helpful to see a pinhole camera being converted into a lens camera to get a feel for image
formation before proceeding to experiments to measure power and focal length.
Class experiment
Activity 170E Experiment 'Converging lenses: power and focal length'

It is useful to discuss how a lens forms an image using:
Activity 180D Demonstration 'Where are the parts of an object in its image?

   Display Material 180O
   OHT 'Formation of an image'

                   How a lens makes 'a little picture'
                                                                       The lens does not alter the direction of
                                              Image of traffic light
  Traffic light                                 image of red lamp      the light, it just alters the curvature of
  red light on                                       at the bottom     the wavefronts.

                                                                       It is useful to recall work on rays of light
                                                                       passing through a rectangular glass
                                                                       block to see why this happens.

                                                                       When drawing ray diagrams for lenses
                                                                       it is usually a good idea to draw a ray
                                                                       form the object that passes through the
                                              image of green lamp
                                                         at the top    centre of the lens as this ray never
  green light on                                                       changes direction.

                                          images of red and amber
                                              lamps: amber in the
  red and amber                                             middle
  both on

The rule for how lenses shape light (p 23)

curvature of waves coming out = curvature of waves coming in + curvature added by lens

Since curvature is 1/r where r is the radius of the wavefronts, we have:

1/v = 1/u + 1/f

Distances to the right are positive; distances to the left are negative. All measurements are made
from the centre of the lens.

The Cartesian convention is used; a lens adds curvature 1 / f to the wave fronts going through it.

 Display Material 190O                        OHT 'Action of a lens on a wavefront'

Lenses change wave curvature

                                                                                                                image of
source                                                                                                          source

                distance from lens to source = u                       distance from lens to image = v
                (for example minus 0.2 m)                              (for example plus 0.1 m)

                                                     1 1                             power of lens     = 15 dioptre
  curvature of wave front after leaving the lens =    =    = 10                                      f
                                                     v 0.1
                                                   1   1                             focal length of lens f =
  curvature of wave front on reaching the lens =     =    = –5                                                  power
                                                   u –0.2
  curvature added by lens = power of lens = curvature after – curvature before
                                            1 1                                                           = 0.067 m
                                          = –
                                            v u
                                            = 10– (–5) = 15 dioptre                                       = 67 mm

 Display Material 200O                        OHT 'Where object and image are to be found'

Lenses add constant curvature 1/f

 1      1
   =0 +
 v      f             zero curvature                         1
                      before                     curvature     after
 v=f                                                         f

 very distant
                                                                       image of


 1 1 1
  = +                       1                           1
 v u f          curvature     before        curvature     after
                            u                           v

                                                                       image of

                u (negative)                     v (positive)

  1 1                    –1
0=   +       curvature      before          zero curvature
  u f                     f                 after
 u= –f

 source at                                                         very distant
 the focus                                                         image of


A camera lens with a focal length of 5 cm forms a clear image of an object on a film when the
object is 3 m from the lens. Calculate the distance between the image and the lens.

(a) Will the image be more or less than 5 cm from the lens?

(b) Calculate the distance between the image and the lens.
[Hint: Is the object distance positive or negative?]

Class experiment to show that a lens does in fact add constant curvature to the wave fronts.

Activity 190E                          Experiment 'A converging lens adds constant curvature 1 / f '
uses File 130S                         Spreadsheet Model 'Lens action on a spreadsheet'

(a) Calculate the magnification of the image obtained in the camera example above.

(b) What is meant by the negative sign?

(c) Why is the magnification less than 1?

Illustrate magnification, particularly from the point of view of getting the largest possible image on
the retina using a cheap ccd video camera (WebCam):
Activity 210D           Demonstration 'Modelling the eye with a video camera'

Summary video on imaging:
‘Whatever became of X-rays?’ (University of Leeds)
This programme explains and illustrates some computer advances in body imaging.

Section C questions
In Section C there are 30 marks available from two open-ended questions. They have been
designed for candidates to have time and freedom to show what they have learned independently.
Quality of communication is assessed in these questions.

In Module 2860: Physics in Action candidates should demonstrate evidence of initiative and
independence in learning by giving and explaining their own example of:

1      an application of image formation (Chapter 1),

2      an application of signal transmission (Chapter 3),

3      choice and use of a sensor for an application (Chapter 2),

4      the relationship between uses, properties and structures of one material (Chapters 4&5).

In your studies of Imaging you will have covered enough material to write out a side of A4 on item
1 above.

A typical question might start as follows:

‘This question is about an imaging system, producing data that can be stored and displayed by a

Candidates might have written about one of the following: a satellite imaging system using infra-red
radiation or microwaves; the ultrasound medical scanner using ultrasound; a radio telescope
system that detects radio waves; various electron microscopes. However, any imaging system that
involves computer storage and display would be valid.

It is usually important to be able to explain how the image is produced and to consider how it is of
human benefit or scientific interest. You must understand the nature of the waves or radiation
(such as electromagnetic radiation or ultrasound) which carries the information needed to create
the image and you should be able to how the information for the image is obtained with the aid of a
labelled diagram. You will sometimes be asked how to manipulate the image to enhance its
appearance on the computer screen so you should understand about manipulating the pixels. You
might also be asked about the resolution of the image or how long it would take to transmit an
image of given size.

You should choose your own example and carry out some research of your own so that you are
well prepared. If you are interested in the medical scanner you should make sure that you know
and understand the information available in the textbook and the A-Z of the CD-ROM before doing
more research. The A-Z also has detailed information about various electron microscopes. You will
have to look further afield for other examples of imaging.

We will of course discuss various cases in class to make sure that you understand what is

Questions and activities
Section   Essential                                   Optional
1.1         Qu 1-6 AS text p 5
            Qu 1-6 AS text p 9
            Qu 1-6 AS text p 12

            Question 10S Short answer 'Speed,
            wavelength and frequency' Qu 1-17

            Question 20E Estimate 'Large and small
            distances and times' Qu 1-6

            Question 30X Explanation-Exposition
            'Different kinds and uses of images'
1.2         Qu 1-6 AS text p 18                       Question 60S Short Answer 'A scanning
                                                      electron microscope image of Velcro'

            Question 40S Short Answer 'Smoothing      Question 70S Short Answer 'Image processing
            pixels using mean or median'              by brightness and contrast control'

            Question 110S Short Answer 'Bits and      Question 80S      Short Answer 'The Horsehead
            bytes in images'                          nebula'

            Question 120S Short Answer 'Logarithms    Question 90C      Comprehension 'Saturn's aurorae
            and powers'                               in ultraviolet'

            Activity 120S Software Based 'How big     Question 100C Comprehension 'Betelgeuse in
            are your computer files?'                 ultraviolet'

                                                      Question 130C Comprehension 'X-ray image of
                                                      the Kepler supernova remnant'

                                                 Activity 90S 'Spreadsheet models: Image
                                                 processing’. This provides an insight into the
                                                 calculations behind image processing.

                                                 Activity 100S 'Mapping the South Atlantic sea

                                                 Activity 110H Home Experiment 'Logarithmic
                                                 ('times') ladders of quantities'

                                                 Reading 20T     Text to read 'A collection of visual

1.3       Qu 1-6 AS text p 26                    Question 160C Comprehension 'Satellite
          Question 140S Short Answer 'Response
          of the human eye to differences in     Display Material 140O    OHT 'Ouchi illusion'
          brightness'                            Display Material 150S    Computer screen 'A
                                                 pointillist painting'
          Question 150S Short Answer 'Cameras    Display Material 210O    OHT 'The lens maker's
          and eyes'                              equation'
                                                 Display Material 220O    OHT 'Thin lens: proof of
                                                 1 / v = 1 / u + 1 / f'

                                                 File 130S      Spreadsheet Model 'Lens action
                                                 on a spreadsheet'
Summary   Qu 1-6 AS text p 28


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